L9-free groups

Abstract In this article we classify all L9-free finite groups.


Introduction
There are some algebraic laws that hold in a lattice L if and only if L does not have a sublattice of a specific shape.For example, a lattice is modular if and only if it does not have a sublattice isomorphic to the so-called pentagon L 5 .
If L is a lattice, then we call a group L-free if and only if its subgroup lattice does not contain a lattice isomorphic to L. For example, the finite L 5 -free groups are exactly the modular groups, and these have been classified by Iwasawa in 1941, see [4].The subgroup lattice of the dihedral group of order 8, often denoted by L 10 , and some of its sublattices are of particular interest.One reason is that, if p is a prime number, then a finite p-group is L 5 -free if and only if it is L 10 -free.
There are several sublattices of L 10 containing L 5 : In 1999 Baginski and Sakowicz [2] studied finite groups that are L 6 -free and L 7 -free at the same time, and later Schmidt [8] classified the finite groups that are L 6 -or L 7 -free.Together with Andreeva and the first author he also characterized, in [1], all finite groups that are L 8 -free or M 8 -free.Finally, the finite M 9 -free groups have been classified by Pölzing and the second author in [6].Furthermore, there is a general discussion of L 10 -free groups by Schmidt, which can be found in [9] and [10].
In this paper we investigate finite L 9 -free groups.Since L 9 is a sublattice of L 10 , the groups that we consider are L 10 -free and therefore we can use Corollary C in [9] as a starting point for our analysis: Every finite L 10 -free group G has normal Hall subgroups N 1 ≤ N 2 such that N 1 = P ∈ Syl(G) | P ¢ G , N 2 /N 1 is a 2-group and G/N 2 is meta-cyclic.
Our strategy is to choose N := N 1 maximal with respect to the above constraints, and then we show that N has a complement K that is a direct and coprime product of groups of the following structure: cyclic groups, groups isomorphic to Q 8 or semi-direct products Q R, where Q has prime order and R is cyclic of prime-power order such that 1 = (R) = C Q (R).Furthermore, [N, K] ∩ C N (K) is a 2group and C O 2 (N) (K) is cyclic or elementary abelian of order 4 and every Sylow subgroup of [N, K] is elementary abelian or isomorphic to Q 8 .If the action of K on N satisfies some more conditions, then we say that NK is in class L.
The aim of our article is to prove the following theorem: Main Theorem.A finite group is in class L if and only if it is L 9 -free.

Notation and preliminary results
In this article we mostly follow the notation from Schmidt's book [7] and from [5].All groups considered are finite and G always denotes a finite group, moreover p and q always denote prime numbers.We quickly recall some standard concepts: L(G) denotes the subgroup lattice of G, consisting of the set of subgroups of G with inclusion as the partial ordering.The infimum of two elements A, B ∈ L(G) is A ∩ B (their intersection) and the supremum is A ∪ B = A, B (the subgroup generated by A and B).
If L is any lattice, then G is said to be L-free if and only if L(G) does not have any sub-lattice that is isomorphic to L.
A lattice L is said to be modular if and only if for all X, Y, Z ∈ L such that X ≤ Z, the following (also called the modular law) is true: (X ∪ Y) ∩ Z = X ∪ (Y ∩ Z).We say that a group G is modular if and only if L(G) is modular.
The modular law is similar to Dedekind's law (see 1.1.11of [5]).For all X, Y, Z ≤ G such that X ≤ Z it says that XY ∩ Z = X(Y ∩ Z).We will use Dedekind's law frequently throughout this article without giving an explicit reference each time.
If N ≤ G, then we say that an element g ∈ G induces power automorphisms on N if and only if U g = U for all subgroups U of N. Furthermore Pot G (N) Lemma 1.1.Let K be a finite group that acts coprimely on the p-group P. Then P = [P, K]C P (K).If P is abelian, then this product is direct.If [P, K] ≤ (P), then [P, K] = 1.Furthermore [P, K] = [P, K, K] and for all K-invariant normal subgroups N of P we have that C P/N (K) = C P (K)N/N.
Proof.Let g ∈ P \ C P (K) and R := [g, K], K .Then we first remark that R ≤ g, K .Lemma 1.1 shows that P = C P (K)P 0 and hence we have elements c ∈ C P (K) and h ∈ P 0 such that g = ch.We note that P 0 ¢ G and therefore (P 0 ) is a normal subgroup of G, moreover P 0 is a p-group and hence P 0 / (P 0 ) is elementary abelian.Let − : G → G/ (P 0 ) denote the natural homomorphism.As P 0 = [ P, K] is abelian and K acts coprimely on it, we see that C [ P, K] ( K) ∩ [ P, K, K] = 1, again by Lemma 1.1.Therefore C [ P, K] ( K) = C [ P, K, K] ( K) = 1.We recall that ch = g / ∈ C P (K) and thus h / ∈ C P (K), and then by hypothesis h / ∈ (P 0 ) and in particular 1 = h ∈ P 0 .It follows that [ h, K] = 1 because C [ P, K] ( K) = 1, see above.We conclude that 1 = [ h, K] = [ḡ, K] = [g, K] ≤ P 0 ∩ R. By hypothesis K acts irreducibly on P 0 , hence K does as well and we see that P 0 = P 0 ∩ R (P 0 ) = (P 0 ∩ R) (P 0 ).The main property of the Frattini subgroup (see for example 5.2.3 in [5]) finally gives that P 0 = P 0 ∩ R. Lemma 1.3.Let Q be a 2-group that is elementary abelian, cyclic or isomorphic to Q 8 .Then Q does not admit power automorphisms of odd order.
Proof.If Q is abelian, then the assertion follows from 2.2.5 of [5].If Q ∼ = Q 8 , then Aut(Q) ∼ = Sym 4 , and any automorphism of order 3 interchanges the maximal subgroups of Q. Lemma 1.4.Suppose that G = NK, where N is a normal Hall subgroup of G and K is a complement.Let N 1 , N 2 ≤ N, Q, R ≤ K and x ∈ N. Then the following hold:

x). (c) If K is abelian and acts irreducibly on the abelian group N/ (N) or if N is abelian and K induces power automorphisms on it, then C
Proof.Suppose that N 1 Q and N 2 R x are subgroups of G.
For (a) we do the following calculation: Then (a) yields that Q∩N 2 R x = (Q∩R)∩N 2 (Q∩R) x .Therefore, if x Q∩R ∩N 2 = 1 and if a, b ∈ Q∩R and y ∈ N 2 are such that a = yb x = yx −1 x b −1 b, then the fact that ab −1 = yx −1 x b −1 ∈ K ∩ N = 1 implies that a = b and that y If K is abelian, then x ∈ C N (C K (x)) and C N (C K (x)) is K-invariant.Thus, if K acts irreducibly on N/ (N) and x / ∈ (N), then C N (C K (x)) (N) = N, and the fact that [C K (x), N] = 1 implies that C K (x) = C K (N).If K induces power automorphisms on the abelian group N, then 1.5.4 of [7] implies that these are universal.If x = 1, then x ∈ (N), and otherwise x = 1 and every element of K that centralizes x also centralizes N. Altogether (c) holds.
Suppose finally that Q ≤ R.Then, for all g ∈ Q, we have that On the other hand (g −1 ) x ∈ Q x ≤ R x for all g ∈ Q and therefore [x, g] = (g −1 )

Battens and batten groups
Definition 2.1.(a) We say that G is a batten if and only if G is a cyclic p-group, or isomorphic to Q 8 , or G = QR, where Q is a normal subgroup of prime order and R is a cyclic p-group of order coprime to |Q| and such that C R (Q) = (R) = 1.(b) We say that G is a batten group if and only if G is a direct product of battens of pairwise coprime order.
(c) If G is a batten group, then we say that B ≤ G is a batten of G if and only if B is a batten that is one of the direct factors of G.
Warning: It is possible for a subgroup of a batten group G to be a batten, abstractly, but not to be a batten of G.This can happen when it is a p-subgroup for some prime p of a batten as in the third case of Definition 2.1.
Example 2.2.(a) Suppose that Q := x is a group of order 19 and that R = y is a subgroup of Aut(X) of order 27.Further suppose that x y := x 7 .Then B := QR is a non-nilpotent batten.For this we calculate that x y 3 = (x 7 ) y 2 = (x 49 ) y = (x 11 ) y = x 77 = x.Then the fact that x y = x 7 = x implies that C R (Q) = y 3 = (R).We note that B is a batten group and that Q is a subgroup of B that is a batten, but not a batten of B because [Q, R] = 1.(b) Let B = QR be as in (a), let T ∼ = Q 8 and let S be a cyclic group of order 625.Then B × T × S is a batten group.
Remark 2.3.Let G be a batten group and let B be a batten of G such that |π(B)| = 2.
(a) B is not nilpotent, but B has a unique normal Sylow subgroup.(b) A Sylow subgroup Q of B is cyclic, and therefore Q is batten.But Q is not a direct factor of G and hence Q is not a batten of G. Definition 2.4.Suppose that G is a non-nilpotent batten.Then there is a unique prime q ∈ π(G) such that G has a normal Sylow q-subgroup Q, and Q is cyclic of order q.In this case we set B(G) := Q.
From the definition we can immediately see that B(G) is a characteristic subgroup of a non-nilpotent batten G and that it has prime order.Lemma 2.5.Suppose that G is a non-nilpotent batten, that r ∈ π(G) and that R ∈ Syl r (G) has order at least r 2 Proof.Since |R| ≥ r 2 , we see that R = B(G).Then Definition 2.1 implies that R is cyclic and that there is a prime q ∈ π(G) \ {r} such that We recall that R is cyclic, and then this implies that This proves all statements.
Lemma 2.6.If G is a batten group and P ≤ G is a Sylow p-subgroup of G for some prime p, then G has a subgroup of order p.In addition, 1 (P) ≤ Z(G) or there is some non-nilpotent batten B of G such that Proof.Since P ≤ G, it follows that P is cyclic or isomorphic to Q 8 .Therefore 1 (P) has order p.
If P is a batten, then 1 (P) ≤ Z(P) ≤ Z(G).In particular 1 (P) is the unique subgroup of G of its order.Otherwise there is a non-nilpotent batten B of G such that P ≤ B. If P has order p, then P = 1 (P) = B(B) is a normal subgroup of B and so of G. Again 1 (P) is the unique subgroup of K of order q.Otherwise we have that 1 (P) ≤ (P) = Z(B) ≤ Z(G) by Lemma 2.5.In particular 1 (P) is the unique subgroup of K of its order.Lemma 2.7.Suppose that H is a batten and that U H. Then U is a cyclic batten group.Furthermore, all subgroups of batten groups are batten groups.
Proof.Assume for a contradiction that U is not a cyclic batten group.Then U is not a cyclic batten, and therefore H is neither cyclic of prime power order nor isomorphic to Q 8 .Thus H is not nilpotent, in particular |π(H)| = 2 and all Sylow subgroups of H are cyclic.This implies that U is not a p-group.Let But then U is a direct product of cyclic groups of prime power order, i.e. a cyclic batten group, and this is a contradiction.
Next suppose that G is a batten group and that U is a subgroup of G. Then U is a direct product of subgroups of the battens of G whose orders are pairwise coprime.Consequently U is a batten group as well, by the arguments above.
We remark that sections of battens, or batten groups, are not necessarily batten groups.For example, We first suppose that Q is not a normal subgroup of K. Since K is a direct product of battens, it follows that Q is not normal in B. Thus B is neither abelian nor hamiltonian (otherwise all subgroups of B would be normal), and it follows that B is not nilpotent.Now Q is a proper subgroup of B because Q ¢B.We conclude from Lemma 2.5 that neither Suppose now that Q ¢ K. Using ( * ) we see that Hence we may suppose that B is not abelian.If B is not nilpotent, then Z(B) and a Sylow q-subgroup of B centralize Q.Thus Lemma 2.5 yields that |K : which completes the proof.

