# 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:HG|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/HG|n for every subgroup H of G; here, HG 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 p2, 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 26 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 D2n 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 D8Q8 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/HG|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 p2 ([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 GA, 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 xpG whenever xG and |E:EZ(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=O2(G). Then Q is abelian of odd order, every subgroup of Q is normal in G and G=QS with SSyl2(G). The group S is core-2, and, since |S:SG|2, CS(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=CS(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 TUS whenever US=TU. 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,tx2=y8=s4=t4=1,[x,y]=y2,s2=t2=[s,t]=y4.

Then SD16Q8 is core-2, but there is no abelian subgroup of index 4 in S that does not contain y. Now let G=SQ, where QC3, CS(Q)=x,y2,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.

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=expB and let U be the unit subgroup of Z/nZ. There is a homomorphism ϵ:XU given by bx=bϵ(x) whenever bB and xN.

Notation

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

j(X)=|{xXx2=1}|.
Lemma 2

## Lemma 2 ([1, Satz 1])

Let P be a 2-group and let 1NΩ1(Z(P)). Let H be the set of maximal subgroups of N. Then

Uj(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 1uZ(P), Z(P) is cyclic and Φ(P)=Ω1(Z(P)). Let s and t be involutions in Φ(P). If [s,t]1, then s,tD8. Now P normalises a maximal subgroup V of s,t, of order 4; this, however, implies that [Φ(P),V]=1, a contradiction. Let Ω1(Φ(P))=Y. We have just seen that Y is elementary abelian, whence |Y:YZ(P)|2, implying |Y|4 and YZ(Φ(P)). Let Ω1(Z(P))=z. If zUΦ(P), then UΦ(P). Since Ω1(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 QQ8. 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]Ω1(Z(P)). If |U:CU(v)|p for all v in V, then |[U,V]|p or |U:CU(V)|=p.

## Proof.

Let vVCV(U) and wV. If [U,w][U,v] and CU(w)CU(v), then [U,vw]=[CU(v)CU(w),vw]=[CU(w),v][CU(v),w]=[U,v]×[U,w], of order p2. It follows that every element of V belongs to CV(CU(v)) or belongs to the subgroup {xV[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]u4. Unless o([u,v])2, there is xU with x2U.

## 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)=2n+2 with n2 and [u,v]=u2n. Observe that this implies γ3(U)=1. Let |U:u|=2m. A generator w of u may be chosen such that v2m is equal to a power w2; as Cu(v)=u4 and Cv(u)=v4, we have m2.

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

x2=w2v-2m[v,w]2m-(22)=1.

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

[x2,v]=[w2,v]=u2n+1

and x2U.

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

y2m=v2mw-2[w,v]2-m(2m2)u2n+-1.

So y2m=1 unless =2=m. Now n2 implies [u2n-1,v]2=1. Hence if =2=m, then (vw-1u2n-1)4=(u2n+1)2=1. Accordingly, vu=xu with o(x)=2m and xu=1. However, [v2,u]1 yields x2U. ∎

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:HG|p is required for all subgroups of G generated by (no more than) two elements. Indeed, suppose |x,y:x,yG|p for all elements x,y of the p-group G. This property carries over to subgroups and quotients. Let QG and let G¯=G/QG. 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 xpG whenever xG and that |E:EZ(G)|p for each elementary abelian subgroup E of G.

Notation

Let p=2. For and yG{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]Ω1(Z(G)) and not every subgroup of A is normal in G. Let pk be the minimal order of a cyclic subgroup of A that is not a normal subgroup of G. Then Ωk-1(A)Z(G).

## Proof.

Let a be any element of A of order pk and let zΩk-1(A). Let t be a complement of a in a,z. Note that o(t)<pk. There is aap satisfying o(a)=o(t). Both t and at are normal in G, while aZ(G). Thus tZ(G), i.e. zap,tZ(G). ∎

Lemma 7

Assume |G|=p. Then |G:Z(G)|=p2 unless p=2, expG=4 and G=UV×E, where E is elementary abelian, [U,V]=1, VQ8, and U=u,v with [u,v]=G, v2=1 and u2G.

## Proof.

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

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|=p2m+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 1sΩ1(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 xGZ(G) of order pn. Since n>1, expΦ(P)=pn-1, so Φ(P)=xp. Since p is odd, raising elements to their pth power is an endomorphism of G, such that G=xΩ1(G) and expΩ1(G)=p. Hence there is tG with tp=1[x,t]. Let Q=CG(x,t), noting that G=x,tQ. For uΩ1(Q)Z(Q), t,u is core-free elementary abelian of order p2. Thus vpxp2 whenever vQZ(Q). Yet if vQ satisfies o(v)=o(x), then Φ(G)=vp and Q=vQ1 with expQ1=p, i.e. Q1Z(Q). Since Q/Q1 is cyclic, Q is abelian after all.

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

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 2n-1, i.e. Φ(G)=x2. Let yGCG(x) and Q=CG(x,y), observing that G=x,yQ and QZ(G) if and only if Q is abelian.

Assume that n3. For each vG, there is a generator w of x such that v2=w2 with some 1. Either (vw-2-1)2=1, or =1, [v,x]1 and (vw-1x2n-2)2=1. In particular, y may be chosen of order 2. Assume that Q is nonabelian. If expQ=2n, then an analogous argument yields an involution s in QZ(Q). However, this would mean s,yG=1. Thus expQ<2n and, for uQ, we have u2x4 and there is an involution su satisfying ux2=sux2. If uQZ(G), then suZ(Q) and su,yG=1.

We turn to the case n=2. Then G=DE with D extraspecial and E abelian. The group D8D8 is not core-2, whence either |G:Z(G)|=4 or DD8Q8 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 |Ω1(Z(G))|>2. Let Ω1(Z(G))=N and sNz. 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 sNz. Thus |G:Z(G)|=16 implies expG=4.

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

S={sGs2z}.

Then S<G, in particular GSZ(G). Let uGZ(G) with u2z, let tGCG(u) and let CG(u,t)=Q.

For qQZ(Q) and r{u,t,ut}, q,r is an abelian subgroup of G with zq2,r2. Suppose that |u2,t2,z|=8. If qQZ(Q), then q2=q2,tq2,u=z. Consequently, we have Q=x,y×E with E elementary abelian, x,yQ8, and x2=y2=z. Now let W=ux,ty. Then WC4×C4 and WZ(G)=Φ(W)=u2z,u2t2z, so WG=Φ(W).

We have seen that u2,t2u2,z whenever tGCG(u). We shall show that Ω1(G)Z(G). Assume otherwise. Let tGCG(u). If t2=u2z, then (ut)2=1, so t2{u2,z}. Since GCG(u)=G>S, it follows that t may be assumed to satisfy t2=u2. Let Q=CG(u,t) and qQZ(Q). Then tqCG(u), while (tq)2=u2 would imply q2=1. Hence we have (tq)2=z, i.e. q2=u2z. Now let qQCQ(q). Then (q)2=u2z=(qq)2=z, a contradiction.

We have confirmed the existence of an involution v in the set GZ(G). Since GSCG(v), the element u may be assumed to satisfy [u,v]=z. Let U=u,v and let Q=CG(U). For wQZ(Q), zw,v, so Q=V×E with VQ8 and elementary abelian E. This completes the proof. ∎

Notation

The elementary abelian group of order pm will, as usual, be denoted by Epm.

Let UG and xG. Following [8], we define bU(x), the breadth of x in U, as bU(x)=logp|U:CU(x)|. We write b(x) instead of bG(x). The breadth b(A) in G of the subgroup A is defined as max{b(a)|aA}.

Let N=Ω1(Z(G)). For a subgroup U of N, let DU be the preimage of Z(G/U) in G.

