Accessible Published by De Gruyter August 11, 2016

More on core-2 2-groups

Bettina Wilkens
From the journal Journal of Group Theory

Abstract

Let G be a finite 2-group with the property that | H : H G | 2 for all subgroups H of G. Then G has an abelian normal subgroup of index at most 4 in G. This result represents an affirmative answer to Question 18.56 from the current edition of the Kourovka Notebook.

1 Introduction

Let n be a positive integer. A group G is called core-n if | H / H G | n for every subgroup H of G; here, H G denotes the largest normal subgroup of G contained in H. In [4], it was shown that a locally finite core-n group has a normal abelian subgroup whose index in the group is bounded by a function of n. In [3] the same authors proved that, for odd primes p, a finite core-pp-group has an abelian normal subgroup of index p 2 , a best possible bound. The paper ([3, Theorem 2]) also contains a proof of the existence of a normal abelian subgroup of index at most 2 6 in a finite core-2 2-group. The authors mention that no examples of finite core-2 2-groups were known that did not possess an abelian subgroup of index at most 4. Later ([5]) it was shown that finite core-2 groups of class 2 indeed must have an abelian subgroup of index at most 4. The present paper is devoted to showing that every finite core-2 2-group has an abelian subgroup of index at most 4. This provides an affirmative answer to Question 18.56 from the 18th edition of the Kourovka Notebook, posed by G. Cutolo.

The dihedral groups D 2 n are core-2, which shows that core-2-ness does not entail bounded class or bounded subgroup breadth. This places core-2 2-groups in contrast to odd-order core-pp-groups, those being of class at most 3 ([6]). The group D 8 Q 8 is core-2, which shows that 4 is the best possible upper bound for the index of an abelian subgroup in a core-2 2-group. Our main result reads as follows:

Theorem

Let G be a finite 2-group. Suppose that | H / H G | 2 for every subgroup H of G. Then G has an abelian subgroup that contains Φ ( G ) and has index at most four in G.

We note that the corresponding result for odd p is an immediate corollary of the results of [3]. Letting p be an odd prime and G a finite core-pp-group, G has a normal abelian subgroup A of index p 2 ([3, Theorem 1]). We may obviously take A to be a maximal abelian subgroup of G. Then, according to [3, Lemma 1.1 (iii)], 1 ( G ) Z ( G ) A and G A , because G / A is too small to be nonabelian.

We conclude this introduction with three very helpful observations: Subgroups as well as quotients of a core-n group are core-n. If G is a core-pp-group, then x p G whenever x G and | E : E Z ( G ) | p if E is an elementary abelian subgroup of G. We also note a corollary.

Corollary

Let G be a finite core-2 group. Then G has an abelian subgroup of index at most 8, and this bound is best possible.

Proof.

Let G be a finite core-2 group of composite order, and let Q = O 2 ( G ) . Then Q is abelian of odd order, every subgroup of Q is normal in G and G = Q S with S S y l 2 ( G ) . The group S is core-2, and, since | S : S G | 2 , C S ( Q ) has index at most 2 in S. A direct product of a core-2 2-group with an abelian group of odd order is core-2; if [ S , Q ] 1 and T = C S ( Q ) , then the corresponding semidirect product SQ is core-2 if and only if S normalises every subgroup of Q, | S : T | = 2 , S is core-2 and T U S whenever U S = T U . By the theorem, T has an abelian subgroup of index at most 4 and the direct product of this with Q has index at most 8 in G. Now let

S = x , y , s , t x 2 = y 8 = s 4 = t 4 = 1 , [ x , y ] = y 2 , s 2 = t 2 = [ s , t ] = y 4 .

Then S D 16 Q 8 is core-2, but there is no abelian subgroup of index 4 in S that does not contain y. Now let G = S Q , where Q C 3 , C S ( Q ) = x , y 2 , s , t and y inverts the elements of Q. This G does not have an abelian subgroup of index four. ∎

The reader interested in the structure of core-p-groups of class 2 may find the results of Lemma 9 somewhat relevant.

2 Preliminaries

Lemma 1

Lemma 1 ([2, Lemma 3.4.1])

Let X be a finite group and let B be an abelian normal subgroup of X. Let N be the subgroup of X consisting of elements inducing power automorphisms on B. Let n = exp B and let U be the unit subgroup of Z / n Z . There is a homomorphism ϵ : X U given by b x = b ϵ ( x ) whenever b B and x N .

Notation

Let X be a finite group. Following Blackburn’s notation, we let

j ( X ) = | { x X x 2 = 1 } | .

Lemma 2

Lemma 2 ([1, Satz 1])

Let P be a 2-group and let 1 N Ω 1 ( Z ( P ) ) . Let H be the set of maximal subgroups of N. Then

U j ( P / U ) = | | j ( P / N ) + j ( P ) .

The fact that core-pp-groups have abelian Frattini subgroup is proved in [3, see Lemma 1.1 (ii)–(iii) for odd p and Proposition 2.11 for p = 2 ]. We include a short proof for p = 2 , as this will be a vital ingredient of the forthcoming proof.

Lemma 3

Lemma 3 ([3, Proposition 2.11])

A finite core-2 2-group P has abelian Frattini subgroup.

Proof.

Let P be a counterexample of minimal order. Since Φ ( P ) u whenever 1 u Z ( P ) , Z ( P ) is cyclic and Φ ( P ) = Ω 1 ( Z ( P ) ) . Let s and t be involutions in Φ ( P ) . If [ s , t ] 1 , then s , t D 8 . Now P normalises a maximal subgroup V of s , t , of order 4; this, however, implies that [ Φ ( P ) , V ] = 1 , a contradiction. Let Ω 1 ( Φ ( P ) ) = Y . We have just seen that Y is elementary abelian, whence | Y : Y Z ( P ) | 2 , implying | Y | 4 and Y Z ( Φ ( P ) ) . Let Ω 1 ( Z ( P ) ) = z . If z U Φ ( P ) , then U Φ ( P ) . Since Ω 1 ( x ) Z ( P ) for every x in P of order greater than 2, a subgroup U of Φ ( P ) not containing z must be contained in Y, hence in Z ( Φ ( P ) ) . Thus Φ ( P ) is Hamiltonian, i.e. Φ ( P ) = Q × E with E elementary abelian and Q Q 8 . At least one maximal subgroup V of Q must be normal in P, such that [ Φ ( P ) , V ] = 1 = Q , a contradiction. ∎

The following is a variation of [8, Lemma 8] and its proof is analogous to [8, Lemma 2]. We provide a short proof.