L 10 and its sublattices
Throughout this article we will use the notation from the next definition whenever we refer to L 10 and its sublattices: Proof.Let ≡ be a congruence relation on L 9 and suppose that E ≡ D.
First we assume that X 0 ∈ L 9 \ {F} is such that F ≡ X 0 .Then there is some X ∈ {A, B, C} such that X 0 ≤ X, and then X We have seen that E is not congruent to any of the elements A, B, C, D, F. Next we assume that X ∈ {S, T, U} is such that E ≡ X and we choose Y ∈ {A, C} such that X Y. Then Y = Y, E ≡ Y, X = F, which gives another contradiction.We conclude that {E} and {F} are singleton classes with respect to ≡.
As this is impossible, we conclude that such elements X, Y do not exist.
In particular A, B, C are pairwise non-congruent and D, S, T, U are also pairwise non-congruent.We assume that D ≡ X ∈ {A, B, C}.Then we choose Y ∈ {T, U} such that Y X, and this gives the contradiction F = Y, D ≡ Y, X = F. Hence {D} is a singleton as well.
Finally, we assume that there are X ∈ {A, B, C} and Y ∈ {T, S, U} such that X ≡ Y.We have seen that X ∩ Y = E and then it follows that Y ≤ X by the structure of L 9 .Now D = X ∩ D ≡ Y ∩ D = E, which is impossible.In conclusion, for all X, Y ∈ L 9 , we have that X ≡ Y if and only if X = Y.This means that ≡ is equality.
In the following lemma we argue similarly to Lemma 2.2 in [8].Lemma 3.4.Suppose that n ∈ N and that G 1 , . .., G n are normal subgroups of G of pair-wise coprime order such that G = G 1 × • • • × G n .Then G is L 9 -free if and only if, for every i ∈ {1, . .., n}, the group G i is L 9 -free.
Proof.Since subgroups of L 9 -free groups are L 9 -free we just need to verify the "if" part.
Suppose that G 1 ,…,G n are L 9 -free.Then Lemma 1.6.4 of [7] implies that L(G) that is isomorphic to L 9 as in Definition 3.1.Then the projections ϕ 1 and ϕ 2 of L into L(G 1 ) and L(G 2 ), respectively, are not injective, because L(G 1 ) and L(G 2 ) are L 9 -free.Let i ∈ {1, 2} and define, for all X, Y ∈ L: Then ≡ i is a congruence relation on L, because ϕ i is a lattice homomorphism, but it is not equality because φ i is not injective.Then Lemma 3.3 implies that ϕ 1 (D) = ϕ 1 (E) and ϕ 2 (D) = ϕ 2 (E), and hence D = E.This is a contradiction.
Proof.We first remark that L 9 satisfies the relations given in L9 (i) -L9 (v).Suppose conversely that a lattice L = {A, B, C, D, E, F, S, T, U} satisfies the relations given in L9 (i) -L9 (v).Then we see that E ≤ A, B, C, D, S, T, U ≤ F and D ≤ A, B, C as well as S, T ≤ A and U ≤ C. If these are the unique inclusions and |L| = 9, then L ∼ = L 9 .
For all X, Y ∈ L such that X ≤ Y we have that X ∩ Y = X and X, Y = Y.Thus L9 (ii) shows that S, T, and D are pair-wise not subgroups of each other.
Using L9 (iii) we obtain that D U and U D, and L9 (iv) gives that also S, T and U are pair-wise not subgroups of each other.In addition, by L9 (v), we have that A, B and C are pair-wise not subgroups of each other.Together with the fact that S, T ≤ A and U ≤ C, this implies that A U, C S, T and B S, T, U. Moreover we have that D ≤ A, B, C and S, T ≤ A and U ≤ C and A = F = B and C = F. Together with L9 (ii) and L9 (iii), this information yields that A U and C S, T as well as B S, T, U.
We conclude that there is a lattice homomorphism ϕ from L 9 to L. Hence we obtain a congruence relation ≡ on L 9 by defining that X ≡ Y if and only if ϕ(X) = ϕ(Y), for all X, Y ∈ L 9 .
If ϕ is not injective, then Lemma 3.3 implies that E = D.This contradicts L9 (i).Consequently ϕ is injective and L ∼ = L 9 .
The next lemma gives an example of a group that is not L 9 -free.
Altogether it follows, with Lemma 3.5, that L is isomorphic to L 9 , and then G is not L 9 -free.
The next lemma shows how we can construct an entire class of groups that are not L 9 -free.
Lemma 3.7.Suppose that p = q and that G = PQ, where P is an elementary abelian normal Sylow psubgroup of G and Q is a cyclic Sylow q-subgroup of G. Suppose that Q acts irreducibly on [P, Q] = 1 and that |C P (Q)| ≥ 3. Then G is not L 9 -free.
Proof.Since Q is abelian, we see that E := C Q (P) ¢ G.We claim that G/C Q (P) is not L 9 -free.Therefore we may suppose that E = 1.Our hypotheses imply that [P, Q] is not centralized by Q, and in particular |[P, Q]| ≥ 3. Moreover |C P (Q)| ≥ 3 by hypothesis.Since P is elementary abelian, Lemma 1.1 gives that P = [P, Q] × C P (Q).
We let V ≤ [P, Q] and D ≤ C P (Q) be subgroups of minimal order such that |V| ≥ 3 and |D| ≥ 3 and we set A := V × D. If p is odd, then A has order p 2 , and if p = 2, then |A| = 2 4 = 16.In the first case A has We recall that A is elementary abelian, and then it follows that L9 (i) and L9 (ii) hold and that A = S, V = T, V ( * ).
We further set If X ∈ {T, S}, then X C P (Q) and then the irreducible action of Q on [P, Q] and Lemma 1.2 yield that V ≤ [P, Q] ≤ X, U .Using ( * ) it follows that D ≤ A ≤ V, X, U = X, U .Combining all this information gives that X, U = [P, Q]DQ.Now if we set F := [P, Q]DQ, then we have L9 (iv).To prove our claim, it remains to show that property L9 (v) of Lemma 3.5 is satisfied.We set B := DQ x for some x ∈