Lemma 8

Suppose that p=2 and GN. Then |G:DU|{24,22,1} for UN. For any subgroup Wof N such that |N:W|=4 and WG=N, there is at most one maximal subgroup M of N with WM and |G:DM|=24.

## Proof.

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

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

Let M be a maximal subgroup of N with |G:DM|=24. 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=UV×E, where EE2n-5, VQ8 and U=u,v with u2Φ(V) and v2=1. Letting u2=s and V=x,y, we have Ω1(UV)=z,s,vE8. Accordingly, j(X)=2n-2. Certainly j(G/M)<2n+1 for any maximal subgroup M of N not containing G. So if |G:DM|=16 for at least two maximal subgroups of N containing W, then

W<MNj(G/M)<2n-1+2n+1<32n=3|G/N|,

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

We note: For 4n, j(Q8D8×E2n-4)=32n-2, j(D8×E2n-2)=32n-1, j(Q8×E2n-2)=2n-1, and j(Q8C4×E2n-3)=2n.

Let MN. From Lemma 7 and Φ(G)=16, we deduce that |G:DM|=16 if and only if G/MQ8D8×E2n-4. From Φ(G)N it also follows that, if MN satisfies |G:DM|=4, then G/M is isomorphic to one of the groups D8×E2n-2, Q8×E2n-2, or Q8C4×E2n-3. Suppose that |G:DM|=16 for at least two maximal subgroups M containing W. Then, as we have seen,

W<MNj(G/M)32n=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 GN. Let A be an abelian subgroup of G of maximal order. Then the following hold:

• If p is odd, then |G:A|p2 and if |G:A|=p2, 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,dZ(G) with b(c)=b(d)=1, in particular |[A,G]|4.

• If p=2 and Ω2(A)Z(G), then |G:A|2 or |A:Z(G)|2.

## Proof.

Note that Z(G)Φ(G)AG.

We start with the case p odd. Let [A,G]=W of order pn, and let A/Z(G)=V of order pm. For any maximal subgroup U of W, there is a maximal subgroup U1 of N satisfying U1W=U. Applying Lemma 7 to the quotient G/U1, we obtain |A/U:(A/UZ(G/U))|=p. Note that this implies that mn.

Let 𝒮 be the set of pairs (v,U), where 1vV, UW and v=aZ(G) with aA and [a,G]U. Let r=pm-1p-1. We list the subgroups of V of order p as v1,vr. For i1,,r, let vi=aiZ(G) and let pi=|[G,ai]|. We assume the vi listed as to satisfy ii+1 for all i. We count the elements of 𝒮 in two ways, to obtain

|𝒮|=1(p-1)2(pn-1)(pm-1-1)=1p-1i=1r(pn-i-1),

i.e.

i=1r(pn-i-1)=1p-1(pn-1)(pm-1-1).

Observe that

1p-1(pn-1)(pm-1-1)=1p-1(pn-1-1)(pm-1)-pn-1+pm-1.

Suppose that 2>1. Then

i=1r(pn-i-1)1p-1(pn-1-1)(pm-1)-2pn-1+pn-1+pn-2.

However, pn-1>2pn-2pn-1+pn-2, whence

-2pn-1+pn-1+pn-2+pn-1-pm-1=-pn-1+pn-1+pn-2-pm-1<0,

We have established that either b(A)=1 or there is aA such that a,Z(G) contains every element of A of breadth greater than 1. Assume the latter together with m2. Let bAaZ(G). Then

b(b)=b(ab)=1,

and [a,G]=[b,G]×[ab,G], while CG(b)CG(ab)=G. Let xCG(b)CG(ab) and yCG(ab)CG(b). Then [ab2,x]=[ab,x] and [ab2,y]=[b,y], i.e. [a,G]=[ab,x]×[b,y]=[ab2,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,bZ(G)|=4 and b(a)2b(b). First suppose that there are subgroups s,t of [G,a] and s,t of [G,b] such that |s,ts,t|=16. Let U=s,t,s,t and let V be a complement of U in N. Let X1=ss,tt, X2=ss,tst, and X3=st,st. For i=1,2,3, let Wi=V×Xi. For each index i, we have U=s,t×Xi=s,t×Xi, while X1X2X3=1 and W1W2W3=V. Let i{1,2,3}. Let WiXN; then neither a nor b belongs to DX. Now Lemma 8 may be invoked to yield that abDX for at least two out of three different choices of X; thus

[ab,G]W1W2W3=V.

The intersection over all complements of U in N being trivial, we obtain that abZ(G) in contrast to our assumptions on a,b. If [a,G][b,G] has a subgroup U of order 4, then, similarly, [ab,G]V whenever N=U×V, which results in abZ(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, W1=Vtt, W2=Vtst. Lemma 8 again yields

[ab,G]W1W2=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=aB, where Z(G)B and b(b)=b(ab)=1 for every b in B that satisfies |b,aZ(G):Z(G)|=4. In particular, b(B)1.

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

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

G=CG(b)CG(bc)=CG(c)CG(bc).

For d{b,c,bc}, we have [ad,G]=[a,CG(d)] because of [CG(d),a]1 and b(ad)=1. Furthermore, [ad,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 [ab,G]=[ac,G]. However, this yields [bc,G]=[B,G]=[b,G]=[ab,G], a contradiction. Consequently, |[B,G]|=2 entails that A=a,b with b(b)=1=b(ab), while CG(a)=CG(b)CG(ab)=CG(A)=A, i.e. |G:A|=4.

We turn to the case |G:CG(B)|=2. Let bBZ(G). We have CG(B)=CG(b) and G=CG(b)CG(ab). Letting xCG(B)CG(a), we find that

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

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

Letting p be any prime, assume |A:Z(G)|=p. Let A=aZ(G), i.e. b(a)=logp|G:A|. It will suffice to derive a contradiction from |G:A|=p3. For xGA, the maximality of |A| implies that CG(x)=Z(G),x is a maximal abelian subgroup of G. Let xGZ(G) and let yGZ(G)x. One of the maximal subgroups of x,y is normal in G, say x,yp,[x,y]. Since Φ(x,y)=xp,yp,[x,y], i.e. d(Φ(x,y))3, we must have [x,G]=Ω1(Φ(x,y)). If o(x)=p, then Φ(x,y)=yp,[x,y], so o(x)>p and Ω1(x)[x,G]. It follows that there is vGx,Z(G) with [v,x]=Ω1(x). This implies that Φ(v,x)=vpxp and there is wv,xZ(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:CG(A)|=|G:A|=p. Only the first case needs further consideration. Let [G,A]=z, of order p. We note that for UA we have zU or |U:UZ(G)|p. Let A=a1××an, where z projects nontrivially into a1 along the given decomposition. Let B~=a2an. Then |B~:B~Z(G)|p; however, this implies that A=aZ(G) and |G:A|=p or A=a,bZ(G) with A=CG(a)CG(b), i.e. |G:A|=p2.

At this point, only the final assertion of the lemma has not been taken care of. By way of contradiction, assume that p=2, Ω2(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=a1××an, either n=2, or some c of the form c=i=1naiεi with εi{0,1} belongs to Z(G). For such a c, c has a complement in E, whence cD has a complement in G. It follows that n=2 and A=D×a×b with DZ(G). Note that o(a)8o(b).

The following observation is going to be quite helpful: Suppose there is zN[A,G] such that Ω2(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 Bz. However, [B,AB]=1, whence BA=G and AB=Z(G)A.

1. (${\ast}$)

Let zN[A,G]. Then Ω2(A/z)Z(G)/z.

Suppose that [A,G]a^,b^. Let zΩ1(D). Any dA satisfying d4=z then belongs to Da2,b2Z(G), and () yields D=1.