Lemma 4

Let P be a finite p-group. Let U and V be subgroups of P satisfying [ S , T ] Ω 1 ( Z ( P ) ) . If | U : C U ( v ) | p for all v in V, then | [ U , V ] | p or | U : C U ( V ) | = p .

Proof.

Let v V C V ( U ) and w V . If [ U , w ] [ U , v ] and C U ( w ) C U ( v ) , then [ U , v w ] = [ C U ( v ) C U ( w ) , v w ] = [ C U ( w ) , v ] [ C U ( v ) , w ] = [ U , v ] × [ U , w ] , of order p 2 . It follows that every element of V belongs to C V ( C U ( v ) ) or belongs to the subgroup { x V [ U , x ] [ U , v ] } of V. The assertion follows because V cannot be the union of two proper subgroups. ∎

A proof of the following lemma could be compiled from references to the available literature on metacyclic groups.

Lemma 5

Let U = u , v be a 2-group with [ u , v ] u 4 . Unless o ( [ u , v ] ) 2 , there is x U with x 2 U .

Proof.

Assume that o ( [ u , v ] ) > 2 ; it will suffice to establish the corresponding statement in the quotient U / [ u , v ] 4 . Hence assume o ( u ) = 2 n + 2 with n 2 and [ u , v ] = u 2 n . Observe that this implies γ 3 ( U ) = 1 . Let | U : u | = 2 m . A generator w of u may be chosen such that v 2 m is equal to a power w 2 ; as C u ( v ) = u 4 and C v ( u ) = v 4 , we have m 2 .

Assuming < m , let x = w v - 2 m - . Note that the order of x u in U / u is equal to 2 . Since 2 and m - 1 ,

x 2 = w 2 v - 2 m [ v , w ] 2 m - ( 2 2 ) = 1 .

It follows that o ( x ) = o ( x u ) = 2 and x u = 1 . However,

[ x 2 , v ] = [ w 2 , v ] = u 2 n + 1

and x 2 U .

We have established that m . Letting y = v w - 2 - m , we find that

y 2 m = v 2 m w - 2 [ w , v ] 2 - m ( 2 m 2 ) u 2 n + - 1 .

So y 2 m = 1 unless = 2 = m . Now n 2 implies [ u 2 n - 1 , v ] 2 = 1 . Hence if = 2 = m , then ( v w - 1 u 2 n - 1 ) 4 = ( u 2 n + 1 ) 2 = 1 . Accordingly, v u = x u with o ( x ) = 2 m and x u = 1 . However, [ v 2 , u ] 1 yields x 2 U . ∎

Notation

As usual, “ ” denotes “maximal subgroup of” and d ( P ) the minimal number of generators of the p-group P.

Observe that the core-p-ness of a p-group G is already guaranteed if | H : H G | p is required for all subgroups of G generated by (no more than) two elements. Indeed, suppose | x , y : x , y G | p for all elements x , y of the p-group G. This property carries over to subgroups and quotients. Let Q G and let G ¯ = G / Q G . Subgroups of Q ¯ are core-free, and if | Q ¯ | > p , then Q ¯ has a 2-generated subgroup of order greater than p, a contradiction. Another equivalent characterisation of core-p-ness may be found in [3, Lemma 1.3]: The group G enjoys core-p-ness if and only if every nontrivial subgroup H of G has a maximal subgroup M with [ M , G ] Φ ( H ) .

For the remainder of this paper, p is a prime and G denotes a finite core-pp-group of order greater than p. The equivalent formulation of core-p-ness of a p-group predominantly used in the proofs to come is a mixture of the two characterisations given in the previous paragraph: Every nontrivial subgroup of a core-pp-group generated by two elements has a maximal subgroup that is normal in the group.

Observe that x p G whenever x G and that | E : E Z ( G ) | p for each elementary abelian subgroup E of G.

Notation

Let p = 2 . For and y G { 1 } , the involution in y is denoted by y ^ .

Although our theorem only concerns the prime 2, a few lemmas in this section will be stated for all primes. This was done where it would not increase the length of the respective proof inordinately and it was felt that the information contained in the lemma could be of some interest beyond the requirements of the present paper.

Lemma 6

Let A be an abelian normal subgroup of G such that [ A , G ] Ω 1 ( Z ( G ) ) and not every subgroup of A is normal in G. Let p k be the minimal order of a cyclic subgroup of A that is not a normal subgroup of G . Then Ω k - 1 ( A ) Z ( G ) .

Proof.

Let a be any element of A of order p k and let z Ω k - 1 ( A ) . Let t be a complement of a in a , z . Note that o ( t ) < p k . There is a a p satisfying o ( a ) = o ( t ) . Both t and a t are normal in G, while a Z ( G ) . Thus t Z ( G ) , i.e. z a p , t Z ( G ) . ∎

Lemma 7

Assume | G | = p . Then | G : Z ( G ) | = p 2 unless p = 2 , exp G = 4 and G = U V × E , where E is elementary abelian, [ U , V ] = 1 , V Q 8 , and U = u , v with [ u , v ] = G , v 2 = 1 and u 2 G .

Proof.

The commutator map induces a symplectic form on G / Φ ( G ) , whence G / Z ( G ) is elementary abelian of even degree. Let exp G = p n .

Assume that p is odd. If n = 1 , then G is a direct product Q × E with Q extraspecial and E elementary abelian. Letting | Q | = p 2 m + 1 , m is the degree of a maximal elementary abelian core-free subgroup of Q, i.e. m = 1 .

Now assume n > 1 . If Φ ( P ) is noncyclic, then there is 1 s Ω 1 ( Z ( P ) ) such that Z ( P / s ) = Z ( P / s ) and the theorem follows by induction. Hence Φ ( P ) may be supposed to be cyclic. There is x G Z ( G ) of order p n . Since n > 1 , exp Φ ( P ) = p n - 1 , so Φ ( P ) = x p . Since p is odd, raising elements to their pth power is an endomorphism of G, such that G = x Ω 1 ( G ) and exp Ω 1 ( G ) = p . Hence there is t G with t p = 1 [ x , t ] . Let Q = C G ( x , t ) , noting that G = x , t Q . For u Ω 1 ( Q ) Z ( Q ) , t , u is core-free elementary abelian of order p 2 . Thus v p x p 2 whenever v Q Z ( Q ) . Yet if v Q satisfies o ( v ) = o ( x ) , then Φ ( G ) = v p and Q = v Q 1 with exp Q 1 = p , i.e. Q 1 Z ( Q ) . Since Q / Q 1 is cyclic, Q is abelian after all.

Now let p = 2 , let G = z and let x G Z ( G ) have order 2 n .