Group orders with few prime divisors
Much of our analysis will focus on non-nilpotent groups with a small number of primes dividing their orders.The next lemma sheds some light on why this situation naturally occurs.
Proof.This lemma follows from Theorem 2.1.2 in [7] and Lemma 2.1 in [9], since L 9 is a sublattice of L 10 containing L 5 .
Lemma 4.3.Suppose that p = q and that G is an L 9 -free {p, q}-group.Let P be a normal Sylow p-subgroup of G and let Proof.First we note that all subgroups and sections of G are L 9 -free and that G is L 10 -free.
Let G be non-nilpotent and assume for a contradiction that Q is neither cyclic nor isomorphic to Q 8 .Given that G is L 10 -free, we may apply Theorem B of [9] and we see that neither (a), (b) nor (c) hold.Therefore p = 3 and q = 2. Now there are a, b ∈ Q such that a, b is not cyclic and b is an involution.If, for all choices of b, we have that C P (b) = C P (Q), then 1 (Q) acts element-wise fixed-point-freely on P/C P (Q), contradicting 8.3.4(b) of [5].Therefore we may choose b such that a does not centralize C P (b), and we also choose a of minimal order under these constraints.Then a 2 centralizes P and a inverts an element x ∈ C P (b) by a result of Baer (e.g.6.7.7 of [5]).If follows that a inverts 1 ( x ) and we may suppose that x has order 3. Now x, a, b /C a (x) is isomorphic to D 12 , contrary to Lemma 3.6.Lemma 4.4.Suppose that p = q and that G is an L 9 -free {p, q}-group.Furthermore, let P be a normal Sylow p-subgroup of G and let Q ∈ Syl q (G) be cyclic such that 1 = [P, Q] is elementary abelian.
Then every subgroup of Q acts irreducibly or by inducing (possibly trivial) power automorphism on [P, Q].Moreover, C P (Q) is a cyclic 2-group and P is abelian.
Proof.First we note that all subgroups and sections of G are L 9 -free and that all subgroups and sections of [P, Q] are elementary abelian, by hypothesis.In addition Lemma 2.2 of [9] as G is also L 10 -free.In particular P is abelian if Assume that the lemma is false and let G be a minimal counterexample.Since [P, Q] is elementary abelian, we introduce the following notation with Maschke's theorem: Let n ∈ N and let M 1 , . . ., M n ≤ [P, Q] be Q-invariant and such that [P, Q] = M 1 × • • • × M n and that Q acts irreducibly on M 1 , . . ., M n , respectively.Lemmas 2.3.5 of [7] and 4.2 yield that 1 (C P (Q)) is elementary abelian.Now there are r ∈ N and cyclic subgroups M n+1 , . . ., M n+r of 1 (C P (Q)) such that Then for every i ∈ {1, 2} the group O p (H i ) is elementary abelian.Moreover H i is a proper subgroup of G and then the minimal choice of G implies that every subgroup of Q either induces (possibly trivial) power automorphisms on [O p (H i ), Q] or acts irreducibly on it. ( Proof.We assume for a contradiction that n + r ≥ 3. Then Q does not act irreducibly on both O p (H 1 ) and O p (H 2 ), and it follows that Q induces (possibly trivial) power automorphisms on We suppose first that for both i ∈ {1, 2}.Therefore Lemma 1.5.4 of [7], together with the fact that 1 But this means that Q, and hence every subgroup of Q, induces (possibly trivial) power automorphism on O p (H 1 )O p (H 2 ) = [P, Q] in this case.Thus G is not a counterexample, which is a contradiction.
We conclude that C P (Q) = 1 and now there is some i ∈ {1, 2} such that 1 Then H i satisfies the hypotheses of our lemma and it follows that is an non-trivial 2-group.In particular p = 2.But then Lemma 1.3 provides the contradiction that For the proof of (1), we assume for a further contradiction that r = n = 1.Then Q acts irreducibly on the elementary abelian group [P, Q] and Lemma 3.7, applied to In particular we have that p = 2. Thus the minimal choice of G and Lemma 1.3 yield, for every proper subgroup U of Q, that U centralizes P or acts irreducibly on Since G is a counterexample, it follows that C P (Q) is not cyclic.But | 1 (C P (Q))| = 2 and therefore C P (Q) is a generalized quaternion group.It follows that C P (Q) ∼ = Q 8 by Lemma 4.2.In this case 1 = Z := Z(C P (Q)) ¢ G and G/Z satisfies the hypotheses of our lemma, but not the conclusion.Thus G is not a minimal counterexample, contrary to our choice.
Proof.Assume for a contradiction that n = 2.By hypothesis Q is cyclic, and then we may suppose that [9] implies that Q induces power automorphisms on P. Thus G is not a counterexample, which is a contradiction. Therefore The minimal choice of G yields that Q 0 acts irreducibly or by inducing power automorphisms on [P, Q 0 ] and that C P (Q 0 ) is a cyclic 2-group.Now we notice that M 2 ≤ C P (Q 0 ), but M 2 1 = C P (Q), whence we deduce a contradiction from 2.2.5 of [5].
Since G = PQ is a counterexample to the lemma, Q has a proper subgroup U that does not act irreducibly on [P, Q] = P and it also does not induce power automorphisms on [P, Q].In particular it does not act trivially.Since PU is a proper subgroup of our minimal counterexample G, it follows that Lemma 4.5.Suppose that q is odd and that G is an L 9 -free {2, q}-group.Suppose further that P is a normal Sylow 2-subgroup of G such that [P, Q] is hamiltonian and let Q ∈ Syl q (G).Then one of the following holds: (a) G is nilpotent or (b) [P, Q] ∼ = Q 8 and there exists a group I of order at most 2 such that P = Proof.We suppose that G is not nilpotent.
Then Q is not normal in G and Lemma 4.3 implies that Q is cyclic.Furthermore, P is L 9 -free and hence it is modular by Lemma 4.2.Since [P, Q] is hamiltonian, Theorem 2.3.1 of [7] provides subgroups P 0 , I ≤ P such that P 0 ∼ = Q 8 , I is elementary abelian and P = P 0 × I.
We recall that the automorphism group of Q 8 is isomorphic to Sym 4 .Thus, if We conclude that our assertion holds if I = 1.Now suppose that I = 1.We recall that P ¢ G and therefore is elementary abelian of order 4. In particular P is elementary abelian, hence it is a non-hamiltonian 2-group of order at least 8. Lemma 4.4 states that Q acts irreducibly on [ P, Q] = 1 or induces power automorphisms on it.The second case is not possible by Lemma 1.3.
Hence Q ∼ = Q acts irreducibly on [ P, Q] = [P, Q] and, by Lemma 4.4, we see that C P( Q) is a cyclic 2-group.Since I = 1 and 1 (P) = (P 0 ) × I, we have that 1 = Ī = 1 (P) is Q-invariant and P0 is a non-cyclic complement of Ī in P.This implies that Ī = C P( Q) is cyclic and elementary abelian at the same time.Thus I ∼ = Ī has order 2 and with Lemma 1.1 we deduce that C P (Q) (P 0 ) = I (P 0 ) = 1 (P) is elementary abelian of order 4. Then we deduce that C P (Q) = 1 (P) and then In conclusion, [P, Q] is a subgroup of order at most 8 admitting an automorphism of odd order that centralizes 1 ([P, Q]).It follows that [P, Q] ∼ = Q 8 and then, together with the fact that I ≤ Z(G), our assertions follow.Definition 4.6.Suppose that Q is a cyclic q-group that acts coprimely on the p-group P. We say that the action of Q on P avoids L 9 (and we indicate more technical details by writing "of type (•)") if and only if one of the following is true: Every subgroup of Q acts irreducibly or by inducing (possibly trivial) power automorphisms on the elementary abelian group [P, Q] = P. (cent) Every subgroup of Q acts irreducibly or trivially on the elementary abelian group [P, Q], P is abelian and Lemma 4.7.Suppose that p is an odd prime and that G is an L 9 -free {2, p}-group.Let further P be a normal Sylow p-subgroup of G and let Q ∈ Syl 2 (G) be isomorphic to Q 8 and such that 1 = [P, Q] is elementary abelian.Then p ≡ 3 mod 4, |P| = p 2 and Q acts faithfully on P.
Proof.We set 3 is applicable and we see that Ḡ is nilpotent.But then G is also nilpotent, contrary to our hypothesis that [P, Q] = 1.Thus 1 (Q) is not normal in G and Q acts faithfully on P. Now, for all y ∈ Q of order 4, we apply Lemma 4.4 on [P, Q] y to deduce that y either induces power automorphisms on [P, Q] or acts irreducibly on it.Theorem 1.5.1 of [7] states that Pot G (P) is abelian, but Q ∼ = Q 8 is not, which means that we may choose y such that y does not induce power automorphisms on P. In particular P is not cyclic of prime order.Moreover p is odd and therefore 4 divides (p + 1)(p − 1) = p 2 − 1, and Satz II 3.10 of [3] yields that |P| ≤ p 2 .It follows that |P| = p 2 .More precisely, as |P| = p, the result implies that p ≡ 3 (mod 4), and then the proof is complete.Definition 4.8.Suppose that Q ∼ = Q 8 acts coprimely on the p-group P. We say that the action of Q on P avoids L 9 if and only if p ≡ 3 mod 4, |P| = p 2 and Q acts faithfully on P. Lemma 4.9.Suppose Q ∼ = Q 8 and that P is a p-group on which Q acts avoiding L 9 .Then P is elementary abelian, 1 (Q) inverts P, and every subgroup of Q of order at least 4 acts irreducibly on P.
Proof.Since a cyclic group of order p 2 has an abelian automorphism group by 2.2.3 of [5], it follows that P is elementary abelian.If 1 = R is a cyclic subgroup of P, then |Aut(R)| = p−1 and therefore R does not admit an automorphism of order 4. Additionally, [P, 1 (Q)] is Q-invariant and, since Q acts faithfully on P, we see that [5] states that 1 (Q) inverts P.
In particular 1 (Q) inverts every cyclic subgroup R of P.
In addition, these arguments show that every subgroup U of order 4 of Q does not normalize any nontrivial proper subgroup of the elementary abelian group P.This means that U acts irreducibly on P.
Corollary 4.10.Suppose that p = q and that G is an L 9 -free {p, q}-group such that P Then either G is nilpotent and P and Q are modular or Q is a batten and it acts on P avoiding L 9 .
In particular, if G is not nilpotent, then Q is isomorphic to Q 8 or cyclic and [P, Q] is elementary abelian or isomorphic to Q 8 , where in the second case q = 3.
Proof.By hypothesis G is L 9 -free, hence P and Q are, too.Then Lemma 4.2 implies that P and Q are modular.
Suppose that G is not nilpotent.Then Lemma 4.3 applies: Q is cyclic or isomorphic to Q 8 and hence it is a batten.Moreover Lemma 2.2 of [9] states that [P, Q] is a hamiltonian 2-group or elementary abelian.In the first case Lemma 4.5 gives the assertion.In the second case our statement follows from Lemmas 4.4 and 4.7.
Lemma 4.11.Let Q be a nilpotent batten that acts on the p-group P avoiding L 9 , and suppose that U is a subgroup of Q.
Then U induces power automorphisms on P or it acts irreducibly on Proof.First suppose that Q ∼ = Q 8 .Then Lemma 4.9 implies that every subgroup of order at least 4 of Q, and in particular Q itself, acts irreducibly on P = [P, Q]/ ([P, Q]).Moreover, the involution of Q inverts P by Lemma 4.9, and then the statement holds.
Next we suppose that Q is cyclic.Then Definition 4.6 gives the assertion unless the action of Q on P avoids L 9 of type (hamil).In this case every proper subgroup of Q centralizes P, while Q acts irreducibly on Next we investigate groups of order divisible by more than two primes.This needs some preparation.Lemma 4.12.Suppose that P and R are distinct Sylow subgroups of G, that Q ∈ Syl q (G) is cyclic and that it normalizes P and R, but does not centralize them.Suppose further that R normalizes every Q-invariant subgroup of P.
We claim that G/E is not L 9 -free, and for this we may suppose that E = 1.Then Q = 1 acts faithfully on P and R. Now we need a technical step before we move on: There are a Q-invariant subgroup D of P and elements x, y , and for all integers n we have the following: , u] and we can use the information from the end of the previous paragraph: This concludes the proof of ( * ).
We use ( * ) and its notation and, similarly, we find a Q-invariant subgroup R 0 of R and an element and F := DR 0 Q, and we claim that {A, B, C, D, E, F, S, T, U} is isomorphic to L 9 .The properties L9 (i) and L9 (iii) of Lemma 3.5 follow from the choice of D, since h and Q normalize D. For L9 (ii) we first note that Using Lemma 3.5 we conclude that G/E is not L 9 -free, and hence G is not L 9 -free.
Corollary 4.13.Suppose that p, q and r are pairwise distinct primes and that G is a directly indecomposable L 9 -free {p, q, r}-group.Suppose further that P Proof.Since G is directly indecomposable, we see that Q acts non-trivially on both P and R.Moreover, PQ and RQ are L 9 -free by hypothesis, and then we conclude that Q is cyclic or isomorphic to Q 8 .
In the first case, our assertion follows from Lemma 4.12, and in the second case, we choose a maximal subgroup Q 1 of Q.Then Q 1 acts irreducibly on P and R by Lemma 4.9, and the same Lemma shows that (Q) inverts P and R. Thus Q 1 acts on P and on R avoiding L 9 , respectively, and it acts faithfully.This contradicts Lemma 4.12.
We explain another example where a subgroup lattice contains L 9 .
Lemma 4.14.Suppose that p, q and r are pairwise distinct primes and that G is a {p, q, r}-group.Suppose If R acts irreducibly on P, but non-trivially, and if 1 = [P, Q] is elementary abelian, then G is not L 9 -free.
Proof.We first remark that G is soluble, because P ¢ PR ¢ PRQ = G.We will construct the lattice L 9 in L(G) using Lemma 3.
Assume for a contradiction that X := R c , Q has odd order for some c ∈ {a, b}.
In both cases X is a p -Hall subgroup of the soluble group G = PRQ and therefore R c = O r (X).It follows that Q normalizes R c and then that Q c −1 and Q normalize R.
Since N P (R) is R-invariant and R acts irreducibly, but nontrivially on P, we conclude that N G (R) = RQ.Thus Sylow's theorem provides some y ∈ R such that and this contradicts the fact that c It follows that X has even order and since R c acts irreducibly on P, we conclude that P ≤ X.This implies that We finally set B := PQ z for some z ∈ R # .Then R = z and Lemma 1.4 (b) and (c), together with our hypothesis, show that Altogether Lemma 3.5 gives the assertion.
Proposition 4.15.Suppose that p, q, and r are pairwise distinct primes and that G is a non-nilpotent L 9free {p, q, r}-group with normal Sylow p-subgroup P. Suppose further that R ∈ Syl r (G) and Q ∈ Syl q (G) Then RQ is a batten, P is elementary abelian of order p r , R and Q act irreducibly on P and (Q) induces non-trivial power automorphisms on P.
Proof.We proceed in a series of steps.
(1) The groups PQ and PR are not nilpotent, Q is cyclic and and in particular PR is not nilpotent.But PR is L 9 -free, because G is.Moreover, RQ is non-nilpotent and L 9 -free, again by hypothesis.Then Corollary 4.10 implies that R and Q are battens, that R acts on P avoiding L 9 and that Q acts on R avoiding L 9 .More specifically, R and Q are cyclic or isomorphic to Q 8 , and [P, R] as well as [R, Q] are elementary abelian or isomorphic to Q 8 .
It follows that R ∼ = Q 8 or that R is cyclic of order r.In both cases Lemma 1.1 yields R = [R, Q] and the avoiding L 9 action of Q on R gives that Q is cyclic.
In addition and then the Three Subgroups Lemma (see for example 1.5.6 of [5] If it was true that Proof.Since [P, R] = 1 by (1), we can apply Corollary 4.10 to PR, and this shows that R acts on P avoiding L 9 .Otherwise (1) implies that R ∼ = Q 8 and then Definition 4.8 gives that R acts faithfully on P. If |R| = r, then R acts faithfully of P because [P, R] = 1.In both cases we see that C R (P) = 1, and then the last statement of (1) implies that [R, C Q (P)] ≤ C R (P) = 1.Then C Q (P) centralizes P and R, and Q is cyclic by (1), and therefore it follows that C Q (P) ≤ Z(G).
Proof.We suppose that p = 2. Let − : G → G/Z(G) be the natural homomorphism.We show that Ḡ satisfies the hypotheses of our lemma.
From (1) we see that none of the groups P, Q or R is contained in Z(G).We even have that R∩Z(G) = 1 by (1).In particular p, q, r ∈ π( Ḡ) and Ḡ is L 9 -free.We see that P is a normal Sylow p-subgroup of G and that and X is a characteristic Sylow subgroup of XZ(G) and hence normal in G.This is a contradiction.
We deduce that all hypotheses of the lemma hold for Ḡ and that then the minimal choice of G implies that ( Q) induces non-trivial power automorphisms on the elementary abelian group P. Then Lemma 1.3 yields that p = 2.
Proof.As PR and PQ are not nilpotent by (1), Corollary 4.10 implies that X acts on P avoiding L 9 and that [P, X] is elementary abelian or isomorphic to Q 8 for both X ∈ {Q, R}.
Assume for a contradiction that [P, X] is isomorphic to Q 8 for some X ∈ {Q, R}.Then X acts on P of type (hamil).Hence we obtain a group I of order 1 or 2 such that P ∼ = Q 8 × I.It follows that Aut(P) is a {2, 3}-group.But this is impossible because Q and R both act coprimely and non-trivially on P by (1), and p = 2, q and r are pairwise distinct.
We conclude that [P, Q] and [P, R] are elementary abelian, and then it follows that P is abelian, by Lemma 4.4, applied to PR.
Assume for a further contradiction that C P (R) = 1.As R avoids L 9 in its action on P, it follows from Lemma 4.9 that R is not isomorphic to Q 8 .Now (1) yields that R is cyclic and we may apply Lemma 4.4 to PR, because [P, R] is elementary abelian.It follows that C P (R) is a cyclic 2-group and thus q and r are odd.
Furthermore C P (R) is normalized by Q, because Q normalizes P and R. Since q is odd, it follows that C P (R) is centralized by Q.We recall that P is abelian, and then (3) yields that C P (R) ≤ Z(G) = 1.This is a contradiction.
Altogether C P (R) = 1 and Lemma 1.1 gives that P = [P, R] is elementary abelian.
(5) C P (Q) = 1 and Q acts on P avoiding L 9 of type (std).
Proof.Since PQ is not nilpotent and Q is cyclic by (1), Corollary 4.10 implies that the action of Q on P avoids L 9 .But P is elementary abelian by (4), and therefore the action is not of type (hamil).
We assume for a contradiction that C P (Q) = 1.Then it follows that PQ is not of type (std).We consequently have type (cent) and we see that C P (Q) is a cyclic 2-group.In particular p = 2, and thus q and r are odd.Then |R| = r by (1).
Next we claim that G satisfies the hypotheses of Lemma 4.14.First, q and r are pairwise distinct odd primes and G is a finite {2, q, r}-group.From above, (1) and our assumption we see that P ∈ Syl 2 (G) is normal in G and that Q ∈ Syl q (G) and R ∈ Syl r (G) are cyclic groups such that R ¢ RQ.We have shown that R has order r.
We recall that C P (Q) is a cyclic 2-group (first paragraph).As r is odd and C P (R) = 1 by (4), we see that and now the irreducible action of R on P and the fact that 1 By (4) P is an elementary abelian 2-group, and P = [P, R] = 1 by ( 1) and ( 4).In particular R does not induce power automorphism on P by Lemma 1.3.As C P (R) = 1 by (4), we deduce that R acts irreducibly on P (using Lemma 4.4).Finally, (1) and ( 4) yield that 1 = [P, Q] is elementary abelian.
All hypotheses of Lemma 4.14 are satisfied now, and we infer that G is not L 9 -free.This is a contradiction.
Thus C P (Q) = 1 and we deduce that the action of Q on P is not of type (cen).It remains that Q acts on P avoiding L 9 of type (std).( 6) R and Q act irreducibly on P. If X ≤ Q induces power automorphisms on P, then X centralizes R.
Proof.We recall from (2) that and RQ/C Q (P) is isomorphic to a subgroup of Out(P).Since RQ is not nilpotent, we see that RQ/C Q (P) is not nilpotent.
The group P is an elementary abelian p-group by ( 4) and hence, if X ≤ RQ induces power automorphisms on it, then it follows that XC Q (P)/C Q (P) ≤ Z(RQ/C Q (P)), see page 177 of [3].We denote this fact by ( * ).Then we deduce that In addition, the fact ( * ) shows that neither R nor Q induces power automorphisms on P. It follows from ( 5) and Definition 4.6 (std) that Q acts irreducibly on P. Furthermore (1), together with Corollary 4.10, yields that R acts on P avoiding L 9 .Since C P (R) = 1 by (4), this action has type (std) or (hamil).In the first case, the irreducible action follows from Definition 4.6 (std), and in the second case, it follows from Lemma 4.9.
Proof.We set Q 0 := C Q (R) and we assume that Q 0 acts irreducibly on P. Then II 3.11 of [3] implies that RQ = C RQ (Q 0 ) is isomorphic to a subgroup of the multiplicative group of some field of order |P|, and it follows that RQ is cyclic.This is a contradiction.
Thus (6) and Lemma 4.11 show that Q 0 induces power automorphisms on P. Since R normalizes every Q-invariant subgroup of P by (7), we see from Lemma 4.12 and (2 By (5) and Lemma 4.11, it follows that every subgroup of y either acts irreducibly on P or induces power automorphism on it (in particular normalizing every subgroup of P).Then P y satisfies (b) of Lemma 3.1 in [8], which implies that it satisfies one of the possibilities 3.1 (i)-3.1 (iii).By (5) and the choice of y, we see that 3.1 (i) is not true.Further (7) provides some x ∈ C Q (R) that induces a power automorphism of order q on P.This implies that q divides p−1 and therefore P y satisfies (ii) of Lemma 3.1 (b) in [8].It follows that |P| = p q and that, if k is the largest positive integer such that q k divides p − 1, then y induces an automorphism of order q k+1 on P. We conclude that q k+1 = | y : C y (P)| = | y : C R (P)|, because Q is cyclic.Finally, we deduce that o(y) is uniquely determined, that Q = y and thatC By (6), we see that R and Q act irreducibly on P, and (1) gives that Q is cyclic.Then ( 8) and (7) say that (Q) induces nontrivial power automorphisms on P. In addition P is elementary abelian by (4), and it has order p r by (8).If |R| = r, then R is a cyclic group of order r and RQ is a batten.In particular G satisfies the assertion of our lemma.
But G is a counterexample, and then it follows that R ∼ = Q 8 and r = 2. Then (4) yields that PR fulfills the hypothesis of Lemma 4.7, and consequently p r = p 2 = |P| = p q by (8).This is our final contradiction, because q = r.Definition 4.16.Suppose that B is a non-nilpotent batten that acts coprimely on the p-group P. We say that the action of B on P avoids L 9 if and only if [P, Z(B)] = 1 and if one of the following occurs: Proof.If B ∼ = Q 8 , then Definition 4.8 and Lemma 4.9 imply that P is elementary abelian and that B acts irreducibly on it.We conclude that P = [P, B] is elementary abelian and we deduce from Lemma 1.1 that C P (B) = 1.Hence, in this case, all statements of our lemma hold.Now suppose that B is not nilpotent and that it acts of type (NN).Then Definition 4.16 states that, once more, P is elementary abelian and B acts irreducibly on it.Again we see that P = [P, B] is elementary abelian, and as before all statements hold.
Next we suppose B is not nilpotent and that it acts of type (Cy), or that B is cyclic.In the first case B has a cyclic Sylow subgroup Q that acts on P avoiding L 9 such that C P (Q) = C P (B) and [P, B] = [P, Q] by Definition 4.16.In the second case we set Q := B.
Then, in both cases, Q is a cyclic group that acts on P avoiding L 9 such that C P (Q) = C P (B) and If Q acts of type (std), then P = [P, Q] is elementary abelian by Definition 4.6.Again we deduce the statements of our lemma.
Suppose that Q acts of type (cent).Then Definition 4.6 yields that [P, Q] = [P, B] is elementary abelian, that P is abelian and that C P (Q) = C P (B) is a cyclic 2-group.In particular P = [P, B] × C P (B) by Lemma 1.1.It also follows that C P (B) is centralized by every automorphisms of P of odd order that leaves C P (B) invariant.These are the statements of our lemma.
Finally, suppose that Q acts of type (hamil).Then Definition 4.6 yields that [P, Q] ∼ = Q 8 and P = [P, Q] × I, where I is a group of order at most 2. In particular statement (a) is true.Moreover, we deduce that C P (Q) ≤ 1 (P) = ([P, Q]) × I, where ([P, Q]) is cyclic of order 2. In particular ([P, Q]) is centralized by B and by every automorphisms of P. We conclude that C P (Q) is elementary abelian of order at most 4 and that every automorphism of P centralizes a cyclic subgroup of order 2.This implies (b).