Assume that |[A,G]|=2. Since there is c{a,b,ab} 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 wGA. There is cA satisfying w2=c2, i.e. (wc-1)2=[w,c]. Since o(a)8o(b), [w,c]Φ(Z(G)) and the coset wA contains an involution. Letting [A,G]=s, it follows that G=Ax,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,xy is in Z(G). Accordingly, we have x,yD8, 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 1vG, let Av be the preimage of Z(G/v)A/v in A. Since expG/v>4, Lemma 7 yields |A:Av|=2 for each v. Since |A:Z(G)|=4, AvAv whenever v,v=G, and it follows that each of the three maximal subgroups of A that contain Z(G) is equal to some Av, i.e. b(A)=1. Now Lemma 4 yields |G:CG(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,bE. If 1zΩ1(E), then Ω2(A/z)Z(G)/z. As previously mentioned, there is an element c{a,b,ab} 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=Ω1(Φ(G))=a^,b^,d^.

Suppose that o(d)=4. There is an element ta^,b^ satisfying td2[A,G]. Since td22(A)=a^,b^, condition () yields a contradiction. Hence o(d)8 and N2(G).

Let VG. We have seen that expG/V8, and Lemma 7 says that the subgroup A/VZ(G/V) is of index at most 2 in A/V. Only two of the three elements a,b,ab have breadth 1 in G, so there is an element s[A,G] with Z(G/s)A/s=Z(G)/s. Letting tG[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)4o(d), o(b)>expE<o(d),

2. (b)

EZ(P), Φ(P)B, [P,B]Ω1(Z(P)),

3. (c)

b(a)=1 and bPd,

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,wB 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 bd and abd being normal in P, so a,(bd)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 zΩ1(Z(P))b^,d^, P/z therefore satisfies (a)–(d) and induction yields a contradiction. Accordingly,

(2.2)Ω1(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 u21, for u,aP=1 otherwise. Certainly u2CB(u)=a,b2,d. Set u2=aib2jd.

Assume that o(b)>o(d). It follows that o(b)>4 and there is bb such that (ub)2=aid. Replace u by ub. If both i and are odd, then [u2,w]=b^ and [u2,v]=d^, i.e. u2P. If i is odd and is not, then ud has an element whose square is a, making a normal in P, which it is not. Accordingly, we may take u2d. Since [u,b]=[ua,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 [u2,w]=b^d^, i.e. must be even; yet that implies there is sud with s2ab, i.e. [s2,w]s2. Hence u2=b2jd. If o(b)>4 or j is even, then there is sub satisfying s2d, 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 u2=b2d. If is odd, then, as before, u,aP=u2, so, replacing u by an appropriate element of ud, we obtain u2=b2. Since this means u2,aP, one of ua and u is normal and there is no loss in taking uP. By (2.3), the situation under consideration is symmetric in b and d, so we may also assume v2=d2 and vP. Let Q=u,b,v,d. Then QQ8×Q8 and uv,bdQ=b2d2, 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=Ω1(Q)=Ω1(Z(P)).

2. (b)

Q=x1x2x3, where o(x1)o(x2)8 and o(x2)o(x3)4, while x1x2=x1x2x3=1.

3. (c)

P=Qx with uxu-1Ω1(Q) for uQ.

4. (d)

[Ω1(QΦ(P)),Q]=1.

Then P is not core-2.

## Proof.

It will be helpful to be aware of the fact that (uv)2=u2v2 whenever u,vQ and 2. Let Y=Ω1(Q). Combining conditions (a) and (b), we see that Φ(Q)=x12×x22×x32, while Q=Y=[a,b],[a,c],[b,c] whenever Q=a,b,c; in particular bQ(u)=2 for every uQΦ(Q). The condition placed on the orders of the xi in (b) entails that Y=x^1×x^2×x^3={1}{s^|sQΦ(Q)}. Let y1,y2,y3Q. Then (y1,y2,y3) will be called an admissible triple if, for i=1,2,3, yi has each of the properties ascribed to xi in conditions (a) and (b). Note that this implies o(yi)=o(xi) for i=1,2,3. Next, let P=Qx. By (c), x2Z(P) and u2x=u-2 whenever uQ, so x inverts every element of the abelian group Φ(Q). It follows that xQCΦ(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 Ω2(x)=z and let aQΦ(Q) satisfy a^=x^. If o(a)=4, then zaΩ1(QΦ(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 za2. By (d), [s,Q]=1, whence [s,P]=[s,x]=x^. For tQΦ(Q) with t^x^, we have s,tP{t,st}; either possibility yields bQ(t)1, which contradicts Y=Q. Consequently:

1. (1)*

o(x)4, in particular QP.

For uQ, let zu=u2[u,x], in other words (ux)2=x2zu. By condition (c), Ω1(Φ(u,x))=x2,zu,u^, while Φ(u,x)Y/Y=u2Y/Y. Let yQ. 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,u2,[u,y] and yu,u2,[u,y] can be normal in P. Thus:

1. (2)*

If uQy, then u,y2,u2[u,y]P.

We intend to show that o(x)=2. Since x is an arbitrary element of PQ, that will, Q being nonabelian, suffice to establish a contradiction. So let us assume o(x)=4. Let aQΦ(Q) satisfy a^=x2; by condition (2)*, [Q,a]=a^,za. Let [Q,a]=W. Since bQ(a)=2, we have Q=a,u1,u2 with [u1,a]=za and [u2,a]=a^za. For i{1,2}, xuiNP(a), and condition (2)* implies that xui,aP=a,x2zui=a,zui, while W=a^,zui. Let uu1a,Φ(Q)u2a,Φ(Q). Since Yx,ua^,zu,u^ and [u,Q]W, condition (2)* yields:

1. (3)*

Let uQΦ(Q). Unless uNQ(a), u^[a,Q]=x2,za=x2,zu.

Every element of Q of order less than o(x3) belongs to Φ(Q), so we know o(a)o(x3). Suppose that o(a)=o(x3). Then Q=av1v2, where v1NQ(a) and (v1,v2,a) is an admissible triple. Suppose that v2NQ(a). By condition (3)* this means that v^W whenever vv2Φ(Q)v1Φ(Q). Accordingly, o(v1)=o(v2), since otherwise some w in v2v12 would satisfy w^=v^1v^2. So v2NQ(a) or o(v2)=o(v1). If o(v2)<o(v1), then there is an element wv2v12 with w^W, while of course (v1,w,a) is admissible. If o(v2)=o(v1), then (v1,v1v2,a) is admissible, while, by (3)* and because of v1v2^=v^1v^2, v2NQ(a) if and only if v^2W if and only if v1v2NQ(a) if and only if v1v2^W.

We have found an admissible triple (v1,w,a) with w^W and [w,a]=a^. Since [a,v1w]=a^[a,v1] and (v1w,w,a) is admissible, we may even take [v1,a]=za. Recall that o(w)8. From a,wP=a,w2 we get w^a^. Since a^[w,Q]W, it follows that w^[w,Q]. Suppose that w^=a^za. Applying condition (2)* with u=w and y=xa, we obtain [w,Q]=a^zw,a^za, in particular w^[w,Q]. Thus the only remaining option is w^=za. If (xv1)2=w^ or (xv1w)2=w^, then (2)* (with u=w and y=xv1 or xv1w, respectively) again yields w^[w,Q]=W. Since {zv1,zv1w}{za,a^za}, we are forced to conclude that (xv1)2=a^zv1=a^za=(xv1w)2=a^zv1w. It follows that zv1=zv1w=za. Since zv1w=zv1zw[v1,w], this yields [v1,w]=zw. Yet now we have wxv1=w-1 and (2)* says that xv1,wP=w,(xv1)2=w,a^za. This also implies [w,Q]=W.

Thus:

1. (4)*

If aQΦ(Q) and a^=x2, then o(a)>o(x3).