We first prove the lemma making the additional assumption that Φ ( G ) is cyclic. Then Φ ( G ) is a cyclic group generated by elements of order at most 2 n - 1 , i.e. Φ ( G ) = x 2 . Let y G C G ( x ) and Q = C G ( x , y ) , observing that G = x , y Q and Q Z ( G ) if and only if Q is abelian.

Assume that n 3 . For each v G , there is a generator w of x such that v 2 = w 2 with some 1 . Either ( v w - 2 - 1 ) 2 = 1 , or = 1 , [ v , x ] 1 and ( v w - 1 x 2 n - 2 ) 2 = 1 . In particular, y may be chosen of order 2. Assume that Q is nonabelian. If exp Q = 2 n , then an analogous argument yields an involution s in Q Z ( Q ) . However, this would mean s , y G = 1 . Thus exp Q < 2 n and, for u Q , we have u 2 x 4 and there is an involution s u satisfying u x 2 = s u x 2 . If u Q Z ( G ) , then s u Z ( Q ) and s u , y G = 1 .

We turn to the case n = 2 . Then G = D E with D extraspecial and E abelian. The group D 8 D 8 is not core-2, whence either | G : Z ( G ) | = 4 or D D 8 Q 8 and E is elementary abelian, a scenario covered by the assertion.

From now on, Φ ( G ) is assumed to be noncyclic; observe this implies that | Ω 1 ( Z ( G ) ) | > 2 . Let Ω 1 ( Z ( G ) ) = N and s N z . Then Z ( G / s ) = Z ( G ) / s , and | G : Z ( G ) | { 4 , 16 } is immediate by induction on the group order. Induction also yields that if | G : Z ( G ) | = 16 , then 2 ( G ) s whenever s N z . Thus | G : Z ( G ) | = 16 implies exp G = 4 .

Now assume that exp G = 4 , | G : Z ( G ) | = 16 and d ( Φ ( G ) ) 2 . Let

S = { s G s 2 z } .

Then S < G , in particular G S Z ( G ) . Let u G Z ( G ) with u 2 z , let t G C G ( u ) and let C G ( u , t ) = Q .

For q Q Z ( Q ) and r { u , t , u t } , q , r is an abelian subgroup of G with z q 2 , r 2 . Suppose that | u 2 , t 2 , z | = 8 . If q Q Z ( Q ) , then q 2 = q 2 , t q 2 , u = z . Consequently, we have Q = x , y × E with E elementary abelian, x , y Q 8 , and x 2 = y 2 = z . Now let W = u x , t y . Then W C 4 × C 4 and W Z ( G ) = Φ ( W ) = u 2 z , u 2 t 2 z , so W G = Φ ( W ) .

We have seen that u 2 , t 2 u 2 , z whenever t G C G ( u ) . We shall show that Ω 1 ( G ) Z ( G ) . Assume otherwise. Let t G C G ( u ) . If t 2 = u 2 z , then ( u t ) 2 = 1 , so t 2 { u 2 , z } . Since G C G ( u ) = G > S , it follows that t may be assumed to satisfy t 2 = u 2 . Let Q = C G ( u , t ) and q Q Z ( Q ) . Then t q C G ( u ) , while ( t q ) 2 = u 2 would imply q 2 = 1 . Hence we have ( t q ) 2 = z , i.e. q 2 = u 2 z . Now let q Q C Q ( q ) . Then ( q ) 2 = u 2 z = ( q q ) 2 = z , a contradiction.

We have confirmed the existence of an involution v in the set G Z ( G ) . Since G S C G ( v ) , the element u may be assumed to satisfy [ u , v ] = z . Let U = u , v and let Q = C G ( U ) . For w Q Z ( Q ) , z w , v , so Q = V × E with V Q 8 and elementary abelian E. This completes the proof. ∎

Notation

The elementary abelian group of order p m will, as usual, be denoted by E p m .

Let U G and x G . Following [8], we define b U ( x ) , the breadth of x in U, as b U ( x ) = log p | U : C U ( x ) | . We write b ( x ) instead of b G ( x ) . The breadth b ( A ) in G of the subgroup A is defined as max { b ( a ) | a A } .

Let N = Ω 1 ( Z ( G ) ) . For a subgroup U of N, let D U be the preimage of Z ( G / U ) in G.

Lemma 8

Suppose that p = 2 and G N . Then | G : D U | { 2 4 , 2 2 , 1 } for U N . For any subgroup Wof N such that | N : W | = 4 and W G = N , there is at most one maximal subgroup M of N with W M and | G : D M | = 2 4 .

Proof.

The first assertion is a straightforward application of Lemma 7 applied to G / M , where M N .

For the second assertion, let | G / N | = 2 n and let W N with | N / W | = 4 and N = W G . Note that W is contained in exactly three maximal subgroups of N.

Let M be a maximal subgroup of N with | G : D M | = 2 4 . Let G / M = X and let X = z . By Lemma 7, 2 ( G ) M and | Φ ( G ) N / N | 2 . Assume that Φ ( G ) N . By Lemma 7, X = U V × E , where E E 2 n - 5 , V Q 8 and U = u , v with u 2 Φ ( V ) and v 2 = 1 . Letting u 2 = s and V = x , y , we have Ω 1 ( U V ) = z , s , v E 8 . Accordingly, j ( X ) = 2 n - 2 . Certainly j ( G / M ) < 2 n + 1 for any maximal subgroup M of N not containing G . So if | G : D M | = 16 for at least two maximal subgroups of N containing W, then

W < M N j ( G / M ) < 2 n - 1 + 2 n + 1 < 3 2 n = 3 | G / N | ,

which Blackburn’s formula (Lemma 2) makes impossible. Thus Φ ( G ) N .

We note: For 4 n , j ( Q 8 D 8 × E 2 n - 4 ) = 3 2 n - 2 , j ( D 8 × E 2 n - 2 ) = 3 2 n - 1 , j ( Q 8 × E 2 n - 2 ) = 2 n - 1 , and j ( Q 8 C 4 × E 2 n - 3 ) = 2 n .

Let M N . From Lemma 7 and Φ ( G ) = 16 , we deduce that | G : D M | = 16 if and only if G / M Q 8 D 8 × E 2 n - 4 . From Φ ( G ) N it also follows that, if M N satisfies | G : D M | = 4 , then G / M is isomorphic to one of the groups D 8 × E 2 n - 2 , Q 8 × E 2 n - 2 , or Q 8 C 4 × E 2 n - 3 . Suppose that | G : D M | = 16 for at least two maximal subgroups M containing W. Then, as we have seen,

W < M N j ( G / M ) 3 2 n = 3 | G / N | .