Avoiding L 9
We now work toward a classification of arbitrary L 9 -free groups, and therefore we need to understand in more detail the group structures that appear when "L 9 is avoided" in the sense of the previous section.Definition 5.1.Suppose that K is a batten group that acts coprimely on the p-group P. We say that the action of K on P avoids L 9 if and only if [P, K] = 1 and every batten of K either centralizes P or avoids L 9 in its action on P. Lemma 5.2.Let K be a batten group that acts coprimely on the p-group P avoiding L 9 .Suppose further that L ≤ K and L 0 ¢ L such that [P, L] = 1 = [P, L 0 ].Then L/L 0 acts on P avoiding L 9 .In particular L/L 0 is a batten group.
Proof.By induction we may suppose that K is a batten and that either L 0 = 1 and L is a maximal subgroup of K or that L 0 is a minimal normal subgroup of K = L. Thus either |L 0 | has order q or |K : L| = q.Since L 0 ≤ C K (P), we first remark that L/L 0 induces automorphisms on P.
If K ∼ = Q 8 , then K acts faithfully on P by Definition 4.8.Thus L 0 ≤ C K (P) = 1 and it follows that L is a cyclic group of order 4. Thus Lemma 4.9 yields that 1 (L) inverts P and that L acts irreducibly on the elementary abelian group P = [P, L].Then we see that L ∼ = L/L 0 acts on P avoiding L 9 of type (std).
Next suppose that K is cyclic.Then L/L 0 is cyclic.If K acts of type (std) on P, then L and every subgroup of L act irreducibly or via inducing power automorphisms on the elementary abelian group P = [P, K].Since [P, L] = 1 and power automorphisms are universal, by Lemma 1.5.4 of [7], it follows that P = [P, L].Moreover, the action of L on P is equivalent to that of L/L 0 , and then it follows that L/L 0 acts on P avoiding L 9 of type (std).
If K acts on P of type (cent), then L and all its subgroups act irreducibly or trivially on the elementary abelian group [P, K].Again the fact that [P, L] = 1 implies that P = [P, L], and then C P (L) = C P (K) by Lemma 1.1.Since the action of L on P is equivalent to that of L/L 0 , it follows that L/L 0 acts on P avoiding L 9 of type (cent).
We suppose now that K acts of type (hamil).Then K is a cyclic 3-group and K/C K (P) has order 3.It follows that L = K.But again, the action of L/L 0 = K/L 0 on P is equivalent to the action of K on P, which means that it has type (hamil).
We finally suppose that K is a non-nilpotent batten.Let R be a Sylow subgroup of K such that K = B(K) • R. Suppose first that L/L 0 is a q-group.Then our choice of L and L 0 implies that L/L 0 ∼ = R.If K acts on P of type (Cy) in this case, then it follows that L/L 0 ∼ = R is cyclic and that it acts on P avoiding L 9 , according to Definition 4.16.Otherwise, if K acts of type (NN), then B(K) C K (P) and then L 0 = 1.It follows that L = R acts irreducibly on the elementary abelian group P, whence P = [P, L] = [P, L/L 0 ].In addition (L) and all of its subgroups induce power automorphism on P. Altogether the cyclic group L/L 0 ∼ = L acts on P of type (std).Now we suppose that L/L 0 does not have prime power order.Then We have already proven that R acts on P avoiding L 9 , and then Z(K) also acts on P avoiding L 9 .In addition B(K) either centralizes P or it acts irreducibly on the elementary abelian group P = [P, B(K)].Since B(K) has prime order, it follows that the cyclic group B(K) acts on P avoiding L 9 of type (std).
Finally, suppose that L/L 0 is not nilpotent.Then L is not nilpotent and hence Lemma 2.7 implies that L = K.We conclude that L 0 = 1.Since [P, Z(K)] = 1 by Definition 4.16, it follows that L 0 is a proper subgroup of Z(K) and that L/L 0 = K/L 0 ∼ = B(K) R/L 0 is a non-nilpotent batten.If [P, B(K)] = 1, then our investigation above imply that the cyclic group R/L 0 acts on P avoiding L 9 , and then K/L 0 acts on P avoiding L 9 .Otherwise 1 = [P, B(K)] is elementary abelian of order p |K:B(K)Z(K)| = p |K/L 0 :B(K)Z(K)/L 0 | , moreover B(K) ∼ = B(K)L ) /L 0 and R/L 0 act irreducibly on P. At the same time Z(K/L 0 ) = Z(K)/L 0 induces power automorphisms on P. Altogether K/L 0 acts on P avoiding L 9 of type (NN).Lemma 5.3.Let K be a batten group that acts coprimely on the p-group P avoiding L 9 .Then the following assertions are true: Proof.Let L ¢ K be such that [P, L] = 1.Then L is a batten group by Lemma 2.7 and therefore there is a batten B of L such that [P, B] = 1.Assume for a contradiction that [P, B] ≤ [P, L] [P, K] ≤ P. Then the fact that P = [P, B] implies that C P (B) = 1 by Lemma 1.1.In addition B avoids L 9 in its action on P, by Lemma 5.2.Since B is a batten of L, it is characteristic in L, and therefore B ¢ K.In particular C P (B) is K-invariant and hence it is centralized by K by Lemma 4.17 (c).This implies that which is a contradiction.In particular (a) is true.
Together with Lemma 4.17 (b), the statement in (b) follows from (a).
Lemma  ).In the first case B = C B (P)Q induces power automorphism on P and in the second case B acts irreducibly on [P, K]/ ([P, K]).This is a contradiction.

Lemma 5.5. Let K be a batten group that acts on the p-group P avoiding L 9 , and suppose that H is a subgroup of K that acts non-trivially on R ≤ P. Then C H (R) = C H (P), C P (H) = C P (K) and [P, H] = [P, K].
Proof.From Lemma 5.4 we see that [P, H] = [P, K].In addition C P (K) ≤ C P (H) ≤ C P (Q) for every q-subgroup Q of H and every prime q.Let Q be a q-subgroup of H for some prime q such that C P (Q) = P. Then Q is a batten by Lemma 2. Finally, suppose that R ≤ P is H-invariant, but not centralized by H, and set H 0 := C H (R) ≥ C H (P). Assume for a contradiction that H 0 does not centralize P. Then we deduce, as above, that C P (K) = C P (H 0 ) ≥ R.This is a contradiction, because H does not centralize R. For the final comment we just use that p is odd and then apply (a).
Lemma 5.7.Let B be a batten that acts on the p-group P avoiding L 9 .Let R ≤ P be B-invariant and R 0 ≤ C P (B).
Then B avoids L 9 in its action on R/R 0 .
Proof.Since B centralizes R 0 , the action of B on R/R 0 is well-defined.We first suppose that B ∼ = Q 8 .Then Lemma 4.9 yields that B acts irreducibly on P, and then it follows that R = P and R 0 = 1.Thus our assertion is true in this case.
Next suppose that is cyclic.If B acts of type (std) on P, then R 0 ≤ C P (B) = 1.If B acts irreducibly on P, then again P = R and there is nothing left to prove.Otherwise B and all of its subgroups induce power automorphisms on P, and hence on R = [R, B] as well.It follows that B also acts of type (std) on R ∼ = R/R 0 .
Suppose now that B acts of type (cent).Then, since B does not centralize R and B acts irreducibly on [P, B], it follows that [P, B] ≤ R. Moreover P is abelian and then we use the fact that R 0 ≤ C P (B).This gives that R/R 0 = [R/R 0 , B]×C R/R 0 (B) ∼ = [R, B]×C R (B)/R 0 , where C R (B)/R 0 is a cyclic 2-group.Since every subgroup of B that does not centralize P acts irreducibly on [P, B] ∼ = [R/R 0 , B] in this case, there are two possibilities for the action of B on R ∼ = R/R 0 : If R 0 = C R (B), then B acts of type (cent), and otherwise it acts of type (std).
Suppose now that B acts of type (hamil).Then the cyclic 3-group B acts irreducibly on [P, B]/ ([P, B]), by Lemma 5.4, and we see again that [P, B] ≤ R. It follows that R ∼ = Q 8 × I, where I is a group of order at most 2, and then Otherwise we have that R 0 ≥ ([P, B]) and therefore R/R 0 is elementary abelian of order 4 or 8. Moreover B acts irreducibly on [R/R 0 , B], which is a group of order 4. In addition every proper subgroup of B centralizes R/R 0 .Consequently, if R/R 0 = [R/R 0 , B], then B acts on R/R 0 of type (std) or of type (cent).
The final case is that B is not nilpotent, and we suppose that B acts of type (NN) on P. Then Definition 4.16 yields that B acts irreducibly on P. Hence there is nothing left to prove.
Suppose that B acts of type (Cy).Then we choose a Sylow subgroup

R, B(B)] = [P, B(B)]
= 1 and Q acts on P avoiding L 9 in such a way that (Q) = Z(B) does not centralize P. Then Lemma 5.5 yields that (Q) does not centralize R. In particular, we have that [R/R 0 , Z(B)] = 1.In addition Q acts on R/R 0 avoiding L 9 , by our arguments above.Altogether B acts on R/R 0 avoiding L 9 of type (Cy) in this final case.

