On Finite Quasi-Core-p p-Groups

Abstract

Given a positive integer n, a finite group G is called quasi-core-n if $⟨x⟩/{⟨x⟩}_G$ has order at most n for any element x in G, where ${⟨x⟩}_G$ is the normal core of $⟨x⟩$ in G. In this paper, we investigate the structure of finite quasi-core-p p-groups. We prove that if the nilpotency class of a quasi-core-p p-group is $p+m$, then the exponent of its commutator subgroup cannot exceed $p^{m+1}$, where p is an odd prime and m is non-negative. If $p=3$, we prove that every quasi-core-3 3-group has nilpotency class at most 5 and its commutator subgroup is of exponent at most 9. We also show that the Frattini subgroup of a quasi-core-2 2-group is abelian.

1. Introduction

Let G be a group and H is a subgroup of G. Then $H_G$ is the normal core of H in G, where $H_G={\bigcap }_{g\in G}{g}^{-1}Hg$ is the largest normal subgroup of G contained in H. A group G is called core-n if $|H/H_G|\le n$ for every subgroup H of G, where n is a positive integer. Buckley, Lennox, Neumaan, Smith and Wiegold investigated the core-n groups in [1]. They show that every locally finite group G with $H/H_G$ finite for all subgroups H is core-n for some n. Moreover, G has an abelian normal subgroup of index bounded in terms of n only. In [2], Lennox, Smith and Wiegold show that, for $p\ne 2$, a core-p p-group is nilpotent of class at most 3 and has an abelian normal subgroup of index at most $p^5$. Furthermore, Cutolo, Khukhro, Lennox, Wiegold, Rinauro and Smith [3] prove that a core-p p-group G has a normal abelian subgroup whose index in G is at most $p^2$ if $p\ne 2$. Furthermore, if $p=2$, Cutolo, Smith and Wiegold [4] prove that every core-2 2-group has an abelian subgroup of index at most 16. As a deepening of research in this area, it is interesting to study the following question.

How about the structure of a p-group G in which ${|\langle x\rangle /\langle x\rangle }_G|\le p$, for any $x\in G$?

In this paper we hope to investigate the structure of a p-group G in which ${|\langle x\rangle /\langle x\rangle }_G|\le p$, for any $x\in G$. For convenience, we call this kind of p-groups quasi-core-p p-groups.

2. Preliminaries

For convenience, we first recall some notations.

Let G be a p-group. We use $d\left(G\right)$ and $c\left(G\right)$ to denote the minimal number of generators and the nilpotency class of G respectively. We use $C_{p^m}$ to denote the cyclic group of order $p^m$. Let $G_n=\langle [g_1,g_2,\dots ,g_n]|g_i\in G\rangle$. If H and K are groups, then $H\times K$ denotes a product of H and K. For other notations the reader is referred to [5].

Lemma 1.

([6], Section Appendix 1, Theorem A.1.4) Let G be a p-group and $x,y\in G$.

1.

${\left(xy\right)}^p\equiv x^p{y}^p$ (mod ${\mho }_1\left(G^{\prime }\right)G_p$).

2.

$[x^p,y]\equiv {[x,y]}^p$ (mod ${\mho }_1\left(N^{\prime }\right)N_p$), where $N=\langle x,[x,y]\rangle$.

Lemma 2.

([7], Lemma 2.2) Suppose that G is a finite non-abelian p-group. Then the following conditions are equivalent.

1.

G is minimal non-abelian;

2.

$d\left(G\right)=2$ and $|G^{\prime }|=p$;

3.

$d\left(G\right)=2$ and $\mathsf{\Phi }\left(G\right)=Z\left(G\right)$.

Lemma 3.

([8], Theorem) Let p be a prime and $d,e$ positive integers. A regular d-generator metabelian p-group G whose commutator subgroup has exponent $p^e$ has nilpotency class at most $e(p-2)+1$ unless $e=1,d>2,p>2$ when the class can be p. These bounds are best possible.

Lemma 4.