However, j ( G / W ) > 0 , and Blackburn’s formula is good for another contradiction. ∎

The assertions of the next lemma, describing finite core-pp-groups G with Φ ( G ) Z ( G ) , are partly contained in [3, Theorem 1] and [5, Theorem]. We shall, however, require more detail than is provided in the referenced sources.

Lemma 9

Suppose that G N . Let A be an abelian subgroup of G of maximal order. Then the following hold:

  • If p is odd, then | G : A | p 2 and if | G : A | = p 2 , then | A : Z ( G ) | = p .

  • If p = 2 , then either | G : A | 2 or | G : A | = 4 | A : Z ( G ) | . If | A : Z ( G ) | = 4 , then A = c , d Z ( G ) with b ( c ) = b ( d ) = 1 , in particular | [ A , G ] | 4 .

  • If p = 2 and Ω 2 ( A ) Z ( G ) , then | G : A | 2 or | A : Z ( G ) | 2 .

Proof.

Note that Z ( G ) Φ ( G ) A G .

We start with the case p odd. Let [ A , G ] = W of order p n , and let A / Z ( G ) = V of order p m . For any maximal subgroup U of W, there is a maximal subgroup U 1 of N satisfying U 1 W = U . Applying Lemma 7 to the quotient G / U 1 , we obtain | A / U : ( A / U Z ( G / U ) ) | = p . Note that this implies that m n .

Let 𝒮 be the set of pairs ( v , U ) , where 1 v V , U W and v = a Z ( G ) with a A and [ a , G ] U . Let r = p m - 1 p - 1 . We list the subgroups of V of order p as v 1 , v r . For i 1 , , r , let v i = a i Z ( G ) and let p i = | [ G , a i ] | . We assume the v i listed as to satisfy i i + 1 for all i. We count the elements of 𝒮 in two ways, to obtain

| 𝒮 | = 1 ( p - 1 ) 2 ( p n - 1 ) ( p m - 1 - 1 ) = 1 p - 1 i = 1 r ( p n - i - 1 ) ,

i.e.

i = 1 r ( p n - i - 1 ) = 1 p - 1 ( p n - 1 ) ( p m - 1 - 1 ) .

Observe that

1 p - 1 ( p n - 1 ) ( p m - 1 - 1 ) = 1 p - 1 ( p n - 1 - 1 ) ( p m - 1 ) - p n - 1 + p m - 1 .

Suppose that 2 > 1 . Then

i = 1 r ( p n - i - 1 ) 1 p - 1 ( p n - 1 - 1 ) ( p m - 1 ) - 2 p n - 1 + p n - 1 + p n - 2 .

However, p n - 1 > 2 p n - 2 p n - 1 + p n - 2 , whence

- 2 p n - 1 + p n - 1 + p n - 2 + p n - 1 - p m - 1 = - p n - 1 + p n - 1 + p n - 2 - p m - 1 < 0 ,

a contradiction.

We have established that either b ( A ) = 1 or there is a A such that a , Z ( G ) contains every element of A of breadth greater than 1. Assume the latter together with m 2 . Let b A a Z ( G ) . Then

b ( b ) = b ( a b ) = 1 ,

and [ a , G ] = [ b , G ] × [ a b , G ] , while C G ( b ) C G ( a b ) = G . Let x C G ( b ) C G ( a b ) and y C G ( a b ) C G ( b ) . Then [ a b 2 , x ] = [ a b , x ] and [ a b 2 , y ] = [ b , y ] , i.e. [ a , G ] = [ a b , x ] × [ b , y ] = [ a b 2 , G ] , a contradiction. Thus | A : Z ( G ) | = p or b ( A ) = 1 .

Now let p = 2 . Suppose that A has a subgroup a , b such that | a , b : a , b Z ( G ) | = 4 and b ( a ) 2 b ( b ) . First suppose that there are subgroups s , t of [ G , a ] and s , t of [ G , b ] such that | s , t s , t | = 16 . Let U = s , t , s , t and let V be a complement of U in N. Let X 1 = s s , t t , X 2 = s s , t s t , and X 3 = s t , s t . For i = 1 , 2 , 3 , let W i = V × X i . For each index i, we have U = s , t × X i = s , t × X i , while X 1 X 2 X 3 = 1 and W 1 W 2 W 3 = V . Let i { 1 , 2 , 3 } . Let W i X N ; then neither a nor b belongs to D X . Now Lemma 8 may be invoked to yield that a b D X for at least two out of three different choices of X; thus

[ a b , G ] W 1 W 2 W 3 = V .

The intersection over all complements of U in N being trivial, we obtain that a b Z ( G ) in contrast to our assumptions on a , b . If [ a , G ] [ b , G ] has a subgroup U of order 4, then, similarly, [ a b , G ] V whenever N = U × V , which results in a b Z ( G ) . The only remaining possibility is that | [ a , G ] | = | [ b , G ] | = 4 , while | [ a , G ] [ b , G ] | = 2 ; let [ a , G ] = s , t and [ b , G ] = s , t , U = s , t , t , N = U × V , W 1 = V t t , W 2 = V t s t . Lemma 8 again yields

[ a b , G ] W 1 W 2 = V

and a contradiction ensues as before.

Suppose that A has an element a of breadth greater than 1. As seen in the two preceding paragraphs, this implies that A = a B , where Z ( G ) B and b ( b ) = b ( a b ) = 1 for every b in B that satisfies | b , a Z ( G ) : Z ( G ) | = 4 . In particular, b ( B ) 1 .

Assume that B Z ( G ) . Applying Lemma 4, we obtain | [ B , G ] | = 2 or | G : C G ( B ) | = 2 . For b in B Z ( G ) , we have [ a , G ] [ b , G ] [ a b , G ] , which implies [ a , G ] = [ b , G ] × [ a b , G ] and C G ( a ) = C G ( b ) C G ( a b ) .

Suppose that | [ B , G ] | = 2 . If | B : Z ( G ) | > 2 , then there are b and c in B with G = C G ( b ) C G ( c ) . Note that

G = C G ( b ) C G ( b c ) = C G ( c ) C G ( b c ) .

For d { b , c , b c } , we have [ a d , G ] = [ a , C G ( d ) ] because of [ C G ( d ) , a ] 1 and b ( a d ) = 1 . Furthermore, [ a d , G ] is a complement of [ B , G ] in [ a , G ] . Since there are only two such complements, b and c may be chosen such that [ a b , G ] = [ a c , G ] . However, this yields [ b c , G ] = [ B , G ] = [ b , G ] = [ a b , G ] , a contradiction. Consequently, | [ B , G ] | = 2 entails that A = a , b with b ( b ) = 1 = b ( a b ) , while C G ( a ) = C G ( b ) C G ( a b ) = C G ( A ) = A , i.e. | G : A | = 4 .