The first implication
We now investigate the general case.

Proposition 6.1. Let G be a finite L 9 -free group. Then G = NK, where N is a nilpotent normal Hallsubgroup of G with modular Sylow subgroups and K is a batten group. Moreover, for all p ∈ π(N), every batten of K acts on O p (N) avoiding L 9 or it centralizes O p (N).
Proof.We first remark that G is L 10 -free, whence Theorem A of [9] implies that G is soluble.Furthermore, Corollary C of [9] provides normal Hall-subgroups N and M of G such that N ≤ M and such that N is nilpotent, M/N is a 2-group and G/M is metacyclic.We choose N as large as possible with these constraints.From Lemma 4.1 we see that N = 1.We also have that every Sylow subgroup of N is L 9 -free, and hence it is modular by Lemma 4.2.In addition the Schur-Zassenhaus Theorem (see for example 3.3.1.of [5]) provides a complement K of N in G.
(1) If RQ is a non-nilpotent Hall {r, q}-subgroup of K, where R is a normal Sylow r-subgroup of RQ and Q ∈ Syl q (RQ), then RQ is a batten.For all p ∈ π(N), the group RQ centralizes O p (N) or acts on it avoiding L 9 .
Proof.If there is some p ∈ π(N) such that [O p (N), R] = 1, then we set P := O p (N). Otherwise the maximal choice of N implies that R is not a normal subgroup of K.Then, using the solubility of G, we find a prime s ∈ π(K) \ {r} and a normal s-subgroup In this case we set P := T.
In both cases p, r and q are pairwise different primes and PRQ is a non-nilpotent {p, r, q}-subgroup that satisfies the hypothesis of Proposition 4.15.For this we note that (2) Every Sylow subgroup S of K is a batten, and for all p ∈ π(N) it is true that S centralizes O p (N) or acts on O p (N) avoiding L 9 .
Proof.Let S be a Sylow subgroup of K.If S centralizes N, then the choice of N provides some Sylow subgroup R of K such that RS is not nilpotent.Then (1) implies that RS is a batten, then that S is cyclic and hence that S is a batten.
Let p ∈ π(N/C N (S)).Then O p (N)S is an L 9 -free {p, q}-group for some prime q.Hence Corollary 4.10 implies the assertion.
(3) K is a batten group.
Proof.Let 1 = B ≤ K be such that there is some and B is not a direct product of nontrivial subgroups of coprime order.In particular B is a Hall subgroup of K.If B is nilpotent, then B is a Sylow q-subgroup of K for some prime q.In this case (2) implies that B is a batten.
Assume for a contradiction that B is not a batten of K. Then B is not nilpotent and therefore (1) yields that |B| is divisible by at least three different primes.Since B ≤ G is L 9 -free, Lemma 4.1 provides a normal Sylow r-subgroup R of B for some prime r ∈ π(B).We remark that R is a normal subgroup of K = K 1 × B. In addition B is is not a direct product of non-trivial subgroups with coprime order and hence there are a prime q ∈ π(B) and some Q ∈ Syl q (B) such that RQ is not nilpotent.Now B is a Hall subgroup of K and thus RQ is a Hall subgroup of K.In particular (1) implies that RQ is a batten and it follows that |R| = r and 1 = C Q (R) = (Q).We further see, from Definition 4.16 and (1), that for every p ∈ π(N) with the property [O p (N), R] = 1 we have that |O p (N)| = p q .Since R is a normal Sylow subgroup of K, the maximal choice of N provides some p ∈ π(N) such that R does not centralize P := O p (N).In particular we have that |P| = p q .
Let s ∈ π(B) \ {q, r} and let S be a Sylow s-subgroup of B such that QS = SQ.Such a subgroup exists by Satz VI. 2.3 in [3].If S does not centralize R, then RS is not nilpotent and therefore our arguments above show that p s = |P| = p q .This is impossible because r = s.Consequently [R, S] = 1.
Since B is directly indecomposable, we conclude that SQ is not nilpotent.But SQ is a Hall subgroup of B and then it is a Hall subgroup of K.In particular (1) yields that SQ is a batten.From the fact that 1 = C Q (R) = (Q) we conclude that |Q| = q, and then |S| = s and S ¢ SQ.In addition Q is an L 9 -free group, and this situation contradicts Corollary 4.13.
We conclude that, by construction, G = NK, where N is a nilpotent normal subgroup of G with modular Sylow subgroups and K is a batten group, by (3), such that for all p ∈ π(N) it is true that every batten of K centralizes O p (N) or acts on O p (N) avoiding L 9 , by ( 1) and ( 2).
The converse of Proposition 6.1 is false, as can be seen in the following example and subsequent lemma. Then Moreover, N := H × J is a nilpotent normal subgroup of G with modular Sylow subgroups.Since and , it follows that x and y have coprime order.
Thus K := x × y is cyclic, which means that it is a batten group.We see that x and y induce non-trivial power automorphisms on H. Thus every batten of K acts on H = O 19 (N) avoiding L 19 of type (std).In addition x 2 and y 3 induce power automorphisms on J. Since x and y act irreducibly on J, it follows that every batten of K acts on J = O 5 (N) avoiding L 19 of type (std), too.
Altogether G satisfies the conclusion of Proposition 6.1.
On the other hand we observe that π If g ∈ HJ centralizes x or y , then g = 1.Thus the following lemma yields that G is not L 9 -free.

Lemma 6.3. Let H be a non-trivial abelian group where all non-trivial Sylow subgroups are non-cyclic elementary abelian, let L be a cyclic group inducing power automorphism on H such that π(L) = π(L/C L (H))
and let 1 = J be an abelian group admitting L as a group of automorphisms such that the action of L on O p (J) avoids L 9 for every p ∈ π(J) and such that (|H|, |J|) = 1.
Then (H × J) L is not L 9 -free.
Proof.We first set L 1 := O π (L) and L 2 := O π (L).Since none of the groups L 1 nor L 2 is centralized by any element of HJ \ {1} = ∅, it follows that L 1 and L 2 are both non-trivial.For every odd prime p ∈ π(H) there is an elementary abelian subgroup H p of H that has order p 2 .In particular there are elements a p and b p of H such that H p = a p × b p .
We set a := These elements are well-defined (in the sense that the ordering of the primes does not matter) because H is abelian.For every p ∈ π(J), we further see that L acts on O p (J) avoiding L 9 .Since J is abelian, we deduce from Lemma 5.3(b) that [O p (J), L] is elementary abelian.In addition L acts irreducibly on [O p (J), L] or it induces power automorphisms on O p (L), by Lemma 5.4.We choose x p ∈ [O p (J), L] # .Then L acts irreducibly on P := x L p .Next we choose l ∈ L such that L = l .Then 1 = x l p ∈ P ≤ HJ and 1 = [l, x p ] ∈ P ≤ HJ.Thus our hypothesis implies that 1 = [[l, x p ], L 1 ] ≤ P and 1 = [x l p , L 2 ] ≤ P. Altogether we have that For every p ∈ π(L) we choose 1 = x p ∈ [O p (J), L], and then we set x := p∈π(J) x p .
Since J is abelian, it follows that Next we set y := x l .Then our previous arguments show that We will construct a subgroup lattice L 9 using Lemma 3.5.For this we set E := C L (HJ) ¢ HJL and D := C L (J).
For every q ∈ π we have In particular E = D and hence (L9(i)) of Lemma 3.5 holds.
Next we set A := a L x 1 E, S := L ax 1 E and T := L a −1 x 1 E. Then we have that A contains the subgroups S, T and In particular, since D centralizes x, it follows that D, T = D, S = A.Moreover, we have that = A and therefore we conclude that S, T = A as well.Next Lemma 5.5 gives that 1 is a π -group for all c ∈ {a, a −1 , a 2 }, whence With all these properties, we see that (L9(ii)) of Lemma 3.5 is true.Now we set C := b DL 2 and Let c ∈ {a, a −1 } and let X := L cx 1 , L 2 .Then X contains a π -Hall subgroup as well as a π(L 2 )-Hall subgroup of HJL.Since HJL is soluble, there is a π(L)-Hall subgroup K of X such that L 2 ≤ K and some g ∈ HJ such that Lemma 1.4(b).The hypothesis of our lemma yields that g = 1 and therefore L = K ≤ X.From there we obtain some h ∈ X such that Lemma 1.4(b).This forces cx = h ∈ X, and then c, x ∈ X, because H and J have coprime order and centralize each other.Altogether c = a and x L = J 0 are subgroups of X, and we conclude that X = a J 0 L. Thus We set F := a, b J 0 L in order to obtain Part (L9(iv)) of Lemma 3.5.Moreover, Dedekind's law and Part (a) of Lemma 1.4 gives that We set B := ab L y .Then In a similar way we obtain that A ∩ B ≤ ab D and therefore We further calculate that Altogether {A, B, C, D, E, F, S, T, U} satisfies every condition of Lemma 3.5, which means that it is isomorphic to L 9 .
The previous lemma and Lemma 4.12 motivate the following definition: Definition 6.4.Here we define a class L of finite groups, and each group in L has a type.
We say that G ∈ L has type (N, K) if and only if the following hold: where N is a normal nilpotent Hall subgroup of G with modular Sylow subgroups and K is a batten group.(L2) If p ∈ π(N), then every batten of K centralizes O p (N) or it acts on it avoiding L 9 .(L3) For all Sylow subgroups Q of K and all distinct Sylow subgroups P and R of N that are not centralized by Q, we have that Then there is some g ∈ (HJ) # that centralizes O π (L) or O π (L).
Theorem 6.5.Let G be a finite L 9 -free group.Then G ∈ L.
Proof.From Lemma 6.1 we see that G = NK and that (L1) and (L2) are satisfied.For (L3) we let Q be a Sylow subgroup of K and we let P and R be distinct Sylow subgroups of N that are not centralized by Q.Since N is a nilpotent normal Hall subgroup of G, it follows that [P, R] = 1 and that Q normalizes P and R. Then (P×R)Q is directly indecomposable and L 9 -free, whence Corollary 4.13 gives that C Q Finally, we look at (L4) and we assume that it is not true.Then there is an abelian subgroup H of N such that the nontrivial Sylow subgroups are not cyclic, and we find a cyclic group We note that P does not centralize O π (L) ≤ L. Then we find a prime q ∈ π(L) such that a Sylow q-subgroup Q of L does not centralize P. Using Lemma 5.2, we see that Q acts on O p (N) avoiding L 9 and then Lemma 5.7 yields that Q acts non-trivially on P, and avoiding L 9 .Now we may apply Corollary 5.6: Since L induces power automorphisms on P, Part (c) shows that P is elementary abelian.Then the hypotheses of Lemma 6.3 are satisfied.It says that (H × J) L is not L 9 -free, which is false.We conclude that (L4) holds.Proof.Let G ∈ L be of type (N, K).Then N is nilpotent normal Hall subgroup of G and G/N ∼ = K is a direct product of p-groups or of groups whose order is divisible by exactly two primes.Thus G/N is soluble as well, and it follows that G is soluble.