([9], Theorem 2) Let G be a metacyclic 2-group. Then G has one presentation of the following three kinds:

1.

G has a cyclic maximal subgroup.

2.

Ordinary metacyclic 2-groups $G=\langle a,b|a^{2^{r+s+u}}=1,b^{2^{r+s+t}}=a^{2^{r+s}},a^b=a^{1+2^r}\rangle$, where $r,s,t,u$ are non-negative integers with $r\ge 2$ and $u\le r$.

3.

Exceptional metacyclic 2-groups $G=\langle a,b|a^{2^{r+s+v+t^{\prime }+u}}=1,b^{2^{r+s+t}}=a^{2^{r+s+v+t^{\prime }}},a^b=a^{-1+2^{r+v}}\rangle$, where $r,s,v,t,t^{\prime },u$ are non-negative integers with $r\ge 2,t^{\prime }\le r,u\le 1,t{t}^{\prime }=sv=tv=0,$ and if $t^{\prime }\ge r-1$, then $u=0$.

Groups of different types or of the same type but with different values of parameters are not isomorphic to each other.

Lemma 5.

([5], Theorem 10.3) Let G be a regular 3-group. Then $G^{\prime }$ is abelian.

Lemma 6.

Let G be a quasi-core-p p-group. If H is a subgroup of G and N is a normal subgroup of G, then H and $G/N$ are quasi-core-pp-groups.

Proof. 

The proof of the lemma comes immediately from the definition of quasi-core-p p-groups. □

Lemma 7.

Let G be a p-group. Then G is quasi-core-p if and only if $\langle x^p\rangle ⊴G$, for any element x in G.

Proof. 

Obviously, G is quasi-core-p if and only if ${|\langle x\rangle }_G/\langle x^p\rangle |\le p$, for any $x\in G$, and this holds if and only if $\langle x^p\rangle ⊴G$, for any element x in G. □

Lemma 8.

Let G be a quasi-core-p p-group. Then $[G^{\prime },{\mho }_1\left(G\right)]=1$.

Proof. 

For any $x\in G$, according to Lemma 7, we see $\langle x^p\rangle ⊴G$. Thus $G/C_G\left(x^p\right)$ is abelian and so $G^{\prime }\le C_G\left(x^p\right)$, which implies $[G^{\prime },{\mho }_1\left(G\right)]=1$. □

3. Quasi-Core-p p-Groups with p > 2

In this section we investigate the quasi-core-p p-groups for $p>2$.

Theorem 1.

Let G be a quasi-core-p p-group and $p>2$. If $G^{\prime }$ is cyclic, then $|G^{\prime }|\le p$.

Proof. 

Suppose the result is not true and G is a counterexample of minimal order. Then there exist $a,b\in G$ such that $o\left([a,b]\right)\ge p^2$. Thus we may assume $G=\langle a,b\rangle$, $[a,b]=c$ and $L=\langle a,c\rangle$. Since G is regular, we may assume $\langle a\rangle \cap \langle b\rangle =1$. By Lemma 1, we see $[a^p,b]=c^{p}x$, where $x\in {\mho }_1\left(L^{\prime }\right)L_p$. Since $L<G$, ${\mho }_1\left(L^{\prime }\right)L_p=1$. So $x=1$ and $[a^p,b]=c^p$. Similarly, $[a,b^p]=c^p$. It follows from Lemma 7 that $c^p\in \langle a\rangle \cap \langle b\rangle =1$, in contradiction to the hypothesis. Thus the theorem is true. □

Corollary 1.

Let G be a quasi-core-p p-group with $p>2$. Then ${\mho }_1\left(G\right)$ is abelian and ${\mho }_2\left(G\right)\le Z\left(G\right)$.

Proof. 