We turn to the case | G : C G ( B ) | = 2 . Let b B Z ( G ) . We have C G ( B ) = C G ( b ) and G = C G ( b ) C G ( a b ) . Letting x C G ( B ) C G ( a ) , we find that

[ x , a ] = [ x , a b ] = [ G , a b ] .

Given b , c B with | [ G , b , c ] | = 4 , we obtain that [ G , a b ] = [ G , a b c ] = [ G , c ] , a contradiction. Accordingly, | B : Z ( G ) | = 2 and if B = b Z ( G ) , then once again C G ( a b ) C G ( b ) = C G ( a ) = C G ( A ) = A , i.e. | G : A | = 4 .

Letting p be any prime, assume | A : Z ( G ) | = p . Let A = a Z ( G ) , i.e. b ( a ) = log p | G : A | . It will suffice to derive a contradiction from | G : A | = p 3 . For x G A , the maximality of | A | implies that C G ( x ) = Z ( G ) , x is a maximal abelian subgroup of G. Let x G Z ( G ) and let y G Z ( G ) x . One of the maximal subgroups of x , y is normal in G, say x , y p , [ x , y ] . Since Φ ( x , y ) = x p , y p , [ x , y ] , i.e. d ( Φ ( x , y ) ) 3 , we must have [ x , G ] = Ω 1 ( Φ ( x , y ) ) . If o ( x ) = p , then Φ ( x , y ) = y p , [ x , y ] , so o ( x ) > p and Ω 1 ( x ) [ x , G ] . It follows that there is v G x , Z ( G ) with [ v , x ] = Ω 1 ( x ) . This implies that Φ ( v , x ) = v p x p and there is w v , x Z ( G ) with [ w , G ] Φ ( v , x ) , a contradiction.

The final case left for us to consider is b ( A ) = 1 . Then Lemma 4 yields | [ G , A ] | = p or | G : C G ( A ) | = | G : A | = p . Only the first case needs further consideration. Let [ G , A ] = z , of order p. We note that for U A we have z U or | U : U Z ( G ) | p . Let A = a 1 × × a n , where z projects nontrivially into a 1 along the given decomposition. Let B ~ = a 2 a n . Then | B ~ : B ~ Z ( G ) | p ; however, this implies that A = a Z ( G ) and | G : A | = p or A = a , b Z ( G ) with A = C G ( a ) C G ( b ) , i.e. | G : A | = p 2 .

At this point, only the final assertion of the lemma has not been taken care of. By way of contradiction, assume that p = 2 , Ω 2 ( A ) Z ( G ) and | G : A | = 4 = | A : Z ( G ) | . Recall that | [ A , G ] | 4 .

Let D be a subgroup of Z ( G ) of maximal order subject to possessing a complement in A and let A = D × E . Setting E = a 1 × × a n , either n = 2 , or some c of the form c = i = 1 n a i ε i with ε i { 0 , 1 } belongs to Z ( G ) . For such a c, c has a complement in E, whence c D has a complement in G. It follows that n = 2 and A = D × a × b with D Z ( G ) . Note that o ( a ) 8 o ( b ) .

The following observation is going to be quite helpful: Suppose there is z N [ A , G ] such that Ω 2 ( A / z ) Z ( G ) / z . Since Z ( G / z ) A / z = Z ( G ) / z , induction yields that | A : Z ( G ) | = 2 or there is a maximal subgroup B of G with B z . However, [ B , A B ] = 1 , whence B A = G and A B = Z ( G ) A .

  1. (${\ast}$)

    Let z N [ A , G ] . Then Ω 2 ( A / z ) Z ( G ) / z .

Suppose that [ A , G ] a ^ , b ^ . Let z Ω 1 ( D ) . Any d A satisfying d 4 = z then belongs to D a 2 , b 2 Z ( G ) , and ( ) yields D = 1 .

Assume that | [ A , G ] | = 2 . Since there is c { a , b , a b } with [ c , G ] a ^ , b ^ , this implies that [ A , G ] a ^ , b ^ . As we just saw, this means A = a , b , in particular Z ( G ) = Φ ( A ) . Let w G A . There is c A satisfying w 2 = c 2 , i.e. ( w c - 1 ) 2 = [ w , c ] . Since o ( a ) 8 o ( b ) , [ w , c ] Φ ( Z ( G ) ) and the coset wA contains an involution. Letting [ A , G ] = s , it follows that G = A x , y with involutions x and y satisfying [ a , x ] = 1 , [ b , y ] = s , [ a , y ] = s , and [ b , y ] = 1 . Now [ x , y ] 1 , for none of the elements x , y , x y is in Z ( G ) . Accordingly, we have x , y D 8 , and, since a maximal subgroup of x , y is normal in G, we must have [ x , y ] = s . This implies G = s and Lemma 7 delivers a contradiction. So | [ A , G ] | = 4 .

Assume that G = [ A , G ] . For 1 v G , let A v be the preimage of Z ( G / v ) A / v in A. Since exp G / v > 4 , Lemma 7 yields | A : A v | = 2 for each v. Since | A : Z ( G ) | = 4 , A v A v whenever v , v = G , and it follows that each of the three maximal subgroups of A that contain Z ( G ) is equal to some A v , i.e. b ( A ) = 1 . Now Lemma 4 yields | G : C G ( A ) | = 2 = | G : A | , a contradiction.

We have found that G [ A , G ] . In particular, A > a , b , which we have seen to imply [ A , G ] a ^ , b ^ . Write D = d × E with [ A , G ] a , b E . If 1 z Ω 1 ( E ) , then Ω 2 ( A / z ) Z ( G ) / z . As previously mentioned, there is an element c { a , b , a b } with 1 [ G , c ] a ^ , b ^ . Since | [ A , G ] | = 4 , it follows that z [ A , G ] , and ( ) is contradicted. Thus E = 1 , and since [ A , G ] < G , we have G = Ω 1 ( Φ ( G ) ) = a ^ , b ^ , d ^ .

Suppose that o ( d ) = 4 . There is an element t a ^ , b ^ satisfying t d 2 [ A , G ] . Since t d 2 2 ( A ) = a ^ , b ^ , condition ( ) yields a contradiction. Hence o ( d ) 8 and N 2 ( G ) .