Note that (4)* entails o(x1)>o(x3). From now on, we take a to be of maximal order subject to aQΦ(Q) and a^=x2. Let o(a)=2 and write a=x1ix2jx3k. It follows from (4)* that >2 and

a^=x12-1ix22-1jx32-1kx^1x^2.

Hence there is sx1x2 with o(a)o(s) and s^=a^. There is thus no loss of generality in assuming ax1x2.

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

1. (5)*

If vQNQ(a), then o(v)o(a).

Since o(x3)<o(a) and CQ(a)=aΦ(Q)x1,x2Φ(Q), condition (5)* implies that [x3,a]=a^ and NQ(a)=a,x3Φ(Q).

Suppose that o(x2)<o(x1). Then the current restrictions on a entail a^=x^1 only if ax1Φ(Q), in particular o(a)>o(x2). But x2NQ(a), so this is made impossible by (5)*. Accordingly, a^{x^2,x^1x^2}; since (x1,y,x3) is admissible whenever yx2x12,x22 satisfies o(y)=o(x2), we may fix notation such that a=x2. If o(x1)=o(x2), then every triple (y1,y2,x3) that satisfies y1,y2Φ(Q)=x1,x2Φ(Q) is admissible and notation may again be arranged such that a=x2.

Let W=[a,Q] as before. By (3)*, we have x^1W. If yx3x1 satisfies o(y)=o(x3), then (x1,x2,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{zx1,a^zx1}. Since o(a)>o(x3), there is wx3a2 with w^=a^zx1. Since this means (xx1)2=w^, (1)* says that xx1,wP=w,zw[x1,w] and [w,Q]=w^,zw[x1,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=Ω1(Q)=Ω1(Z(P)).

2. (b)

|Q|=8.

3. (c)

Q=x1,x2,x3, where o(x1)8 and o(x2)=o(x3)=4.

4. (d)

P=Qx, where uxu-1Ω1(Q) for uQ.

5. (e)

[Q,Ω1(QΦ(P))]=1.

Then P is not core-2.

## Proof.

Let Y=Ω1(Q), let o(x1)=2n and let x12n-2=z. Like in the previous proof, a triple (y1,y2,y3) of elements of Q will be called admissible whenever (c) continues to be true after replacing xi by yi for i=1,2,3. Observe that (b) implies Q=[u,Q][v,Q] whenever uQΦ(Q) and vQuΦ(Q), in particular bQ(u)=2 whenever uQΦ(Q). Let yPQ. Combining conditions (a) and (d), we obtain uyu-1Y, in particular uy2=u whenever uQ. Thus y2Z(P) and (a) implies that y^Y or o(y)=2. Since YZ(P), we have x1x2=x1x3=1.

Assume that P is core-2. Our first goal is showing that some admissible triple (y1,y2,y3) satisfies y3(y1y2)=1. Assuming that such a triple does not exist, we have x32x1x2. Upon replacing x3 by x3z if necessary, we may take x32=x22. If [x2,x3]=x22, then there is w{x2,x3,x2x3} with bQ(w)1. If [x2,x3]x^1,x22, then (x1,x2,x2x3) is an admissible triple endowed with the extra feature we are currently seeking. If [x2,x3]=x^1, then x2x3z is an involution in QΦ(Q), which leaves [x2,x3]=x22x^1 the only undiscarded option. Yet that yields (x2x3)2=x22x^1=(x2z)2=[x2x3,x2z]; letting R=x2z,x2x3, we have RQ8 with RΦ(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, x3x1x2=1.

Observe that (1)* entails xk(xixj)=1 for {i,j,k}={1,2,3}, while Y=Q=x^1,x22,x32 and Q=x1x2x3.

Since Φ(Q)=x12,x22,x32Q=x12Y, either every element of Y is a square in Q, or x22x32 is not. Assume the latter. If [x2,x3]=x^1, then x22x32=(x2x3z)2 and if [x2,x3]x22x32x^1, then there is an involution in x2x3s and condition (a) is contradicted. Accordingly, [x2,x3]x22x^1x32x^1. Interchanging x2 and x3 and replacing x2 by x2z if necessary, we may assume [x2,x3]=x22. Since (x1,x2,x2x3) is admissible and (x2x3)2=x32, one of the subgroups x3,x22 and x2,x32 may be taken to be normal. Let {i,j}={2,3} with xi,xj2P.

Assume that o(x)>4. By the previous paragraph’s results, x^ is a square in Q or P=Qy with [y,xi]=1 and xi,y2P. Since bQ(xi)=2, this implies that y^{xj2,xi2xj2}. Suppose that y^=x22x32. Setting s=Ω2(y), there is an element vx2,x3s with [v,P]=x22. Since sZ(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 Ω2(x)=s and s2=q2 with qQ. Then (e) says that [Q,sq]=1 and qZ(Q)=Φ(Q), which means that [x,sq]=q2=x^1. Letting sq=t, we have t,x2P=x22, a contradiction. Accordingly:

1. (2)*

|P:Q|=2.

By (1)*, Z(Q)=Φ(Q)=x12,x22,x32 and (2)* implies that x2CΦ(Q)(x)=Y. Let V=x2,x3,x12n-2,x; observe that expV=4.

Assume that V=Y. Let U1,U2,U3, and U4 be the distinct complements of x^1 in Y. For 1i4, let Di denote the preimage of Z(V/Ui) in V. Let 1i4. If |V:Di|=4, then x12n-2DiQ because of x12n-2Z(Q) and [x12n-2,x]=x^1. Thus Q=x1Di and Di=bi,ciY, where bi{x2,x12n-2x2} and ci{x3,x12n-2x3}. On the other hand, Lemma 8 says that |V:Di|=16 implies |V:Dj|=4 whenever ji. It follows that the commutator [x2,x3] lies in the intersection of at least three complements of x^1 in Y, i.e. [x2,x3]=1, contradicting (b).

We have seen that |V|<8. Assume |V|=4. Let V=x^1,s and Y=V×t. By Lemma 8, s may be assumed to satisfy |V/s,t:Z(V/s,t)|=4. It follows that Q=x1D with Dx1=x^1 and [V,D]s. Perhaps upon replacing x2 and x3 by other elements of x2x12n-2 and x3x12n-2, respectively, we may take D=x2,x3Y; bearing in mind that [x2,x3]1, we obtain that [x2,x3]=s, which implies V=x2,x3CV(x2,x3).

Let zCV(x2,x3)Q. For w{x2,x3,x2x3}, we have w,zP=w,z2, and [w,Q]=w2,z2, in particular [x2,x3]=[x2,Q][x3,Q]=z2=s. For each u{x1x2,x1x3,x1x2x3}, (u,x2,x3) is an admissible triple satisfying ux2x3=1. We are hence free to assume [x1,xi]=xi2 for i=2,3. Let V1=x2,x3,x12n-2,x1z. Then V1x^1,x22,x32=Y. Replacing V by V1 in the preceding paragraph yields the desired contradiction.

Since x^1V, only the case |V|=x^1 is left. Then [x2,x3]=x^1 and there is zCV(x2,x3)Q. The argument is continued exactly as in the previous paragraph. ∎

## 3 Proof of the theorem

For the remainder of the paper, G denotes a minimal counterexample to the theorem. In particular, p=2. In view of Lemma 3, the following definition makes sense:

Definition

Let 𝒜 be the set of abelian subgroups of G of maximal order subject to containing Φ(G). Let A𝒜. The map xx2 induces a G-isomorphism A/Ω1(A)Φ(A), i.e. every subgroup of A/Ω1(A) is normal in G/Ω1(A). Let M=2(A)Ω1(A) and let HA be the preimage of CG/M(A/M) in G.