For any $a,b\in G$, we assume $H=\langle a^p,b\rangle$. By the hypotheses, we see $\langle a^p\rangle ⊴G$ and so H is metacyclic. By Theorem 1, $|H^{\prime }|\le p$ and so H is abelian or minimal non-abelian. Thus ${\mho }_1\left(H\right)\le \mathsf{\Phi }\left(H\right)\le Z\left(H\right)$ by Lemma 2. It follows that $[a^{p^2},b]=[a^p,b^p]=1$, which implies ${\mho }_1\left(G\right)$ is abelian and ${\mho }_2\left(G\right)\le Z\left(G\right)$. □

Corollary 2.

Let G be a quasi-core-p p-group with $p>2$. Then $G/C_G\left(a^p\right)\lesssim C_p$, for any $a\in G$.

Proof. 

We may assume $a^p\notin Z\left(G\right)$ and $o\left(a\right)=p^n$. Then $n\ge 3$ and there exists an element $b\in G$ such that $b\notin C_G\left(a^p\right)$. By Theorem 1, we may assume $[a^p,b]=a^{p^{n-1}}$. Take $x\in G\backslash C_G\left(a^p\right)$. Assume $[a^p,x]=a^{i{p}^{n-1}}$, where $(i,p)=1$. Then $[a^p,b^{-i}x]=1$, which implies $x\in C_G\left(a^p\right)\langle b\rangle$ and so $G=C_G\left(a^p\right)\langle b\rangle$. It follows from $b^p\in C_G\left(a^p\right)$ that $G/C_G\left(a^p\right)\lesssim C_p$. □

Corollary 3.

Let G be a quasi-core-p p-group with $p>2$. If $c(G/{\mho }_1\left(G\right))\le n$, then $c\left(G\right)\le n+2$.

Proof. 

Set $\overline{G}=G/{\mho }_1\left(G\right)$. Then ${\overline{G}}_{n+1}=\overline{1}$ and so $G_{n+1}\le {\mho }_1\left(G\right)$. It follows from Theorem 1 that $[G_{n+1},G]\le [{\mho }_1\left(G\right),G]\le Z\left(G\right)$, which implies $c\left(G\right)\le n+2$. □

According to Lemma 3 and Corollary 3, we get the following theorem.

Theorem 2.

Suppose that G is a quasi-core-p p-group and $G^{\prime }$ is abelian with $p>2$. If $d\left(G\right)=2$, then $c\left(G\right)\le p+1$. If $d\left(G\right)>2$, then $c\left(G\right)\le p+2$.

If $p=3$, then, according to Lemma 5 and Corollary 3, we get the theorem below.

Theorem 3.

Let G be a quasi-core-3 3-group. If $d\left(G\right)=2$, then $c\left(G\right)\le 4$. If $d\left(G\right)>2$, then $c\left(G\right)\le 5$.

Theorem 4.

Let G be a quasi-core-3 3-group with $d\left(G\right)=2$. Then $\mathsf{\Phi }\left(G\right)$ is abelian.

Proof. 

We may assume $G=\langle x,y\rangle$ and $[x,y]=z$. Then $G^{\prime }=\langle z,[z,g]|g\in G\rangle$. For any $g_1,g_2\in G$, it follows from Theorem 3 that $[z,[z,g]]\in [G_2,G_3]=1$ and $[[z,g_1],[z,g_2]]=1$, which implies $G^{\prime }$ is abelian. So, according to Lemma 8 and Corollary 1, $\mathsf{\Phi }\left(G\right)$ is abelian. □

Now, we investigate the exponent of commutator subgroups of the quasi-core-p p-groups.

Lemma 9.

Let G be a quasi-core-p p-group with $G_{p+1}=1$ and $p>2$. Then $exp\left(G^{\prime }\right)\le p$.

Proof. 

Suppose the result is not true and G is a counterexample of minimal order. For any $g_1,g_2\in G^{\prime }$, let $H=\langle g_1,g_2\rangle$. By Lemma 1, ${\left(g_1{g}_2\right)}^p=g_1^p{g}_2^{p}x$, where $x\in {\mho }_1\left(H^{\prime }\right)H_p$. Since $c\left(H\right)<c\left(G\right)$, $H_p=1$. By induction, $exp\left(H^{\prime }\right)\le p$ and so $exp\left({\mho }_1\left(H^{\prime }\right)\right)=1$. Thus $x=1$. It follows that there exist $a,b\in G$ such that $o\left(\right[a,b\left]\right)>p$ and $exp\left(G_3\right)\le p$.