Let V G . We have seen that exp G / V 8 , and Lemma 7 says that the subgroup A / V Z ( G / V ) is of index at most 2 in A / V . Only two of the three elements a , b , a b have breadth 1 in G, so there is an element s [ A , G ] with Z ( G / s ) A / s = Z ( G ) / s . Letting t G [ A , G ] and V = s , t , we obtain a final contradiction. ∎

The following three auxiliary lemmas each dispense with some scenario that makes repeated appearances during coming proofs.

Lemma 10

Let P be a finite 2-group having a maximal abelian normal subgroup B = a × b × d × E . If

  1. (a)

    o ( a ) = 2 , o ( b ) 4 o ( d ) , o ( b ) > exp E < o ( d ) ,

  2. (b)

    E Z ( P ) , Φ ( P ) B , [ P , B ] Ω 1 ( Z ( P ) ) ,

  3. (c)

    b ( a ) = 1 and b P d ,

  4. (d)

    | P : B | = 8 ,

then P is not core-2.

Proof.

Assume otherwise. The four conditions (a)–(d) entail Z ( P ) = Φ ( B ) E and P = u , v , w B with [ u , a ] = [ u , d ] = 1 , [ u , b ] = b ^ , [ v , a ] = [ v , d ] = 1 , [ v , d ] = d ^ , and [ w , b ] = [ w , d ] = 1 [ w , a ] . Without loss, o ( b ) o ( d ) . Neither of the subgroups b d and a b d being normal in P, so a , ( b d ) 2 is, consequently,

[ a , w ] = b ^ if o ( b ) > o ( d ) ,
(2.1) [ a , w ] = b ^ d ^ if o ( w ) = o ( d ) , in particular [ B , P ] = b ^ , d ^ .

For z Ω 1 ( Z ( P ) ) b ^ , d ^ , P / z therefore satisfies (a)–(d) and induction yields a contradiction. Accordingly,

(2.2) Ω 1 ( Z ( P ) ) = b ^ , d ^ .

Since E is a complement of a , b , d in B and contained in Z ( P ) , condition (2.2) yields B = a , b , d . Note that u 2 1 , for u , a P = 1 otherwise. Certainly u 2 C B ( u ) = a , b 2 , d . Set u 2 = a i b 2 j d .

Assume that o ( b ) > o ( d ) . It follows that o ( b ) > 4 and there is b b such that ( u b ) 2 = a i d . Replace u by u b . If both i and are odd, then [ u 2 , w ] = b ^ and [ u 2 , v ] = d ^ , i.e. u 2 P . If i is odd and is not, then u d has an element whose square is a, making a normal in P, which it is not. Accordingly, we may take u 2 d . Since [ u , b ] = [ u a , b ] = [ a , w ] = b ^ , none of the maximal subgroups of u , a is normal in P. This proves

(2.3) o ( b ) = o ( d ) .

If i is odd, then (2.1) yields that [ u 2 , w ] = b ^ d ^ , i.e. must be even; yet that implies there is s u d with s 2 a b , i.e. [ s 2 , w ] s 2 . Hence u 2 = b 2 j d . If o ( b ) > 4 or j is even, then there is s u b satisfying s 2 d , and none of the maximal subgroups of a , s is normal in P.

The only possibility left standing is o ( b ) = o ( d ) = 4 and u 2 = b 2 d . If is odd, then, as before, u , a P = u 2 , so, replacing u by an appropriate element of u d , we obtain u 2 = b 2 . Since this means u 2 , a P , one of u a and u is normal and there is no loss in taking u P . By (2.3), the situation under consideration is symmetric in b and d, so we may also assume v 2 = d 2 and v P . Let Q = u , b , v , d . Then Q Q 8 × Q 8 and u v , b d Q = b 2 d 2 , a final contradiction. ∎

Lemma 11

Let P be a finite 2-group with a normal subgroup Q such that the following conditions are satisfied:

  1. (a)

    Q = Ω 1 ( Q ) = Ω 1 ( Z ( P ) ) .

  2. (b)

    Q = x 1 x 2 x 3 , where o ( x 1 ) o ( x 2 ) 8 and o ( x 2 ) o ( x 3 ) 4 , while x 1 x 2 = x 1 x 2 x 3 = 1 .

  3. (c)

    P = Q x with u x u - 1 Ω 1 ( Q ) for u Q .

  4. (d)

    [ Ω 1 ( Q Φ ( P ) ) , Q ] = 1 .

Then P is not core-2.

Proof.

It will be helpful to be aware of the fact that ( u v ) 2 = u 2 v 2 whenever u , v Q and 2 . Let Y = Ω 1 ( Q ) . Combining conditions (a) and (b), we see that Φ ( Q ) = x 1 2 × x 2 2 × x 3 2 , while Q = Y = [ a , b ] , [ a , c ] , [ b , c ] whenever Q = a , b , c ; in particular b Q ( u ) = 2 for every u Q Φ ( Q ) . The condition placed on the orders of the x i in (b) entails that Y = x ^ 1 × x ^ 2 × x ^ 3 = { 1 } { s ^ | s Q Φ ( Q ) } . Let y 1 , y 2 , y 3 Q . Then ( y 1 , y 2 , y 3 ) will be called an admissible triple if, for i = 1 , 2 , 3 , y i has each of the properties ascribed to x i in conditions (a) and (b). Note that this implies o ( y i ) = o ( x i ) for i = 1 , 2 , 3 . Next, let P = Q x . By (c), x 2 Z ( P ) and u 2 x = u - 2 whenever u Q , so x inverts every element of the abelian group Φ ( Q ) . It follows that x Q C Φ ( Q ) ( x ) = Y . Thus condition (a) entails that o ( x ) = 2 or x ^ Y .

Suppose that P is a core-2-group. First assume o ( x ) > 4 . Let Ω 2 ( x ) = z and let a Q Φ ( Q ) satisfy a ^ = x ^ . If o ( a ) = 4 , then z a Ω 1 ( Q Φ ( P ) ) , and (d) yields [ Q , a ] = 1 , which is not compatible with (a) and (b). Thus o ( a ) 8 and there is an involution s in z a 2 . By (d), [ s , Q ] = 1 , whence [ s , P ] = [ s , x ] = x ^ . For t Q Φ ( Q ) with t ^ x ^ , we have s , t P { t , s t } ; either possibility yields b Q ( t ) 1 , which contradicts Y = Q . Consequently:

  1. (1)*

    o ( x ) 4 , in particular Q P .