Note that HA={hG[b2,h]b8 for bA}. Also observe that HAAZ(HA), while Lemma 1 entails |G:HA|2.

Notation

We fix an element A of 𝒜 and let H=HA. Let Y=Ω1(A) and let Z=Ω1(Z(H)). The characteristic epimorphism GG/Y will be denoted by a bar. Reviving earlier notation, we let N=Ω1(Z(G)).

Since every subgroup of G of order 4 intersects Z(G) nontrivially, |Y:N|2. In particular, Z=Y or Z=N. We start off the proof by stating two slight, but useful, observations.

Lemma 1

Let cA. Then the following hold:

1. (a)

[c,NH(c)]Ω1(c2).

2. (b)

Let xGH. If [x,Y]=1, then c¯x¯=c¯-1.

## Proof.

Let hNH(c). The assertion being certainly true if o(c)4, we take o(c)>4. Now [c,h]=c2 would imply [c2,h]=c4, yet c4c8 because of o(c)>4. Thus [c,h]c4 and Lemma 5 yields (a).

Let xCG(Y)H. Let W be a complement of c^ in Y. Since x centralises Y, W is normalised by x. Since x2A, conjugation by x induces an automorphism of cW/W of order 2. Since xH, o(c)4 or conjugation by x is nontrivial on c¯. This proves (b). ∎

Lemma 2

Suppose that YZ. Let Y=Z×s. Then A=s×b×D, where bD[s,H]D, while every subgroup of D is normal in G.

## Proof.

Since Ω1(Φ(A))Z(G), the subgroup s has a complement C in A. Let C=c1××cr. Select b{c1,,cr} with the property that [s,H] projects onto b^ along the decomposition. Let E=cjbcj. For eE, we have e,sG{e,es}, and s,b therefore possesses a complement D in A that is the direct product of cyclic factors that are normal subgroups of G. In particular, DG. Given d1,d2 in D such that diG for i=1,2, we have s,d1d2G{sd1d2,d1d2}. Now Φ(sd1d2)D, while, for xH, we have [sd1d2,x][s,x](modD), which shows that sd1d2 cannot be normal in G. Hence d1d2 is. Induction on the length of d as a product of powers of the cj establishes the assertion. ∎

Lemma 3

[H,A]Z.

## Proof.

We start by establishing [A,H]Y. By way of contradiction, assume there is yH such that [A¯,y¯]1. Lemma 1 yields the existence of n satisfying [b¯,y¯]=b¯2n whenever bA. Note that the definition of H implies n>1. Applying Lemma 5 to the groups b¯,y¯, bA, we obtain expA=2n+2.

Let eAΩn+1(A). By Lemma 1, yNG(e), so there is aYe with [e,y]=e2na. If [a,y]=1, then ae,y and Lemma 5, applied to e,y,a/a, provides a contradiction. It follows that aZ(H). Applying Lemma 2 yields a direct decomposition A=a×b×C with b,C[a,y]C and all subgroups of C normal in G. Lemma 1 then yields [C¯,H¯]=1, and, according to Lemma 1, expC<2n+2 and H¯=y¯CH¯(A¯).

If a,b=a,d with dG, then Lemma 1 implies [b,y]Y[A,H]; hence the only maximal subgroup of a,b possibly normal in G is a,b2. Thus [a,G]=Ω1(b). Since A=bΩn+1(A) and [A,y,y]1=[Ωn+1(A),y,y] we may, perhaps upon replacing a by a different element of aN, assume that [b,y]=b2na.

Let xCH(a)A. Then [Y,x]=1, and, as seen above, this implies [b¯,x]=1. If [x,b]Z, then [xy,b]Z=b2nZ and Lemma 5 provides a contradiction when applied to the section xy,bΩ1(C)/Ω1(C). Hence CH(a)CH(b2). Thus CH(a), CH(b2) and the preimage of CH/Z(bZ) in H are one and the same – maximal – subgroup.

Let 1cC and let bb2 with o(b)=o(c). Since a,(bc)2G, there is ε{0,1} such that aεbcG. Since [aεb,CH(a)]=1, we obtain [C,CH(a)]=1, in particular [A,x]=[b,x]Z. Since x2CA(x), it follows that x2 is of the form aαb2βc for suitable integers α and β and cC. Thus we have (xb-β)2=aαc[x,b]β. Since 2n+2>8, there is b~b2CA(x) satisfying [x,b]b~2(modaΩ1(C)). It follows that xb=ub with u2aC. Since acG whenever cC, u2C. Next we show u2Φ(C). Suppose otherwise; then, since [u,C]=1, there is cC with (uc)2=1. Letting w=uc, we obtain that a,wE4 with a,wG=1.

Thus u2CΦ(C), in particular o(u)>2. Since o(b)>expC, 2n+2o(u). As a,u2G, a,uG=aεu with ε{0,1}. It follows that [b,u]=[A,u]=u^. Pick bb with o(b)=o(u). If o(u)=2n+2, then, since n>1, (ub)2n+1=u^b^[u,b](2n2)=u^b^, if o(u)<2n+2, then ub=bu anyway. Let w=bu and ε{0,1}. We have just seen that w^=u^b^, whence neither [a,G] nor u^=[u,b]=[aεw,b] is contained in w2. It follows that none of the maximal subgroups of a,w is normal in G.

We are done proving [A,H]Y. Assume there is xH with [A,x]Z. Then Y=a,Z with a[A,x]. We apply Lemma 2 to obtain a decomposition A=a×b×C, where [a,H]C and every subgroup of C is normal in G. Note that [a,C,x]Z, and [b,x]=az with zZ. Since both CH(a) and the preimage of CH/Z(bZ) in H are proper subgroups of H, we may additionally assume [a,x]1. This yields [b,x2]=[b,x,x]=[a,x]1, contradicting Φ(G)A. ∎

If every subgroup of A were normal in H, then |H:A|=2 by Lemmas 1 and 3. This justifies the following:

Definition

Let k be minimal with the property that there is B𝒜 in which not every subgroup of Ωk(B) is normal in HB.

We shall always assume A to be chosen to actually possess a cyclic subgroup of order 2k that is not normal in H (=HA). It may be useful to keep in mind that Ωk-1(A)Z(H) (see Lemma 6).

Lemma 4

expH4 and expH2 unless k2.

## Proof.

Let u,vH. According to Lemma 3, [u¯,v¯2]=1=[u¯,v¯]2[u¯,v¯,v¯]. It follows that [u¯,v¯] is an element of A¯ inverted by v¯, which is possible only if o([u,v])4.

Suppose that k3. Together with Lemma 6, the above implies that [u,v]Ω2(A)Z(H), i.e. [u,v]2=[u2,v]=[v2,u]u^v^. Thus o([u,v])>2 necessitates u^=v^. Suppose that o([u,v])=4. We have seen that u2Z(u,v)v2. Lemma 6 consequently yields o(w)2k+124 for w=u,v. Without loss, there are positive integers m and satisfying m2, u2=v2m and uv=u2. It follows that

(vu-2-m)2m=v2mu-2[u,v]2-m(2m2).

Thus (vu-2-m)2m=1 unless =m=2. Assume the latter. Let o(u)=2n. Then (vu-1)4=[u,v]2=u2n-1. Since n4,

(vu-1u2n-3)4=u2n[u2n-3,v]2=1.

Thus u,v=u,y with uy=1 and [u,y]=[u,v]. We have seen this to be incompatible with o([u,y])=o([u,v])=4. ∎

The cases k2 and k=1 have been allocated their own subsection each. It will soon turn out to be the case that k=2 is the most difficult case by far.

Lemma 5

k2.