By induction, we may assume $G=\langle a,b\rangle$, $[a,b]=c$ and $L=\langle a,c\rangle$. Then, according to Lemma 1, we see $[a^p,b]=c^{p}y$, where $y\in {\mho }_1\left(L^{\prime }\right)L_p$. Since $c\left(L\right)<c\left(G\right)$, $L_p=1$ and $exp\left(L^{\prime }\right)\le p$. Thus $y=1$. Since G is a quasi-core-p p-group, $\langle a^p\rangle ⊴G$. So $c^p\in \langle a\rangle$. It follows from Theorem 1 that $o\left(c\right)=p^2$. Similarly, we see $c^p\in \langle b\rangle$.

Without loss of generality, we may assume $\langle a\rangle \cap \langle b\rangle =\langle a^{p^s}\rangle =\langle b^{p^t}\rangle$, $a^{p^s}=b^{p^t}$ and $s\ge t\ge 2$. If $s>t$, then, by letting $b_1=a^{-p^{s-t}}b$, we see $[a,b_1^p]=c^p$ and $c^p\notin \langle b_1^p\rangle$, in contradiction to the hypothesis. So $s=t$. Let $b_2=a{b}^{-1}$. Then, by Lemma 1, we see $b_2^p=a^p{b}^{-p}z$, where $z\in {\mho }_1\left(G^{\prime }\right)G_p$. Since $G^{\prime }=\langle c,[c,g]|g\in G\rangle$, we see ${\mho }_1\left(G^{\prime }\right)=\langle c^p\rangle$. Then ${\mho }_1\left(G^{\prime }\right)G_p\le Z\left(G\right)$ and $exp\left({\mho }_1\left(G^{\prime }\right)G_p\right)\le p$. Thus $o\left(z\right)\le p$ and $o\left(b_2\right)=p^s$. Noticing that $[a,b_2^p]=c^p$, we see $c^p\in \langle b_2^p\rangle$. If $s=2$, then $\langle c^p\rangle =\langle b_2^p\rangle$, which implies $b_2^p=a^p{b}^{-p}z\in Z\left(G\right)$, a contradiction. If $s>2$, then $\langle c^p\rangle =\langle b_2^{p^{s-1}}\rangle =\langle a^{p^{s-1}}{b}^{p^{s-1}}\rangle$. It follows that $\langle a\rangle \cap \langle b\rangle =\langle a^{p^{s-1}}\rangle$, another contradiction. □

Corollary 4.

Let G be a quasi-core-p p-group and $exp\left(G_{p+1}\right)=p^n$ with $p>2$ and $n\ge 0$. Then $exp\left(G^{\prime }\right)\le p^{n+1}$.

Proof. 

If $n=0$, then the conclusion holds by Lemma 9. Thus we may assume $n\ge 1$. Set $\overline{G}=G/G_{p+1}$. Then ${\overline{G}}_{p+1}=\overline{G_{p+1}}=\overline{1}$. It follows from Lemma 9 that $exp({\overline{G}}^{\prime })\le p$, which implies $exp(G^{\prime })\le p^{n+1}$. □

Corollary 5.

Let G be a quasi-core-p p-group and $c\left(G\right)=p+n$ with $p>2$ and $n\ge 0$. Then $exp\left(G^{\prime }\right)\le p^{n+1}$.

Proof. 

If $n=0$, then the conclusion holds by Lemma 9. Thus we assume $n\ge 1$. Set $\overline{G}=G/G_{p+n}$. Then $c\left(\overline{G}\right)=p+n-1$. By induction, we see $exp({\overline{G}}^{\prime })\le p^n$. Since $G_{p+n}=[G_{p+n-1},G]\le Z\left(G\right)$, by Lemma 9, we see $exp\left(G_{p+n}\right)\le p$. It follows that $exp\left(G^{\prime }\right)\le p^{n+1}$. □

Theorem 5.