For u Q , let z u = u 2 [ u , x ] , in other words ( u x ) 2 = x 2 z u . By condition (c), Ω 1 ( Φ ( u , x ) ) = x 2 , z u , u ^ , while Φ ( u , x ) Y / Y = u 2 Y / Y . Let y Q . Then it follows from (b) and (c) that [ Q , y ] Y / Y is abelian of degree at least 2 and neither of the groups y , u 2 , [ u , y ] and y u , u 2 , [ u , y ] can be normal in P. Thus:

  1. (2)*

    If u Q y , then u , y 2 , u 2 [ u , y ] P .

We intend to show that o ( x ) = 2 . Since x is an arbitrary element of P Q , that will, Q being nonabelian, suffice to establish a contradiction. So let us assume o ( x ) = 4 . Let a Q Φ ( Q ) satisfy a ^ = x 2 ; by condition (2)*, [ Q , a ] = a ^ , z a . Let [ Q , a ] = W . Since b Q ( a ) = 2 , we have Q = a , u 1 , u 2 with [ u 1 , a ] = z a and [ u 2 , a ] = a ^ z a . For i { 1 , 2 } , x u i N P ( a ) , and condition (2)* implies that x u i , a P = a , x 2 z u i = a , z u i , while W = a ^ , z u i . Let u u 1 a , Φ ( Q ) u 2 a , Φ ( Q ) . Since Y x , u a ^ , z u , u ^ and [ u , Q ] W , condition (2)* yields:

  1. (3)*

    Let u Q Φ ( Q ) . Unless u N Q ( a ) , u ^ [ a , Q ] = x 2 , z a = x 2 , z u .

Every element of Q of order less than o ( x 3 ) belongs to Φ ( Q ) , so we know o ( a ) o ( x 3 ) . Suppose that o ( a ) = o ( x 3 ) . Then Q = a v 1 v 2 , where v 1 N Q ( a ) and ( v 1 , v 2 , a ) is an admissible triple. Suppose that v 2 N Q ( a ) . By condition (3)* this means that v ^ W whenever v v 2 Φ ( Q ) v 1 Φ ( Q ) . Accordingly, o ( v 1 ) = o ( v 2 ) , since otherwise some w in v 2 v 1 2 would satisfy w ^ = v ^ 1 v ^ 2 . So v 2 N Q ( a ) or o ( v 2 ) = o ( v 1 ) . If o ( v 2 ) < o ( v 1 ) , then there is an element w v 2 v 1 2 with w ^ W , while of course ( v 1 , w , a ) is admissible. If o ( v 2 ) = o ( v 1 ) , then ( v 1 , v 1 v 2 , a ) is admissible, while, by (3)* and because of v 1 v 2 ^ = v ^ 1 v ^ 2 , v 2 N Q ( a ) if and only if v ^ 2 W if and only if v 1 v 2 N Q ( a ) if and only if v 1 v 2 ^ W .

We have found an admissible triple ( v 1 , w , a ) with w ^ W and [ w , a ] = a ^ . Since [ a , v 1 w ] = a ^ [ a , v 1 ] and ( v 1 w , w , a ) is admissible, we may even take [ v 1 , a ] = z a . Recall that o ( w ) 8 . From a , w P = a , w 2 we get w ^ a ^ . Since a ^ [ w , Q ] W , it follows that w ^ [ w , Q ] . Suppose that w ^ = a ^ z a . Applying condition (2)* with u = w and y = x a , we obtain [ w , Q ] = a ^ z w , a ^ z a , in particular w ^ [ w , Q ] . Thus the only remaining option is w ^ = z a . If ( x v 1 ) 2 = w ^ or ( x v 1 w ) 2 = w ^ , then (2)* (with u = w and y = x v 1 or x v 1 w , respectively) again yields w ^ [ w , Q ] = W . Since { z v 1 , z v 1 w } { z a , a ^ z a } , we are forced to conclude that ( x v 1 ) 2 = a ^ z v 1 = a ^ z a = ( x v 1 w ) 2 = a ^ z v 1 w . It follows that z v 1 = z v 1 w = z a . Since z v 1 w = z v 1 z w [ v 1 , w ] , this yields [ v 1 , w ] = z w . Yet now we have w x v 1 = w - 1 and (2)* says that x v 1 , w P = w , ( x v 1 ) 2 = w , a ^ z a . This also implies [ w , Q ] = W .

Thus:

  1. (4)*

    If a Q Φ ( Q ) and a ^ = x 2 , then o ( a ) > o ( x 3 ) .

Note that (4)* entails o ( x 1 ) > o ( x 3 ) . From now on, we take a to be of maximal order subject to a Q Φ ( Q ) and a ^ = x 2 . Let o ( a ) = 2 and write a = x 1 i x 2 j x 3 k . It follows from (4)* that > 2 and

a ^ = x 1 2 - 1 i x 2 2 - 1 j x 3 2 - 1 k x ^ 1 x ^ 2 .

Hence there is s x 1 x 2 with o ( a ) o ( s ) and s ^ = a ^ . There is thus no loss of generality in assuming a x 1 x 2 .

Now o ( a ) 8 and a r ^ = a ^ for all r Q of order less than 2 . This combines with (3)* to yield:

  1. (5)*

    If v Q N Q ( a ) , then o ( v ) o ( a ) .

Since o ( x 3 ) < o ( a ) and C Q ( a ) = a Φ ( Q ) x 1 , x 2 Φ ( Q ) , condition (5)* implies that [ x 3 , a ] = a ^ and N Q ( a ) = a , x 3 Φ ( Q ) .

Suppose that o ( x 2 ) < o ( x 1 ) . Then the current restrictions on a entail a ^ = x ^ 1 only if a x 1 Φ ( Q ) , in particular o ( a ) > o ( x 2 ) . But x 2 N Q ( a ) , so this is made impossible by (5)*. Accordingly, a ^ { x ^ 2 , x ^ 1 x ^ 2 } ; since ( x 1 , y , x 3 ) is admissible whenever y x 2 x 1 2 , x 2 2 satisfies o ( y ) = o ( x 2 ) , we may fix notation such that a = x 2 . If o ( x 1 ) = o ( x 2 ) , then every triple ( y 1 , y 2 , x 3 ) that satisfies y 1 , y 2 Φ ( Q ) = x 1 , x 2 Φ ( Q ) is admissible and notation may again be arranged such that a = x 2 .