Lemma 7.2. Let G ∈ L be of type (N, K) and π := π([N, K]). Then every subgroup of O π (N) is normal in N.
Proof.Let U be a subgroup of O π (N).Then U ¢ N if and only if O p (U) ¢ O p (N) for all p ∈ π , since N is nilpotent.Let p ∈ π .We note that this implies that O p (N) is not centralized by K.In particular there is a batten of K that acts non-trivially on O p (N) and avoiding L 9 .If O p (N) is abelian, then O p (U)¢O p (N).If O p (N) is not abelian, then we apply Lemma 4.17 (b) to a batten B of K that acts non-trivially on O p (N).The first possibility described there implies that O p (N) is abelian, which is not the case here.Thus the second possibility holds, and then O p (N) ∼ = Q 8 × I, where I is cyclic of order at most 2. We conclude that O p (N) is hamiltonian and it follows that O p (U) ¢ O p (N).
Proof.We set M := N ∩ X.Then M is a normal Hall subgroup of X, because N is one of G. Then the Schur-Zassenhaus Theorem provides a complement C of M in X, and we notice that C and M have coprime orders.Therefore Since G is soluble by Lemma 7.1, such a complement is conjugate to K, and thus we find g ∈ G such that C ≤ K g .The coprime action of K on N yields, together with Lemma 1.1, that N = C N (K)[N, K], and therefore . By conjugation we may suppose that g = 1 and we set K 1 := U ∩ K. Then Lemma 2.7 yields that K 1 = U ∩ K g ≤ K g ∼ = K is a batten group.Moreover, M := U ∩ N ≤ N is a normal nilpotent Hall subgroup of U with modular Sylow subgroups, by (L1).This means that (L1) holds for U, and now we turn to (L2) and let p ∈ π(M).Suppose that B is a batten of K 1 that does not centralize O p (M).Then it does not centralize O p (N) and therefore Lemma 5.2 implies that B ∼ = B/1 acts on O p (N) avoiding L 9 .Then we may apply Lemma 5.7 to see that B also acts on This gives property (L2) of Definition 6.4 for U, and (L4) follows because M ≤ N and K 1 ≤ K.
For (L3) we let Q 1 be a Sylow subgroup of K 1 and we let p, r ∈ π(M) be different primes such that First we let Q be a Sylow subgroup of ), using Property (L3) for G.In particular, these centralizers cannot both be trivial, and we may suppose that C Q (O p (N)) = 1.Then Q does not act faithfully on O p (N), but the action of Q on O p (N) avoids L 9 .Definition 4.8 immediately gives that Q ∼ = Q 8 .Then it follows that Q is cyclic, which means that the subgroup lattice L(Q) of Q is a chain, and Q 1 is also cyclic.
We assume for a contradiction that ) and that Q is cyclic, and now we may suppose that ) and (L3) holds for U as well.
Lemma 7.5.Let G ∈ L be of type (N, K) and suppose that S is a normal Sylow q-subgroup of K that centralizes N for some prime q.Let K 1 be a Hall q -subgroup of K.
Then G is also of type (N × S, K 1 ).In particular, if we choose (N, K) such that |N| is as large as possible, then π(K) = π(K/C K (N)).
Proof.We show that G = (N × S)K 1 satisfies (L1)-(L4) of Definition 6.4, and we first note that K 1 is a batten group by Lemma 2.7.The structure of K forces all Sylow subgroups of K to be cyclic or quaternion, more specifically S is cyclic or isomorphic to Q 8 .This means that S is modular.Since S is a normal Sylow q-subgroup of K and N is a Hall subgroup of G, by hypothesis, it follows that N × S is a Hall subgroup of G where all Sylow subgroups are modular.
By hypothesis [N, S] = 1 and N is nilpotent, hence N × S is nilpotent, too.This is (L1).For (L2) we let B be a batten of K 1 and p ∈ π(N × S).We keep in mind that B is not necessarily a batten of K -if it is, then it centralizes O p (N × S) or it acts on it avoiding L 9 , because of (L2) for G. Now we suppose that B is not a batten of K and that Then SB is a non-nilpotent batten of K.If p = q, then SB acts on O p (N) avoiding L 9 , and [O p (N), S] = 1.Then Definition 4.16 implies that SB acts of type (Cy) and it follows that B acts on O p (N) avoiding L 9 .Finally suppose that q = p.Then (B) centralizes S = O p (NS), while B induces power automorphisms on the cyclic group S of order p.Thus SB satisfies (std) of Definition 4.6, and we deduce that B acts on S avoiding L 9 .
We turn to (L3).Let Q be a Sylow subgroup of K 1 and let P and R be distinct Sylow subgroups of N × S that are not centralized by Q.First we note that Q is a Sylow subgroup of K because K 1 is a Hall subgroup of K by hypothesis.Therefore, if PR ≤ N, then we immediately have that Without loss suppose that R N, i.e.R = S. Then we recall that Q was chosen not to centralize P and R = S, which means that Q and S cannot come from distinct battens of K, but their product must be a non-nilpotent batten of K.Moreover, [P, QS] = 1.We obtain from Lemma 2.5 and Definition 4.16 that (Q) = Z(Q) = C Q (S) does not centralize P and therefore C Q (R) = C Q (P).This is (L3).
Finally, let H ≤ N × S be such that its nontrivial Sylow subgroups are not cyclic, let L ≤ Pot K 1 (H) be cyclic and such that π(L) = π(L/C L (H)) and let 1 = J be an L-invariant abelian subgroup of M such that As K is a batten group, the set π(K/C K (S)) contains at most one element.We recall that L ≤ K 1 ≤ K, and then it follows that, for every set of primes π , the group S centralizes O π (L) or O π (L).In order to prove (L4) of Definition 6.4, we may thus suppose that H and J are subgroups of N.
Altogether, G = (N × S)K 1 satisfies Definition 6.4.We now suppose that |N| is as large as possible and we assume for a contradiction that S ≤ C K (N) is a Sylow subgroup of K. Then S is not normal in K, hence there is a non-nilpotent batten B of K such Proof.By induction we may suppose that M is a minimal normal subgroup of G.We recall that G is soluble by Lemma 7.1, and we let r be prime such that M is an elementary abelian r-group.If M has a complement C in G, then G/M ∼ = C and Lemma 7.4 gives that C ∈ L and hence G/M ∈ L.
Consequently we may suppose that M does not have a complement in G.We choose N, K ≤ G such that G has type (N, K) and such that |N| is as large as possible.Then π(K) = π(K/C K (N)) by Lemma 7.5.
First suppose that r ∈ π(K).Then M ≤ K and we see that Next we let B be a batten of K that contains M. Then M ≤ C B (N), which means that for all p ∈ π(N), B does not act faithfully on O p (N).
Since there is some p ∈ π(N) such that B acts on O p (N) avoiding L 9 , it follows from Definition 4.8 that B is not isomorphic to Q 8 .In addition M Z(B), if B is not nilpotent, by Definition 5.1.Therefore, in this case, the section B/M is a non-nilpotent batten as well.We conclude that K/M is a batten group.
Assume for a contradiction that r ∈ π(N), but that M C N (K).Then we note that C M (K) ¢ G by Lemma 7.2, and we deduce that C M (K) = 1, because M is a minimal normal subgroup of G.This forces M ≤ [O r (N), K].Then [O r (N), K] is not elementary abelian, because otherwise M would have a complement in this commutator and hence in G.But we are working under the hypothesis that it does not.Now Lemmas 5.4 and 4.17 is the unique subgroup of order 2 of [O r (N), K], which means that it must be centralized by K and contained in M.But this is a contradiction.Thus M ≤ C N (K) and Corollary 5.6 implies that p = 2.We summarise that M ≤ C K (N) and that K/M is a batten group or M ≤ C N (K) and p = 2.
Let − : G → G/M be the natural homomorphism.Then Ḡ = N • K, where N is a normal nilpotent Hall subgroup of G with modular Sylow subgroups, since sections of modular p-subgroups are modular.Moreover K is a batten group.Thus Ḡ satisfies (L1) of Definition 6.4.We further deduce (L2) from Lemmas 5.2 and 5.7.
For (L3) we let Q be a Sylow subgroup of K and we let p, s ∈ π( N) be distinct primes such that We finally let 1 = H ≤ N be abelian with non-cyclic Sylow subgroups and L ≤ Pot K ( H) be cyclic with π( L) = π( L/C L( H)) and we let J be an abelian L-invariant subgroup of N such that (| H|, | J|) = 1 and C H (C L( J)) = 1.We set π : Then we assume for a contradiction that every nontrivial element ḡ ∈ HJ centralizes neither O π ( L) nor O π ( L).Then Lemma 1.1 yields that [ HJ , L] = HJ .We choose pre-images H, L and J in G of smallest possible order.Then HJ = [HJ, L] and π( X) = π(X) for all X ∈ {H, L, J}, because G is soluble by Lemma 7.1.In particular we have that (|H|, |J|) = 1.
If r ∈ π(X) for some X ∈ {H, J}, then r = 2. Then our assumption implies that C O 2 (X) (L) ≤ M. It follows that X ∼ = X or that M ≤ (X) = ([X, L]).In the second case, we apply Lemmas 5.4 and 4.17.Together they show that (L).But then we also have that [O 2 ( X), O σ ( L)] = 1 for some σ ∈ {π , π }, which is a contradiction.We deduce that H ∼ = H and J ∼ = J.In particular H ≤ N is abelian, with non-cyclic Sylow subgroups, and J = 1 is an abelian L-invariant subgroup of N. Since L is cyclic, it follows from our arguments above that L is also cyclic.Moreover We now investigate the action of L on H. Since H ∼ = H and M ∩ L = 1 or M ≤ C L (H), we see that L induces power automorphisms on H.In addition Lemma 1.1 shows that and then the fact that H ∩ M = 1 yields that C H (C L (J)) = 1.Altogether we obtain, by applying (L4) to G, some g ∈ HJ # such that g centralizes O π (L) or O π (L), where π This is a contradiction.
has prime power order.
Proof.We apply Lemma 7.4 and we see that for every r ∈ π(N) \ {p}, and this shows that 1 = [N, Lemma 7.8.Let G ∈ L be of type (N, K) such that C K (N) = 1, let q ∈ π(K) and let Q ∈ Syl q (K).Let p ∈ π(N) be such that 1 (Q) does not centralize P := O p (N).Then the following hold: (e) Suppose that X ≤ G, that q divides |X| and that x ∈ P. Then X = (X ∩ P)N X ( 1 (Q) x ) if and only if Proof.We set P 0 := [P, 1 (Q)].Then Lemma 7.7 implies that Furthermore K acts on P avoiding L 9 and then we have that But we also have that For (b) we recall that, by (a), the subgroups K and O p (N) normalize 1 (Q).Then G = PO p (N)K ≤ PN G (Q) ≤ G. Moreover P ¢ G and Lemma 1.1 implies that P = C P ( 1 (Q))P 0 , where From there we deduce that P 0 acts transitively on 1 (Q) G by conjugation.The second statement of (c) follows because G is soluble (Lemma 7.1), together with the fact that 1 (Q) is the unique subgroup of its order in the Hall subgroup K of G (Lemma 2.6).This means that every subgroup of order q of G is conjugate to 1 (Q), completing (c).
For (d) and (e) we let X ≤ G. Lemma 7.3 provides some g ∈ G such that X = (N ∩ X)(K g ∩ X).Moreover, (a) implies that K g and O p (N)(= O p (N) g ) normalize 1 (Q) g , and then we summarise: Using (b) we see that G = N G ( 1 (Q))P 0 , and then we take y ∈ N G ( 1 (Q)) and x ∈ P 0 such that g = yx.Now we deduce that X = (P ∩ X)N X ( 1 (Q) yx ) = (P ∩ X)N X ( 1 (Q) x ), as stated in (d).
Finally, suppose that q divides |X| and that x ∈ P. Suppose first that X = (X ∩ P)N X (Q ) and, by (d), there is some y Now, conversely, suppose that 1 (Q) x ≤ X.Then (d) provides some z ∈ P 0 such that X = (P ∩ X)(N X ( 1 (Q) z )).In the paragraph above we have shown that 1 (Q) z ≤ X.We apply (c) to X = (X ∩ N)(X ∩ K g ), which is a group in L by Lemma 7.4, and we obtain some Lemma 7.9.Let G ∈ L be of type (N, K) and let p ∈ π(N) such that K induces non-trivial power automorphisms on P := O p (N).
Then for all X, Y ≤ G, there is some In addition i = 0 if and only if there is some g ∈ P such that for both Z ∈ {X, Y} we have Z ≤ (Z ∩ P)O p (N)K g .Proof.We first remark that Lemma 7.2 gives that every subgroup of P is normal in N, and hence in G, because K normalizes every subgroup of P as well.In addition P = [P, K] is elementary abelian by Corollary 5.6 (c).
Let X, Y ≤ G. Then Lemma 7.3 provides x, y ∈ P such that X ≤ (X This implies that X, Y ≤ (X ∩ P)(Y ∩ P) x −1 y O p (N)K x , bearing in mind that X ∩ P and Y ∩ P are normal subgroups of G, and therefore X, Y ∩ P = (P ∩ X)(P ∩ Y) x −1 y .
Since P is elementary abelian, we see that o(xy −1 ) ∈ {1, p} and we deduce the first assertion.
If it is possible to choose x = y, then X, Y ∩ P = (P ∩ X)(P ∩ Y) x −1 y = (P ∩ X)(P ∩ Y) and in particular i = 0 in the statement of the lemma.
For the converse we suppose that i = 0, i.e. | X, Y ∩P| = |(P∩X)(P∩Y)|.Then x −1 y ∈ X, Y ∩P = (P ∩ X)(P ∩ Y) and thus we find x 0 ∈ X ∩ P and y 0 ∈ Y ∩ P such that x −1 y = x 0 y 0 .Then g := yy −1 0 = xx 0 ∈ P ∩ X ∩ Y.We note that x 0 normalizes X, centralizes P ∩ X and normalizes O p (N), which implies that Lemma 7.10.Let G ∈ L be of type (N, K).Suppose that X and Y are subgroups of G such that X, Y = G and let B is a batten of K. Suppose that K has a normal q-complement H. Then one of the following hold: (a) q |G : X|, (b) q |G : Y|, or (c) q = 2, K has a section isomorphic to Q 8 and 4 divides (|X|, |Y|).
Proof.Let Q ∈ Syl q (K) and suppose that q divides neither |G : X| nor q | |G : Y|.Then Lemma 7.3 gives maximal subgroups M X and M . This is impossible.We conclude that Q is not cyclic and then, since K is a batten group, it follows that Q ∼ = Q 8 .We assume for a contradiction that 4 |X|.
Proof.Let X, Y ≤ G be such that X, Y = G and let P := O p (N).We assume for a contradiction that neither X nor Y act irreducibly on [P, K]/ ([P, K]).Lemma 7.2 implies that N normalizes every subgroup of P ( * ).Moreover, by Lemma 7.3, there are x, y ∈ N such that X = (X ∩ N)(X ∩ K x ) and Y = (X ∩N)(Y ∩K y ) and, by assumption, neither X ∩K x nor Y ∩K y act irreducibly on [P, K]/ ([P, K]).It follows from Lemma 5.4 that X ∩ K x and Y ∩ K y both induce power automorphisms on P and that |P| = p.Thus ( * ) yields that X and Y normalize every subgroup of P. Then also G = X, Y normalizes every subgroup of P, which contradicts the irreducible action of K.