Let G be a quasi-core-p p-group with $p>2$. If $G^{\prime }$ is abelian, then $exp\left(G^{\prime }\right)\le p^2$ and $exp\left(G_3\right)\le p$.

Proof. 

Suppose that the result is not true and G is a counterexample of minimal order. Then there exist $a,b\in G$ such that $o\left([a,b]\right)\ge p^3$. We may assume $G=\langle a,b\rangle$, $[a,b]=c$ and $L=\langle a,c\rangle$. By Lemma 1, $[a^p,b]=c^{p}x$, where $x\in {\mho }_1\left(L^{\prime }\right)L_p$. By induction, $exp\left(L^{\prime }\right)\le p^2$ and so $exp\left({\mho }_1\left(L^{\prime }\right)\right)\le p$. On the other hand, since ${[a,c]}^p\in Z\left(G\right)$, it is easy to see that $exp\left(L_3\right)\le p$. So $o\left(x\right)\le p$. According to Theorem 1, we see $o\left(c^{p}x\right)=p$, which implies $o\left(c\right)\le p^2$, in contradiction to the hypothesis. So $exp\left(G^{\prime }\right)\le p^2$. Thus, for any $g\in G^{\prime }$, we see $g^p\in Z\left(G\right)$. It follows that $exp\left(G_3\right)\le p$. □

Theorem 6.

Let G be a quasi-core-3 3-group. Then $exp\left(G^{\prime }\right)\le 9$ and $exp\left(G_3\right)\le 3$.

Proof. 

Take $a,b\in G^{\prime }$ with $o\left(a\right)\le 9$ and $o\left(b\right)\le 9$. Let $K=\langle a,b\rangle$. Then, by Lemma 1, ${\left(ab\right)}^3=a^3{b}^{3}c$, where $c\in {\mho }_1\left(K^{\prime }\right)K_3$. Since $K^{\prime }\le G_4$, we see $c\left(K\right)\le 3$ by Theorem 3. Thus $exp\left(K^{\prime }\right)\le 3$ by Corollary 5, which implies $o\left(c\right)\le 3$. It follows that ${\left(ab\right)}^9=a^9{b}^9=1$. So, we may assume $d\left(G\right)=2$. According to Corollary 5 and Theorem 3, we see $exp\left(G^{\prime }\right)\le 9$.

Take $x\in G^{\prime }$ and $y\in G$. Then $o\left(x\right)\le 9$ and so $\langle x^3\rangle \le Z\left(G\right)$. Assume $[x,y]=z$ and $L=\langle x,z\rangle$. Then, by Lemma 1, $1=[x^3,y]=z^{3}w$, where $w\in {\mho }_1\left(L^{\prime }\right)L_3$. Since $L^{\prime }\le G_5\le Z\left(G\right)$, by Lemma 9, we see ${\mho }_1\left(L^{\prime }\right)L_3=1$. It follows that $z^3=1$. For any $g,h\in G_3$ with $o\left(g\right)\le 3$ and $o\left(h\right)\le 3$, then, by Theorem 3, we see $[g,h]\in G_6=1$. So $o\left(gh\right)\le 3$, which implies $exp\left(G_3\right)\le 3$. □

4. Quasi-Core-2 2-Groups

In this section, we investigate the quasi-core-2 2-groups.

Lemma 10.

Let $G=\langle a,b\rangle$ be a non-abelian metacyclic quasi-core-2 2-group with $\langle a\rangle ⊴G$ and $o\left(a\right)=2^n$. Then $[a,b]=a^{2^{n-1}}$, $a^{-2}$ or $a^{-2+2^{n-1}}$.

Proof. 

Since G is a non-abelian metacyclic 2-group, we see $n\ge 2$ and G is one of the groups listed in Lemma 4.