Let W = [ a , Q ] as before. By (3)*, we have x ^ 1 W . If y x 3 x 1 satisfies o ( y ) = o ( x 3 ) , then ( x 1 , x 2 , y ) is admissible. Thus x ^ 3 may be assumed to be in W, whence W = a ^ , x ^ 3 results. By (3)*, this implies that x ^ 3 { z x 1 , a ^ z x 1 } . Since o ( a ) > o ( x 3 ) , there is w x 3 a 2 with w ^ = a ^ z x 1 . Since this means ( x x 1 ) 2 = w ^ , (1)* says that x x 1 , w P = w , z w [ x 1 , w ] and [ w , Q ] = w ^ , z w [ x 1 , w ] . However, [ a , w ] = a ^ w ^ and it follows that [ w , Q ] = w ^ , a ^ = W , a contradiction that ends this proof. ∎

Lemma 12

Let P be a finite 2-group possessing a normal subgroup Q such that the following conditions are satisfied:

  1. (a)

    Q = Ω 1 ( Q ) = Ω 1 ( Z ( P ) ) .

  2. (b)

    | Q | = 8 .

  3. (c)

    Q = x 1 , x 2 , x 3 , where o ( x 1 ) 8 and o ( x 2 ) = o ( x 3 ) = 4 .

  4. (d)

    P = Q x , where u x u - 1 Ω 1 ( Q ) for u Q .

  5. (e)

    [ Q , Ω 1 ( Q Φ ( P ) ) ] = 1 .

Then P is not core-2.

Proof.

Let Y = Ω 1 ( Q ) , let o ( x 1 ) = 2 n and let x 1 2 n - 2 = z . Like in the previous proof, a triple ( y 1 , y 2 , y 3 ) of elements of Q will be called admissible whenever (c) continues to be true after replacing x i by y i for i = 1 , 2 , 3 . Observe that (b) implies Q = [ u , Q ] [ v , Q ] whenever u Q Φ ( Q ) and v Q u Φ ( Q ) , in particular b Q ( u ) = 2 whenever u Q Φ ( Q ) . Let y P Q . Combining conditions (a) and (d), we obtain u y u - 1 Y , in particular u y 2 = u whenever u Q . Thus y 2 Z ( P ) and (a) implies that y ^ Y or o ( y ) = 2 . Since Y Z ( P ) , we have x 1 x 2 = x 1 x 3 = 1 .

Assume that P is core-2. Our first goal is showing that some admissible triple ( y 1 , y 2 , y 3 ) satisfies y 3 ( y 1 y 2 ) = 1 . Assuming that such a triple does not exist, we have x 3 2 x 1 x 2 . Upon replacing x 3 by x 3 z if necessary, we may take x 3 2 = x 2 2 . If [ x 2 , x 3 ] = x 2 2 , then there is w { x 2 , x 3 , x 2 x 3 } with b Q ( w ) 1 . If [ x 2 , x 3 ] x ^ 1 , x 2 2 , then ( x 1 , x 2 , x 2 x 3 ) is an admissible triple endowed with the extra feature we are currently seeking. If [ x 2 , x 3 ] = x ^ 1 , then x 2 x 3 z is an involution in Q Φ ( Q ) , which leaves [ x 2 , x 3 ] = x 2 2 x ^ 1 the only undiscarded option. Yet that yields ( x 2 x 3 ) 2 = x 2 2 x ^ 1 = ( x 2 z ) 2 = [ x 2 x 3 , x 2 z ] ; letting R = x 2 z , x 2 x 3 , we have R Q 8 with R Φ ( Q ) = Φ ( R ) . One of the three cyclic four-subgroups of R must be normal in P, giving Q an element of breadth 1 and thus clashing with (a). Hence:

  1. (1)*

    Without loss of generality, x 3 x 1 x 2 = 1 .

Observe that (1)* entails x k ( x i x j ) = 1 for { i , j , k } = { 1 , 2 , 3 } , while Y = Q = x ^ 1 , x 2 2 , x 3 2 and Q = x 1 x 2 x 3 .

Since Φ ( Q ) = x 1 2 , x 2 2 , x 3 2 Q = x 1 2 Y , either every element of Y is a square in Q, or x 2 2 x 3 2 is not. Assume the latter. If [ x 2 , x 3 ] = x ^ 1 , then x 2 2 x 3 2 = ( x 2 x 3 z ) 2 and if [ x 2 , x 3 ] x 2 2 x 3 2 x ^ 1 , then there is an involution in x 2 x 3 s and condition (a) is contradicted. Accordingly, [ x 2 , x 3 ] x 2 2 x ^ 1 x 3 2 x ^ 1 . Interchanging x 2 and x 3 and replacing x 2 by x 2 z if necessary, we may assume [ x 2 , x 3 ] = x 2 2 . Since ( x 1 , x 2 , x 2 x 3 ) is admissible and ( x 2 x 3 ) 2 = x 3 2 , one of the subgroups x 3 , x 2 2 and x 2 , x 3 2 may be taken to be normal. Let { i , j } = { 2 , 3 } with x i , x j 2 P .

Assume that o ( x ) > 4 . By the previous paragraph’s results, x ^ is a square in Q or P = Q y with [ y , x i ] = 1 and x i , y 2 P . Since b Q ( x i ) = 2 , this implies that y ^ { x j 2 , x i 2 x j 2 } . Suppose that y ^ = x 2 2 x 3 2 . Setting s = Ω 2 ( y ) , there is an element v x 2 , x 3 s with [ v , P ] = x 2 2 . Since s Z ( P ) and Q does not have elements of breadth 1, this is not possible.

We have found | P : Q | > 2 to imply that x ^ is a square in Q. Let Ω 2 ( x ) = s and s 2 = q 2 with q Q . Then (e) says that [ Q , s q ] = 1 and q Z ( Q ) = Φ ( Q ) , which means that [ x , s q ] = q 2 = x ^ 1 . Letting s q = t , we have t , x 2 P = x 2 2 , a contradiction. Accordingly:

  1. (2)*

    | P : Q | = 2 .

By (1)*, Z ( Q ) = Φ ( Q ) = x 1 2 , x 2 2 , x 3 2 and (2)* implies that x 2 C Φ ( Q ) ( x ) = Y . Let V = x 2 , x 3 , x 1 2 n - 2 , x ; observe that exp V = 4 .

Assume that V = Y . Let U 1 , U 2 , U 3 , and U 4 be the distinct complements of x ^ 1 in Y. For 1 i 4 , let D i denote the preimage of Z ( V / U i ) in V. Let 1 i 4 . If | V : D i | = 4 , then x 1 2 n - 2 D i Q because of x 1 2 n - 2 Z ( Q ) and [ x 1 2 n - 2 , x ] = x ^ 1 . Thus Q = x 1 D i and D i = b i , c i Y , where b i { x 2 , x 1 2 n - 2 x 2 } and c i { x 3 , x 1 2 n - 2 x 3 } . On the other hand, Lemma 8 says that | V : D i | = 16 implies | V : D j | = 4 whenever j i . It follows that the commutator [ x 2 , x