The main result
Proof.Assume for a contradiction that the statement is false and let G be a minimal counterexample.Then there is a sublattice L = {E, S, T, D, U, A, B, C, F} of L(G) isomorphic to L 9 , and in particular L satisfies the relations in Definition 3.1.
We let G be of type (N, K) where, among the minimal counterexamples, we choose G such that |N| is as large as possible and we set Then K has a normal q-complement for every q ∈ π(K) * and Lemma 7.10 is applicable.
We will first analyze how L fits into the subgroup lattice of G.
(1) F = G, C K (N) = 1 and every subgroup of N is normal in N.
Proof.The group F is a subgroup of G that is not L 9 -free, and Lemma 7.4 yields that F ∈ L. Hence the minimal choice of G implies that F = G.Similarly, it follows from Lemma 3.4 that G is not a direct product of two non-trivial groups of coprime order.Let p ∈ π(N).
, where the direct factors have coprime order by (L1).But we just saw above that such a direct decomposition of G is not possible.Therefore We conclude from Lemma 7.2 that every subgroup of O p (N) is normal in N.This implies that every subgroup of N is a normal subgroup of N, because N is nilpotent.
Since we have chosen N as large as possible, Lemma 7.5 implies that π(K) = π(K/C K (N)).Let q ∈ π(C K (N)).Then the previous equation forces q ∈ π(K/C K (N)), and therefore a Sylow q-subgroup of K has order at least q 2 .In particular, for all non-nilpotent battens V of K, we have that B(V) We recall that G is soluble, by Lemma 7.1, hence C K (N) is soluble, and then there must exist a prime q ∈ π(C K (N)) such that O q (C K (N)) = 1.This gives a contradiction, and therefore C K (N) = 1.
We remark that, by (1), we may apply Lemmas 7. Assume for a first contradiction that H∩P is not a normal subgroup of G. Since every subgroup of N is normal in N by (1), it follows that K does not induce power automorphism on P. Then Lemma 5.4 implies that K acts irreducibly on P := [P, K]/ ([P, K]).We apply Lemma 7.11 twice to find some X ∈ M and some Y ∈ M \ {X} such that X and Y act irreducibly on P. In particular , which yields that H ∩ P is normalized by K and hence H ∩ P ¢ NK = G.This is a contradiction.We deduce that P ∩ D ¢ G as stated, in particular N ∩ D ¢ G and also N ∩ E ¢ G.
For the final statement in (2) we use that G/(N ∩ E) is not L 9 -free.Then the minimality of G and Lemma 7.6 give that N ∩ E = 1.
(3) For every q ∈ π(K) such that 1 = Q ∈ Syl q (D), one of the following holds: Proof.We adopt the same notation as in the previous step, which means that H ∈ {E, D}, p ∈ π(N) and P := O p (N).If H = E, then M := {U, T, B}, and otherwise M := {A, B, C}.Whenever X, Y ∈ M are distinct, then X ∩ Y = H and X, Y = F.
Let X and Y be distinct elements of M and assume for a contradiction that X ∩ P and Y ∩ P are subgroups of C P (K).Then which contradicts (1).
For the remainder of this proof we let X and Y in M be such that their intersection with P is not contained in C P (K).We note that C P (K) = C P ( 1 (Q)) by Lemma 5.5.Then it follows that 1 In a similar way we observe that If K acts irreducibly on [P, K]/ ([P, K]), then Lemma 7.11 yields that X or Y, say X, acts irreducibly on Otherwise Lemma 5.4 gives that K and hence 1 (Q) induce power automorphisms on P. Together with (1) this means that every subgroup of P is normal in G. Now, if V, W ∈ M are distinct, then and it follows that P = (V ∩ P)(W ∩ P).We deduce that |P| = |V∩P|•|W∩P| |H∩P| for all W, V ∈ M. In particular |W ∩ P| = |X ∩ P| = 1 for all W ∈ M.This implies all our claims about D, because P = If, still in the power automorphism case, we have that H = E, then we recall that E ∩ P = 1 by (1).Therefore This implies that A ∩ P = T ∩ P = 1.But in this case we may interchange T by S in M. Since 1 = T ∩ P = A ∩ P = S ∩ P, we arrive at the contradiction that 1 = S ∩ T ∩ P = E ∩ P ≤ E ∩ N = 1 (by ( 1)).Thus, we have that π(K) ∩ π(E) = ∅ in this case as well.Again it follows that E = 1.
The remainder of the proof is dedicated to constructing a subgroup of G that violates Property (L4). ( Proof.We assume for a contradiction that A ∩ K centralizes P := O p (N) for some p ∈ π(N).
It follows from Lemma 7.3 that, for every subgroup X of G = PO p (N)K, there is some x ∈ P such that X ≤ (X ∩ P)O p (N)K x .Since [P, A ∩ K] = 1, we further have THAT X ≤ (X ∩ P)O p (N)K y for all X ≤ A and y ∈ P ( * ).
Assume that A ∩ P = 1.Then for both X ∈ {U, B}, we see that P ≤ G Thus A ∩ P = 1 and then (1), together with our assumption at the beginning of the proof, imply that every subgroup of A ∩ P is normal in A. Suppose that X, Y ∈ {S, T, D} are distinct.We recall that A ∩ P ≤ A = X, Y ≤ (X ∩ P)(Y ∩ P)O p (N)K, and then it follows that A ∩ P = (X ∩ P)(Y ∩ P).Since = 1, we know more: T ∩ P ∼ = S ∩ P ∼ = D ∩ P and |T ∩ P| 2 = |A ∩ P|.If K induces power automorphisms on P and if X ∈ {A, T, S}, then (X ∩ P) and (U ∩ P) are normal subgroups of G by (1).In addition there is some u ∈ P such that U ≤ (U ∩ P)O p (N)K u and then X ≤ (X∩P)O p (N)K u by ( * ).We apply Lemma 7.9 to see that |P∩A| = |P : P∩U| = |P∩S| = √ |P ∩ A|.
Now it follows that P ∩ A = 1, which is a contradiction.We conclude that K does not induce power automorphisms on P. Then Lemma 5.4 gives that K acts irreducibly on [P, K]/ ([P, K]).Since P ∩ D ¢ G by (2), we see that either [P, K] ≤ D or that 1 = P ∩ D ≤ C P (K).
In the first case T ∩ P ∼ = D ∩ P ≥ [P, K] and T ∩ D = E (2) = 1.This implies, together with Lemma 1.1, that C P (K) has a subgroup isomorphic to [P, K].We apply Lemma 5.4, in combination with Part (b) of Lemma 4.17, and we deduce that [P, K] is cyclic of order 2 and that, therefore, K centralizes it.This is impossible.
It follows that the second case holds, i.e. 1 = P∩D ≤ C P (K).Then p = 2 by Corollary 5.6 (a).If [P, K] is not elementary abelian, then Part (b) of Lemmas 4.17 and 5.4 give that [P, K] ∼ = Q 8 and P = [P, Q]×I, where I is a subgroup of P of order at most 2. We note that T ∩P ∼ = D∩P and T ∩D = E = 1, we obtain that D ∩ P is cyclic of order 2. It follows that P ∩ D, T ∩ P and P ∩ S have order 2 and hence P ∩ A is elementary abelian of order 4. Moreover (6) The non-trivial Sylow subgroups of H are elementary abelian (in particular N is abelian), but not cyclic, and K ≤ Pot K (H).
Proof.By definition H ≤ N is nilpotent.If, for all p ∈ σ , the group K does not act irreducibly on [O p (N), K]/ ([O p (N), K]), then Lemma 5.4 gives that K ≤ Pot K (H).In particular O p (N) is not cyclic.Moreover, Corollary 5.6 yields that O p (N) = [O p (N), K] is elementary abelian.Therefore, our claim is satisfied in this case.
Let us assume for a contradiction that there is some p ∈ σ such that K acts irreducibly on the group [O p (N), K]/ ([O p (N), K]).We set P := O p (N) and we choose Q ≤ L 1 of order q ∈ π such that [P, Q] = 1, by the definition of σ .By Lemma 7.3 we know that A ∈ L, with type (A ∩ N, A ∩ K).Additionally, since q ∈ π(K) * , the group K has a normal q-complement.Then A ∩ K also has a normal q-complement.We may apply We recall that P = [P, K] is abelian, hence it is contained in Z(N), and then it follows that D centralizes 1 = bc −1 ∈ [P, K].Here we also use Lemma 5.4, i.e. that D centralizes P. We conclude that D = N D (Q h ) for all h ∈ P. Again Lemma 7.8(b) provides t, s ∈ P such that T = (T∩P)N T (Q t ) and S = (T∩S)N C (Q s ).Let X ∈ {T, S}.Then P ∩ A ≤ A = D, X ≤ (X ∩ P)N G (Q x ), which implies that P ∩ X = P ∩ A. But now P ∩ A ≤ P ∩ S ∩ T = P ∩ E = 1, by (3), which gives a final contradiction.by (6), Definition 5.1 gives, for every non-nilpotent batten V of K, that B(V) centralizes P.This implies that r ∈ π(K) * .
We are now able to define L and J. Let u ∈ N be such that U = (U ∩ N)(U ∩ K u ).We set π := {r ∈ π(K) * | [H, R] = 1 for some R ∈ Syl r (U)} and we let L 2 be a Hall π -subgroup of (U u −1 ∩ K).Then L 2 ≤ K and L = L 1 , L 2 is a subgroup of K.In addition (5) and (9) show that π = ∅.
Next  (8) shows that P ∩ S = 1.Moreover P ∩ A = 1 by (7).We apply Lemma 7.9 to obtain some i, j ∈ {1, 0} such that This implies that |P ∩ A|p j = p i , and we obtain that i = 1 and j = 0.
In particular we see that P = (U ∩ P)(A ∩ P).Then Lemma 7.9 and the fact that A ∩ U = E (3) = 1 give some element g ∈ P such that X ≤ (P ∩ X)K g = (P ∩ X)N G (R) g , where X ∈ {A, U}.Assume for a contradiction that r ∈ π(A).Moreover q ∈ π(B)∩π(T) by (8).Since B∩T = E In particular we see that P ∩ U = 1, and this contradicts (8).
We summarize: The definitions of σ and H imply that H ≤ N, and (6) shows that the non-trivial Sylow subgroups of H are not cyclic.In addition L ≤ K induces power automorphisms on H.
We assume for a contradiction that L is not cyclic.Then, since L is nilpotent and a batten group by Lemma 2.7, we deduce that O 2 (L) ∼ = Q 8 .Definition 4.8 implies that, for all p ∈ π(N), the group O 2 (L) does not induce non-trivial power automorphisms on O p (N).In particular O 2 (L) centralizes H, which contradicts ( * ).Thus L is cyclic.
Assume for a contradiction that J is not abelian.Then Lemma 4.17 Hence J is abelian.For every q ∈ π and every subgroup Q ≤ L 1 of order q, the definition of σ and Lemma 7.7 provide some p ∈ σ such that [N, Q] ≤ O p (N) = P. Then we see, using (6) and

Definition 3 . 1 .Definition 3 . 2 .Lemma 3 . 3 .
The lattice L 10 is defined to be isomorphic to L(D 8 ), with notation as indicated in the picture.L 5 := {E, S, U, A, F}, (c) L 7 := L 5 ∪ {D, C}, (e) M 8 := L 10 \ {T, V}, (g) M 9 := L 10 \ {B}, (b) L 6 := L 5 ∪ {T}, (d) L 8 := L 10 \ {B, V}, (f) L 9 := L 10 \ {V}, with the corresponding inclusion relations induced from the lattice L 10 .Let L be a lattice.An equivalence relation ≡ on L is called a congruence relation if and only if, for all A, B, C, D ∈ L such that A ≡ B and C ≡ D, we have that A, C ≡ B, D and A ∩ C ≡ B ∩ D. Let ≡ be a congruence relation on L 9 = {A, B, C, D, E, F, U, T, S} as in Definition 3.1 and suppose that ≡ is not equality.Then E ≡ D.