If G is a group listed in (1) in Lemma 4, then the conclusion holds by the classification of p-groups with a cyclic maximal subgroup.

If G is a group listed in (2) in Lemma 4, then $G=\langle a,b|a^{2^{r+s+u}}=1,b^{2^{r+s+t}}=a^{2^{r+s}},[a,b]=a^{2^r}\rangle$ with $r\ge 2$ and $u\le r$. We may assume $s+u\ge 2$. By calculation, it is easy to see $\langle [a,b^2]\rangle =\langle a^{2^{r+1}}\rangle$. Since G is a quasi-core-2 2-group, we see $a^{2^{r+1}}\in \langle b^2\rangle$, which implies $s\le 1$. Let $a_1=a{b}^{-2^t}$. If $s=0$, then $\langle a_1\rangle \cap \langle a\rangle$ = 1. It follows from G is quasi-core-2 that $a_1^2\in Z\left(G\right)$, which implies $a^2\in Z\left(G\right)$. However, it is impossible. If $s=1$, then $o\left(a_1\right)=2^{r+1}$ and $\langle [a_1^2,b]\rangle =\langle a^{2^{r+1}}\rangle \le \langle a_1^2\rangle$. It follows that $\langle a^{2^{r+u}}\rangle =\langle a_1^{2^r}\rangle$, which implies $b^{2^{r+t}}\in \langle a\rangle$. It is also impossible. So $s+u=1$ and therefore $[a,b]=a^{2^{n-1}}$.

If G is of type (3) in Lemma 4, then $G=\langle a,b|a^{2^{r+s+v+t^{\prime }+u}}=1,b^{2^{r+s+t}}=a^{2^{r+s+v+t^{\prime }}},[a,b]=a^{-2+2^{r+v}}\rangle$ with $r\ge 2$ and $u\le 1$. It follows from $[a,b^2]\in \langle b\rangle$ that $s+t^{\prime }\le 1$ and so $s+t^{\prime }+u\le 2$. We may assume $s+t^{\prime }+u=2$ and so $u=s+t^{\prime }=1$. Then $b^{2^{r+s+t}}=a^{2^{n-1}}$ and $[a,b]=a^{-2+2^{n-2}}$. We assume $o\left(b\right)=2^m$. If $r+s+t=2$, then, since ${\left(ba\right)}^2=b^2{a}^{2^{n-2}}$, we see $o\left(ba\right)=4$. On the other hand, $[a,{(ba)}^2]=a^{2^{n-1}}$. So, by the hypotheses, we see $a^{2^{n-1}}\in \langle {(ba)}^2\rangle =\langle b^2{a}^{2^{n-2}}\rangle$, a contradiction. If $r+s+t\ge 3$, then $o(b^{2^{m-3}}{a}^{2^{n-3}})=4$ and $[b,{(b^{2^{m-3}}{a}^{2^{n-3}})}^2]=a^{2^{n-1}}$. Thus $a^{2^{n-1}}\in \langle {(b^{2^{m-3}}{a}^{2^{n-3}})}^2\rangle =\langle b^{2^{m-2}}{a}^{2^{n-2}}\rangle$, another contradiction. So the conclusion holds. □

Corollary 6.

Let G be a quasi-core-2 2-group. Then $\mathsf{\Phi }\left(G\right)$ is abelian and ${\mho }_2\left(G\right)\le G^{\prime }Z\left(G\right)$.

Proof. 

For any $a,b\in G$, we may assume $H=\langle a^2,b\rangle$ is not abelian and $o\left(a\right)=2^n$. By the hypotheses, we see $\langle a^2\rangle ⊴G$ and so H is metacyclic. It follows from Lemma 10 that $[a^2,b]=a^{2^{n-1}}$, $a^{-4}$ or $a^{-4+2^{n-1}}$. Then, it is easy to see that $[a^2,b^2]=1$, which implies $\mathsf{\Phi }\left(G\right)$ is abelian.