### Proof.

Assume k3. By Lemmas 6, 9, and 4, HZ, |H:A|4, and |A:Z(H)|=2. It follows that GH. Since HZ, we have Φ(H)Z(H), while Lemma 1 says that y2CA(y)Ω2(A)Z(H) whenever yGH. Accordingly, Φ(G)Z(H). Let G=Hx and let A=aZ(H). Note that o(a)8 because of Lemma 6 and that [a,H]=[A,H] is a normal elementary abelian 4-subgroup of G.

Assume there is sYZ(G). Since [a,H]=[as,H], either [a,H]=s,a^, or s,aG=s,a2. In either case [x,s]=a^. Since [x,s] is independent of a, it follows that o(a)=2k=expA and Ωk-1(A)=Z(H), i.e. A=a×D with DΩk-1(Z(H))=Z(H). Let hHA. Since h2CA(h)=Z(H), we have o(h)2k and since k3, there is an element aa satisfying (ha)2D. Yet if h2D, then s,h2G, whence s,hG{h,sh}. It follows that H=Ag,h with g2Dh2 and gGh. Since b(a)=2, this forces the conclusion [a,g]=g^, [a,h]=h^ and gh=1, from which [g,h]=1 results. However, we have seen that Φ(G)Z(H), which, since |Z(H)h,g|>|A|, contradicts the fact that A𝒜. Accordingly,

(3.1)YZ(G).

By (3.1) and part (b) of Lemma 1, b¯x¯=b¯-1 whenever bA. Since expA2k>4, it follows that CG(Ω2(Φ(G)))=H, in particular H=HB whenever B𝒜. Since Z(H)h𝒜 whenever hHZ(H), this implies in conjunction with (3.1) that Ω2(H)Z(H), Y=Ω1(H)=Ω1(Z(G)) and hxh-1Y for all hH. Note that x2CA(x)Ω2(A)Z(H), so o(x)8 and x2Z(G).

Assuming that YH, we can find zY(H{a^}). Induction yields a normal subgroup B of G with |G:B|4, Φ(G)B, and B=z. If BH, then B𝒜, a contradiction. Hence G=HB. However, a2Φ(G)B and if yBA, then a2y=a-2, leaving o(a)=8 and z=a4 the only possibility. Yet za^. Accordingly,

(3.2)Y=H.

For a supplement Q of Z(H) in H, (3.2) yields H=Ω1(H)=Q, in particular QG. We let H=Z(H)a,b,c and Q=a,b,c. Note that Φ(H)Q=Φ(Q)=Z(Q)=Z(H)Q. Let expQ=2m and let uQ be of order 2m. As a cyclic subgroup of maximal order in Φ(Q), u2 has a complement D in Φ(Q). Note that uΦ(Q)=u×D. Since QY, every coset wuΦ(Q) with wQuΦ(Q) contains an element y with y2DY. Since we know that Ω2(H)Z(H), it follows that expD>2.

Let d be an element of D of maximal order and let D=d×E. Then

Qu2×d2EΦ(Q),

and there is vQ with v2u2,d2E, in particular vuΦ(Q). Assume that vu1. Let 2n=|v:uv|. Note that n2. Without loss v2n is a power u2; note that n. It follows that (vu-(2-n))2n=1. There is hence no loss in presuming vu=1. It follows that u2,d×E=(u2×v2)E, in particular |Φ(Q)|=|u2||d||E||u2||v2||E|. If o(v2)>o(d), then o(u2)>o(d) and v^=u^. Accordingly, Φ(Q)=u2×v2×E.

Finally, let wQu,vΦ(Q). Since QY, there is an element yu,v satisfying (wy)2u^,v^×E. Since o(u)8o(v) and Ω2(H)Z(H), there is an element zΩ2(u2,v2) such that (wyz)2E. Hence we may take w2E and it follows that wuv=w2u2,v2=1. Accordingly, H=Q=u^,v^,w^ and Q=uvw. Each of the elements u,v,w has order at least 8, in particular Ω2(Q)Φ(Q). Let P=Qx. We have established that P satisfies conditions (a), (b), and (c) of Lemma 11. Moreover, we know that o(x)8 and x2Z(G). Thus either |P:Q|=2 or o(x)=8, xQ=x4 and any element s of Ω1(Φ(P)Q)Q satisfies s=x2y with yΩ2(Q). Any such s has to centralise Q, which shows that condition (d) in the premise of Lemma 11 applies as well. So Lemma 11 yields a contradiction. ∎

Lemma 6

If k>1, then [u,v]H whenever u,vH.

### Proof.

Suppose otherwise. Due to Lemmas 6 and 4, k=2 and expH=4. Let u,vH be such that [u,v]H and let u,v=V. Since k>1, o([u,v])=4. Let {x,y} be a generating set of V. Then [x,y]Y=[u,v]Y, in particular [x,y]H. Since [x,y,x]=[x2,y][x,y]2x^[x,y]2, the product U=x[x,y] is a subgroup of H. Note that

H1(U)=Φ(U)=x2[x,y]2and[x,y]2=[u,v]2.

Next, yNG(xΦ(U)) – otherwise [x,y]x2,[x,y]2, i.e. [x,y]x2 and [x,y]H. Since x[x,y]Φ(U)=(xΦ(U))y, the only maximal subgroup of U available for normality in G is [x,y],x2. Note that this is possible only if o(x)>2.

If [x,y]2x2, then a power x of x2 satisfies ([x,y]x)2=1. Letting [x,y]x=w, k>1 implies wZ(H) and [x,y,H]=[x,H]x^[x,y]2, a contradiction. We summarise:

1. (1)

Let xVΦ(V). Then

o(x)>2,x[u,v]=1,[u,v,H][u,v]2,x^.

If [u,v,x]=1, then [x2,y]=[u,v]2=x^, contradicting (1). Thus:

1. (2)

CV([u,v])=Φ(V) and [u,v,H]=[u,v,V]=x^,[u,v]2 holds for every xVΦ(V).

By (2), VNG([u,v]). We choose notation such that uNG([u,v])v. Let o(u)=2n, noting that (1) implies n>1. From [u,v]2[u,v,u]=[u2,v]1 and Lemma 6, we deduce

1. (3)

n3 and [u2,v]=[u,v]2[u,v,u]=u^.

Furthermore, (1) implies that [v2,u]=[u,v]2[u,v,v]v^[u,v]=1, i.e.

1. (4)

[u,v2]=1 and [u,v,v]=[u,v]2.

Suppose that o(v)>o(u). Let sv2 be of order 2n. Then (4) implies that [u,s]=1. In particular, (1) may be used to exclude the possibility that u^=v^, for this would yield the existence of in us. Next, (3) implies that

[u2s2,v]=[(us)2,v]=u^us.

But n>1 and, as we have seen, us^=u^v^u^. It follows that 2no(v). Since (uv)2=u2v2[v,u]v=u2v2[u,v], u2v2[u,v]G. By (3), [u2v2[u,v],u]=[u,v,u]=u^[u,v]2. By (1), u^[u,v]2u^, and we obtain that if o(v)<2n, then n=3.

Assume that o(v)=2n. If u^v^, then (2) implies [u,v]2=u^v^, while (3) says [u,v,u]=v^. Thus [u2v2[u,v],u]=v^u4v4[u,v]2. Thus n>3 – otherwise o(u2v2[u,v])2 – and v^=u^v^, impossible. Accordingly, u^=v^. Let |v:uv|=2, observing that 2. Suppose that =2. Then, without loss, u4=v4, and (uv-1)4=[u,v]2, contradicting (1). If >3, then there is a generator v1 of v with the property that

Ω1(u2v2[u,v])=u2-1v12-1.

Combined with (3) and (4), this yields

[u2v2[u,v],v]=u^[u,v]2=u2-1v12-1.

Hence there also is a generator u1 of u satisfying

u12-1v12-1=[u,v]2.