Lemma 3 . 5 .
L 9 = {A, B, C, D, E, F, U, T, D} as in Definition 3.1 is completely characterized by the following:

Lemma 3 . 6 .
D 12 is not L 9 -free.Proof.Let G be isomorphic to D 12 and let a, b ∈ G be such that o(a) = 6, o(b) = 2 and G = a, b .Then we find a sublattice in L(G) isomorphic to L 9 by checking the equations from Lemma 3.5.We let L := {1, b , a 2 b , a 2 , ab , a 2 , b , a , a 2 , ab , G} and we define A := a 2 , b and C := a 2 , ab .): We see that a 2 = 1 and hence a = 1.L9 (ii): We notice that A ≤ G is isomorphic to Sym 3 with cyclic normal subgroup a 2 of order 3 and distinct subgroups b , a 2 b of order 2. Then b , a 2 b = b , a 2 = a 2 b , a 2 = A and b ∩ a 2 b = b ∩ a 2 = a 2 b ∩ a 2 = 1.L9 (iii): The subgroup C is also isomorphic to Sym 3 , the subgroup ab ≤ C has order 2, and moreover a 2 , ab = C and a 2 ∩ ab = 1.L9 (iv): We first see that b , ab = a, b = G and then a 2 b , ab = a 2 b(ab) −1 , ab = a, b = G.L9 (v): a is a cyclic normal subgroup of G of order 6, and the subgroups A and C also have order 6.These three subgroups are maximal in G. Hence A, a = C, a = G.Assume for a contradiction that A = C. Then b ∈ C = {1, a 2 , a 4 , ab, a 3 b, a 5 b}, which is false.Since a 2 is the unique subgroup of order 3 of G, we conclude that

p 2 − 1 p− 1 3 =
= p + 1 ≥ 4 subgroups isomorphic to V. In the second case A has 15•14 70 subgroups isomorphic to V, where 3 • 14 2 − 2 = 19 of these subgroups intersect V non-trivially and 19 of them intersect D non-trivially.In both cases, we find subgroups T and S of A isomorphic to V such that |{D, T, S, V}| = 4 and T

1 [
by Lemma 1.4 (b) and (c), because Q acts irreducibly on [P, Q].In addition, the irreducible action of Q x on [P, Q] and Lemma 1.2 yield that [P, Q] ≤ A, Q x .It follows that A, B = [P, Q]DQ x = F. Finally we deduce from Part (d) of Lemma 1.4 that

5 .
For this we set E := 1 and D := P. Then D = E and we see that L9 (i) is true.Next, we recall that 1 = [P, Q] by hypothesis.Assume that |[P, Q]| = 2. Then Q, which normalizes [P, Q], must centralize it, and then Lemma 1.1 gives a contradiction.Therefore |[P, Q]| 2. As a consequence, we find a, b ∈ [P, Q] # such that a = b, and then we set S := R a and T := R b .Now D ∩ S = 1 = E = D ∩ T and D, T = PR b = PR = PR a = D, S .We set A := PR.In addition, since R acts irreducibly, but non-trivially on P, it follows that C R (ba −1 ) R. Then the fact that |R| = r gives that C R (ba −1 ) = 1 = E. Lemma 1.4 (b) shows that S∩T = (R∩R ba −1 ) a ≤ C R (ba −1 ) a = E, and now we recall that [ba −1 , R] = 1.Moreover, R acts irreducibly on P, and then Part (e) of the same lemma yields the following:

(
Cy) [P, B(B)] = 1 and Q acts on P avoiding L 9 for every Sylow subgroup Q of B different from B(B) or (NN) P is elementary abelian of order p |B:B(B)Z(B)| and the Sylow subgroups of B act irreducibly on P, while Z(B) induces power automorphisms on P. As in Definition 4.6, we specify the type of the L 9 -avoiding action by writing that "B acts on P avoiding L 9 of type (•)".Lemma 4.17.Let B be a batten that acts non-trivially and avoiding L 9 on the p-group P. Then the following hold: (a) If C P (B) = 1, then p = 2. (b) Either P = [P, B] × C P (B), where [P, B] is elementary abelian and C P (B) is cyclic, or P = [P, B] × I, where I is a group of order at most 2 and [P, B] ∼ = Q 8 .(c) C P (B) is centralized by every automorphism of P of order coprime to p that leaves C P (B) invariant.
Similarly to the arguments above, we deduce that [P, K] = [P, H] = [P, B], and Lemma 5.2 gives that B avoids L 9 in its action on P. Therefore Lemma 4.11 yields that B is not nilpotent.From Definition 4.16 we further see that B does not act of type (NN), and thus B acts of type (Cy) on P. Consequently [P, B(B)] = 1 and B has a cyclic Sylow subgroup Q such that B = B(B)Q and Q acts on P avoiding L 9 .Again we have [P, K] = [P, B] = [P, Q] and Lemma 4.11 gives that Q induces power automorphisms on P or acts irreducibly on [P, Q]/ ([P, Q]) = [P, K]/ ([P, K]

Corollary 5 . 6 .
Let K be a batten group that acts non-trivially and avoiding L 9 on the p-group P. Then the following hold: (a) If C P (K) = 1, then p = 2. (b) If C P (K) = 1, then P = [P, K] is elementary abelian.(c) If K induces power automorphisms on P, then P = [P, K] is elementary abelian of odd order.In particular C P (K) = 1 in this case.Proof.Let B be a batten of K that does not centralize P. Then Lemma 5.5 implies that C P (K) = C P (B).Thus Part (a) and (b) of Lemma 4.17 yield the statements (a) and (b) of our lemma.For Part (c) we suppose that K induces power automorphisms on P. Then P is not an elementary abelian 2-group by Lemma 1.3.If C P (K) = 1, then our assertion holds by (b).Otherwise p = 2 by (a), and then Lemma 1.3 implies that [P, K] is neither elementary abelian nor isomorphic to Q 8 , contradicting Part (b) of Lemma 5.3.
by Lemma 1.1.The group L 1 induces power automorphisms on H, which means that it normalizes [c, D].Together with Part (d) of Lemma 1.4 we conclude that by Dedekind's modular law and by Part (b) of Lemma 1.4.Hence ( * * ) implies that (L9(iii)) of Lemma 3.5 holds.
ab D and similarly B ∩ C ≤ b D. Therefore D ≤ B ∩ C ≤ ab D ∩ b D = ( ab ∩ b D)D = D. Finally ( * ) and Part (d) of Lemma 1.4 yield that

Lemma 7 . 1 .
All groups in the class L are soluble.
that B = B(B)S.For all p ∈ π(N) it follows that [O p (N), Z(B)] ≤ [O p (N), S] = 1, by Lemma 2.5.Then Definition 4.16 yields that B does not act on O p (N) avoiding L 9 , whence B centralizes O p (N).But now B(B) ≤ C K (N) and B(B) is a normal Sylow subgroup of K.This contradicts the maximal choice of N. Lemma 7.6.Let G ∈ L and suppose that M ¢ G. Then G/M ∈ L.
7 and 7.8.(2) For all p ∈ π(N) we have that O p (N) ∩ D ¢ G.In particular N ∩ D ¢ G and N ∩ E = 1.Proof.Let H ∈ {E, D}, let p ∈ π(N) and set P := O p (N).If H = E, then we set M := {U, T, B} and otherwise we set M := {A, B, C}.Then for all distinct X, Y ∈ M, we have that X ∩ Y = H and X, Y = F.

≤
(P ∩ X)O p (N)K x and therefore P = P ∩ U = P ∩ B ≤ U ∩ B = E.But this contradicts (3).

= 1 , 1 ( 3 )
and then we conclude that T ∩P and D∩P are cyclic of order 2. Consequently A∩P = (T ∩P)•(D∩P) = 1 (P)¢G.Now = E = U ∩ A ≥ U ∩ 1 (P) and therefore U ≤ O π (N)K u .We arrive at a contradiction:P ≤ G = U, A ≤ (P ∩ A)O π (N)K u , because [P, K] ∼ = Q 8 is not elementary abelian.So we finally have that [P, K] is elementary abelian.Then Lemma 4.17(b) gives that C P (K) is cyclic and Lemma 1.1 shows that T ∩ P ∼ = D ∩ P ≤ C P (K).Together with the fact that T ∩ D = E(3)

Case 1 :
[P, Q] ≤ A. Then Lemma 7.8(c) implies that Q g ≤ A for every g ∈ G.We recall that U ∩ A = E (3)= 1, and then it follows first that q / ∈ π(U) and then that q | |B|, by Lemma 7.10.Here we use that G = F = B, U by (1).Thus we find some g ∈ G such that Q g ≤ D, because D = A ∩ B. We apply (3) to observe that [P, Q] = [N, Q] ≤ D and then D contains every conjugate of Q by Lemma 7.8(c).Recall that q / ∈ π(U), which implies that q ∈ π(T) by Lemma 7.10.Again we use that G = F = T, U .But this contradicts the fact that T ∩ D = E = 1 by (3).

Lemma 7 .
10 to A = T, S = T, D = S, D .It yields that at least two of the groups D, T, S have a subgroup of order |Q|.As |Q| |E| by (2), there is some g ∈ G such that Q g = Q and Q g ≤ A. Then Part (e) of Lemma 7.8 implies that P ∩ A C P (Q) = C P (K) by Lemma 5.5.In the present case we have that [P, A] A, and then Lemma 1.2 gives that A does not act irreducibly on [P, K]/ ([P, K]).Thus A ∩ K induces power automorphism on P by Lemma 5.4, and these automorphisms are not trivial because Q ≤ A ∩ K. Corollary 5.6 (c) implies that P = [P, A ∩ K] 5.4 = [P, K] is elementary abelian and in particular that D ∩ P ≤ C P (K) 5.5 = C P (A ∩ K) = 1.Hence there is some d ∈ [P, K] such that D = N D (Q d ), by Lemma 7.8(b).In addition we see from Lemma 7.11 that B and C act irreducibly on P.If it was true that P ≤ B, then it would follow that 1 = P ∩ A ≤ P ∩ B ∩ A ≤ P ∩ D ≤ C P (K) = 1, which is a contradiction.We conclude that P B and hence P ∩ B = 1 because of the irreducible action of B on P, and similarlyP ∩ C = 1.Then Lemma 7.8(b) provides b, c ∈ P such that B = N B (Q b ) and C = N C (Q c ).If Q b = Q c , then G = F = B, C ≤ N G (Q b ), which is false.Consequently Q b = Q c ,and we can use Lemma 7.8(a), Dedekind's modular law and Lemma 1.4(c).Together this shows thatD ≤ N B (Q b ) ∩ N C (Q c ) = O p (N)K b ∩ O p (N)K c = O p (N)(K b ∩ O p (N)K c ) ≤ O p (N)C K (bc −1 ) ≤ C G (bc −1 ),because N is nilpotent and because (bc −1 ) K ∩ O p (N) ≤ P ∩ O p (N) = 1.
by Dedekind's modular law.Altogether we have that yc ∈ C P (Q).We recall that c ∈ {a, b} ⊆ P, and then y = ycc −1 ∈ P.But we chose y ∈ R and now y ∈ R ∩ P = 1, whence Q c −1 = Q.In other words, c ∈ N P (Q), and this means that [Q, c] ≤ Q ∩ P = 1 and c ∈ C P (Q).We recall that c ∈ [P, Q] # and that [P, Q] is elementary abelian by hypothesis.Then Lemma 1.1 implies that [P Since the action of K on P avoids L 9 , the action of B also does.If B acts of type (Cy), then we obtain the contradiction that [P, Q] = [P, B].Thus B acts of type (NN) and in particular Q acts irreducibly on P.But this is impossible as well.It follows that [P, H] = [P, K].
Proof.Let H ≤ K.If H centralizes P, then it induces power automorphisms on P. We may suppose that[P, H] = 1.Then H has a q-subgroup Q such that [P, Q] = 1.Therefore, if Q ¢ K,then we have that [P, K] = [P, Q] by Lemma 5.3 (a).Then the fact that [P, Q] ≤ [P, H] ≤ [P, K] yields that [P, H] = [P, K].Assume for a contradiction that [P, K] = [P, H].Then Lemma 2.8 implies that K has a non-nilpotent batten B such that B = B(B)Q.We moreover deduce that [P, Q] [P, K] and [P, B] = [P, K] by Lemma 5.3 (a), because B ¢ K.