Since >3, it follows that s=u12-2v12-2[u,v] is now an involution contained in [u,v]Z(H), contradicting k>1. Now =3 is the only option left: then there is a generator v1 of v satisfying

Ω1(u2v2[u,v])=u4v14[u,v]2,

i.e. u^[u,v]2=u4v14[u,v]2, i.e. u^=u4v14. Yet this implies v4u and clashes with =3.

The outcome of the preceding two paragraphs is that o(u)=8 and o(v)=4. If u4=v2, then o(uv)=8, while (uv)4=u4[u,v]2u4 and the pair (uv,v) satisfies every constraint previously placed on (u,v). We summarise:

1. (5)

o(u)=8, o(v)=4, and u and v may be chosen to satisfy uv=1.

We shall assume uv=1 from now on. By (2),

[u,v,H]=[u,v]2,u^=[u,v]2,v^=u4,v2.

Thus, (1) yields [u,v]2=u4v2.

Suppose that A has an element b of order 8. Then (vb)4=b41, while [u,vb][u,v]Z. Every assertion so far made about (u,v) equally applies to (u,vb), whence (5) delivers a contradiction. It follows that

1. (6)

expA=4, in particular G=H.

Together with k>1, (6) implies that YZ(G). Let B=u2,[u,v]. By (1) and (5), BC4×C4, and (6) yields the existence of a complement of B in A, to be called D. For dD, d,[u,v]G{d,[u,v]2,d[u,v],[u,v]2}; thus D may be taken to be a direct product of subgroups d with d,[u,v]2G. Let d,dD be such that d,[u,v]2Gd,[u,v]2. Then dd,[u,v]2G, because [dd[u,v],u]u4Φ(D),[u,v]2. Thus E,[u,v]2G for all subgroups E of D. Let dD. We have seen that some element w of {u,v,uv} is in NG(d). Now Φ(w,d)=w2,d2{u2,d2,v2,d2,u2v2[u,v],d2}, i.e. Φ(w,d)Y[u,v][G,w]Y=[G,wd]Y. This forces the conclusion d,w^G. Now w^{u4,v2} and [d,H]d2,[u,v]2d2,w^, i.e. dG.

1. (7)

A=u2, [u,v]×D with every subgroup of D normal in G.

Assuming expD=4, let dDΩ1(D). Due to [u,vd][u,v]Z, we may replace v by vd in (1)(4), to obtain [u,v,G](vd)2,[u,v]2v2,[u,v]2. If [v,d]=1, this yields d2v2,u4, a contradiction. By (7), this means that v inverts each element of D. It follows that Φ(v,[u,v]d)=v2,u4d2, and, since [u2,v]=[u2,v[u,v]d]=u4, we obtain [[u,v]d,G]v2,u4d2. Now [d,uv]=1 is impossible, as it would imply

[[u,v]d,uv]=[u,v,u][u,v,v]=u4.

Thus [D,u]=1 – according to (7), CG(D)G. Let u2d,[u,v]=W. From [u2d[u,v],u]=[u,v,u]=v2u4d2,u4v2=Φ(W), we infer that [u2d,G]Φ(W). Since both u2 and d are normal in G, this yields

[u2d,G]u4d2,u4v2u4,d2=u4d2,

i.e. u2dG. Lemma 1 yields CG(u2)=CG(D). By (1)(4),

G=uNG([u,v])=u,vCG([u,v]).

Since o(v)=4 and G=H, we have

(uv)2u2[u,v](modZ(G)),

i.e. u2[u,v]G. Thus we obtain that CG([u,v])CG(u2), and

CG([u,v])=CG([u,v],u2,D)=CG(A)=A.

Accordingly, |G:A|=4.

We have seen that expD=2, i.e. A=BZ. By (1), BG, according to (3), [u2,G]=u4=[u2,v] and according to (3) and (4), [u,v,G]=u4,v2=[u,v,u,v]. If sG satisfies [u,v,s]=1[u2,s], then [u2[u,v],s]=u4u4[u,v]2; however, u2[u,v]G. Consequently, G=u,vCG(u2,[u,v])=u,vA, a contradiction that finishes this proof. ∎

Lemma 7

If k=2, then HZ.

### Proof.

Assume otherwise. Pick xH with x2Z(H) and let yHCH(x2). By Lemma 6, Ω1(Φ(G))Z and [x,y]H by Lemma 6, so [x2,y]=x^=[x,y]2[x,y,x]=[x,y]2. In particular, [x,y,x]=1 and Lemmas 1 and 6 yield [H,x,x]=1. The assumption [x,y,y]=1, however, implies that x^=[x,y2]=y^. If this is true, we may assume x2m=y2, where m2 and 2=|y/(xy)|. Either m==2, or w=yx-2m- satisfies wx=1, [x,w]=[x,y], and [x,w,w]=1, a contradiction. If m==2, we have o(x2y-2)=2; yet [y,x2y-2]1 and k>1 is contradicted.

Accordingly, [x,y,y]1; by Lemma 6, this means [x,y]y=[x,y]-1. If o(x)>8, then there is xx4<Z(H) such that o([x,y]x)=2, and k>1 forces the conclusion [x,y,y]=1. Thus o(x)=8. Suppose there is uH such that [u2,x]1. Then, as we have seen, o(u)=8 and u4=x4=[x,u]2. It follows that z=u2x2 is an involution with [z,x]1. Finally, Lemmas 1 and 6 say that [v,y]y=[v,y]-1 whenever vH, implying y2Z(H). We summarise:

1. (1)

If yHCH(Φ(H)), then y2Z(H). Any element v of H with v2Z(H) has order 8 and centralises Φ(H).

Let Φ(H)Z(H)=T and let K=CH(Φ(H)). For all g,hK we have 1=[g2,h]=[g,h]2, i.e. KZ. By (1), the Hughes subgroup of H/T is contained in K/T, whence KH and every element of K is inverted by y modulo T. Suppose there is uK with o(u)>8. By (1), u2Z(H), whence [w,u2]=[w,u]2=1 whenever wH, in particular (xu)2=x2u2[x,u]x2Z(H). However, (xu)8=u81, contradicting (1). Hence expK=8. Let L=Ω2(K), noting that, since KZ, we have expL4. If vKxL, then v4x4 and (vx)2v2x2(modZ). It follows that y normalises every subgroup of v2,x2=v2×x2. Accordingly, [v,y]2=[v2,y]=v41 and Z(H/Z)=L/Z. Applying Lemma 3, we obtain AL. We summarise:

1. (2)

L/Z=Z(H/Z), in particular expA=4 and G=H.

From KH=G and Lemma 9, we further infer:

1. (3)

|K:A|=4 and either |A:Z(K)|=2, or A=c,dZ(K) with bK(c)=bK(d)=1. In particular, |[A,K]|4 and |K|8.

Suppose there is an element zYK. Via induction, G has a normal subgroup B of index 4 satisfying B=z. Unless B is abelian, BK. Given tBK, we have B¯[t¯,x¯]=[y¯,x¯], and z=[x,t,t]=[x,y,y]=x4. It is, however, possible to choose z distinct from x4. Thus:

1. (4)

Y=K.

We consider the possibility that |A:Z(K)|=2 and |Y|=8. First suppose that |K:L|=4, i.e. L=A. Letting A=vZ(K), we obtain K=v,w,xZ(K) with o(w)=o(x)=8 and wx=1. Since o(v)=4 and “v2w4,x4” would necessitate the existence of involutions in AZ, (vw)x=1 and K=wxvZ(K). Let Q=x,w,v. By (4), Y=x4,w4,v2=Ω1(Q). As y inverts the elements of w2×x2, Z(G)Q=Y. Since expA=4, o(y)8. Let P=Qy. We have already noted that P satisfies (a)–(c) in the premise of Lemma 11. Suppose that Φ(P)Q>Q. It follows that o(y)=8 and any tΩ1(Φ(P)Q)Q must be of the form t=y2q with qΩ2(Q). Now Ω2(Q)LQ=AQ so that tA and k=2 implies [Q,t]=1. Hence P satisfies the hypothesis of Lemma 11 and is not core-2. Thus:

1. (5)

If Z(K)|=2 and |Y|=8, then |K:L|=2.

We continue to work from the assumptions |A:Z(K)|=2 and |Y|=8. By (5), we may write L=a,vZ(K). Letting Q=a,v,x, we have K=Z(K)Q and Y=Q. Let G=Ky. Since x¯y=x¯-1 and [L,y]Y, yNG(Q). From (1) and (2) we know that Φ(G)Z(K), so if wLZ(K), then w,Z(K)𝒜 and k>1 implies Ω1(w,Z(K))Z(G). Suppose that o(y)=8. Then we have y2Z(G)L. Any involution in Qy2Q is equal to a product y2q with qΩ2(Q), in particular qLy2 and y2qΩ1(L)Z(G). Letting P=Qy, P therefore satisfies the hypothesis of Lemma 12 and is not core-2. Consequently:

1. (6)

If |A:Z(K)|=2, then |Y|4.

We stick to the hypothesis |A:Z(K)|=2. As |K:Z(K)|=23, it follows that Z(K/W)Z(K)/W whenever WY. By (6), |K|4, and there is vK with bK(v)=1. For such a v, though, CK(v) is an abelian subgroup of K containing Φ(G) and of greater order than A. So:

1. (7)

|A:Z(K)|=4.

By (3), |[A,K]|4. Whenever K=x,uA, we have Y=[A,K][x,u]. Let W be a maximal subgroup of Y not containing x4. By Lemma 7, we have that |K/W:Z(K/W)|4 and A/WZ(K/W) has index at most 2 in A/W.

Assume that |[A,K]|=4. From (2) and k=2, A=a,bZ(K), where b(a)=b(b)=1 and b(ab)=2. By Lemma 6, k>1 implies a,bC4×C4. Note that this excludes K=[A,K]. Indeed, if that was the case, then (4) would imply Y=a2×b2 and there would be cAZ(K) satisfying c2=x4. However, this would make cx2 an involution outside Z(G).

Let [a,K]=s and [b,K]=t. We have seen |Y|=8. So set Y=s,t,r. If x4s,t, then W=x4s,st is a maximal subgroup of Y not containing x4 and satisfying |A/W:A/WZ(K/W)|=4, a contradiction. This places x4 inside [A,K]. Set [A,K]=x4,z. Lemma 8 says that

|A/z,r:Z(K/z,r)A/z,r|2,

which implies [a,K]z or [b,K]z. Thus {s,t}={z,x4z}. However, Lemma 8 also says that |A/W:A/WZ(K/W)|2 for at least two of the three maximal subgroups W of Y that do contain x4, namely s,x4, r,x4, and rs,x4. This is not the case.

1. (8)

|A:Z(K)|=4 and |[A,K]|=2.

By (4) and (7), YE4. The fact that a,bC4×C4 remains unchanged, and we arrive at a contradiction as in the preceding paragraph. The proof is done. ∎

Lemma 8

k=1.

### Proof.

Assume k>1. The proof follows similar lines to those of Lemmas 5 and 7. Lemmas 9, 5, and 7 say that:

1. (1)

k=2, HZ, GH, |H:A|=4, and |A:Z(H)|{2,4}.

From GH we deduce expA>4. Let G=yH.

If sYZ(G) and vH, then either [s,y]=v^ or v,sG{v,vs}, the latter forcing vH. Let C4aA with aH. We have seen that a2=[s,y]. Since k=2 and Φ(A)Z(H), it follows that a22(A) and a has a complement in A. Let A=a×D and let b be an element of maximal order in D. Since GH, o(b)>4. Let Ω2(b)=b, noting that bZ(H). Each of the subgroups b,ab, and ab is normal in H, in particular [a,H]=[ab,H]=a2b^. Let vHCH(a). Then [v,ab]=a2b^[v,b]. It is therefore impossible that both [v,b] and [v,ab] are elements of b^. Accordingly:

1. (2)

YZ(G).

Suppose that Y>H. Let b be an element of A of order 8. Letting zHY, induction yields a normal subgroup B of G with Φ(G)B, |G:B|=4 and Bz. Since HB is abelian, BH. In conjunction with Lemma 1 (b), (2) implies that, for cA, there is zcY with cy=c-1zc. In particular, we have b2x=b-2 for xBH. Now b2B implies that z=b4; yet if YH, then YH{b4}. Hence:

1. (3)

Y=H.

Since b2B𝒜B and b2x=b-2 whenever xH, Lemma 3 says that:

1. (4)

For every B𝒜, CG(b2)=H=HB.

Suppose that |A:Z(H)|=2 and |Y|=8. By condition (3), bH(h)=2 holds for hHZ(H). Since H thus is void of elements of breadth 1, it follows that x2Ω2(Z(H)) whenever xGH. Accordingly, Φ(G)Z(H) and CH(h)=Z(H),h𝒜 whenever hHZ(H). It now follows from (4), from (b) in Lemma 1 and from the definition of k that Y=Ω1(H) and that, for hH, there is an element zhY with hy=h-1zh. Let H=u,vA, where o(u)=2n=expH>4. Because of k>1, v2u4; if u2=v2, then (uv-1)2=[u,v], and, since the coset uv-1Φ(H) is not permitted to contain involutions, [u,v]u. In any case there is wvu with u,v=u,w and uw=1. We replace v by w if necessary.

Since k=2, it follows that A=aZ(H), where o(a)=4. If o(v)>4, then k>1 forces auv=1. Let Q=a,u,v. By (2) and (4), we have Y=Q=Ω1(Q), in particular QG. Let P=Qy. If Φ(P)Q, then o(y)=8 and an involution in Φ(P)QQ is equal to qy2 is for some qQ. If qZ(H), then qy2,Z(H)𝒜, contradicting k>1. Accordingly, [qy2,Q]=1. If o(v)>4, then P satisfies Lemma 11 with x1=u, x2=v, x3=a, if o(v)=4, then P fits the criteria of Lemma 12. Thus G is not core-2. Consequently:

1. (5)

|Y|<8 or |A:Z(H)|>2.

Suppose that |Y|=8. By (5), we have |A:Z(H)|>2. By Lemma 9 and since |H:[A,H]|=2, [A,H]=s,tE4, and H=u,v,a,bZ(H), where a,bZ(H)=A, [u,a]=s, [u,b]=1, [v,a]=1, [v,b]=t, and Y=[A,H]×[u,v]. Let B=u,bZ(H); then B is a maximal normal abelian subgroup of H, though not necessarily an element of 𝒜. Lemma 9 nevertheless yields the existence of w{u,ub} with bH(w)=1, i.e. [w,H]=s. However, this implies that Y=[A,H], a contradiction. Hence

1. (6)

|Y|<8.

Assume that |A:Z(H)|=2, and let aZ(H)=A and H=Au,v. Since Y=[a,u],[a,v], either [u,v]=1 or there is an element w{u,v,uv} such that [u,v]=[a,w]. In either case, H=ACH(w) for some wHA. No element of A has breadth 1 in H, whence Φ(G)Z(H)CH(w). Thus we have CH(w)𝒜, a contradiction. Hence

1. (7)

|A:Z(H)|=4.

Let a be some element of A of order 4 with aH. As previously noted, a has a complement in A of exponent greater than 4. Let A=a×D. By (6) and since d(A)=d(Ω1(A)), A=a×d with o(d)=expA. If xGH, then x2Ω2(A), whence Φ(G)a,d2. According to Lemma 9, we may assume bH(d)=