Take $g\in G$ with $g^4\notin G^{\prime }$. Then $[g^2,h]\in {\mathsf{\Omega }}_1\left(\langle g\rangle \right)$ for any $h\in G$, which implies $[g^4,h]=1$ and therefore $g^4\in Z\left(G\right)$. So ${\mho }_2\left(G\right)\le G^{\prime }Z\left(G\right)$. □

Corollary 7.

Let G be a quasi-core-2 2-group. Then, for any $a\in G$, $G/C_G\left(a^2\right)\lesssim C_2\times C_2$, $G/C_G\left(a^4\right)\lesssim C_2$ and if $G/C_G\left(a^4\right)\cong C_2$, then $a^4\in G^{\prime }$ and $\langle a\rangle \cap Z\left(G\right)={\mathsf{\Omega }}_1\left(\langle a\rangle \right)$.

Proof. 

Without loss of generality, we may assume $a^2\notin Z\left(G\right)$, $o\left(a\right)=2^n$ and $n\ge 3$. By Corollary 6, we see $\mathsf{\Phi }\left(G\right)\le C_G\left(a^2\right)$, which implies $G/C_G\left(a^2\right)$ is elementary abelian. For any $g\in G/C_G\left(a^2\right)$, according to Lemma 10, we see $[a^2,g]=a^{-4},a^{2^{n-1}}$ or $a^{-4+2^{n-1}}$. It is easy to see that $G/C_G\left(a^2\right)\lesssim C_2\times C_2$ and $G/C_G\left(a^4\right)\lesssim C_2$. If $G/C_G\left(a^4\right)\lesssim C_2$, then, there exists an element $b\in G\setminus C_G\left(a^4\right)$ such that $\langle [a^2,b]\rangle =\langle a^4\rangle$. So $a^4\in G^{\prime }$ and $\langle a\rangle \cap Z\left(G\right)={\mathsf{\Omega }}_1\left(\langle a\rangle \right)$. □

Lemma 11.

Let G be a quasi-core-2 2-group with $c\left(G\right)=2$. Then $exp\left(G^{\prime }\right)\le 4$.

Proof. 

If not, then there exist $a,b\in G$ such that $o\left(\right[a,b\left]\right)\ge 8$. We may assume $[a,b]=c$. Then $[a^2,b]=c^2$. By induction, $o\left(c^2\right)\le 4$ and so $o\left(c\right)=8$. It follows from Lemma 10 that $\langle c^2\rangle =\langle a^4\rangle$, which implies $a^4\in Z\left(G\right)$. However, $[a^4,b]=c^4\ne 1$, a contradiction. So the conclusion holds. □

Theorem 7.

Let G be a quasi-core-2 2-group with $c\left(G\right)=n$ and $n\ge 2$. Then $exp\left(G^{\prime }\right)\le 2^{2(n-1)}$.

Proof. 

If $n=2$, then the conclusion holds by Lemma 11. Thus we may assume $n\ge 3$. Set $\overline{G}=G/G_n$. Then $c\left(\overline{G}\right)=n-1$. By induction, we see $exp({\overline{G}}^{\prime })\le 2^{2(n-2)}$. Since $G_n=[G_{n-1},G]\le Z\left(G\right)$, by Lemma 11, we see $exp\left(G_n\right)\le 4$. It follows that $exp\left(G^{\prime }\right)\le 2^{2(n-1)}$. □

Theorem 8.

Let G be a non-abelian quasi-core-2 2-group with $d\left(G\right)=2$. Then ${\mho }_1\left(G^{\prime }\right)$, $G_4$ are cyclic, and either $G^{\prime }\cap Z\left(G\right)\lesssim C_2\times C_2\times C_2$ or $G=\langle a,b|a^8=1,a^4=b^4=c^2,[a,b]=c,[c,a]=[c,b]=1\rangle$.

Proof. 

If G is metacyclic, then the conclusion holds by Lemma 10. So we may assume $G=\langle a,b\rangle$ is non-metacyclic, $[a,b]=c$, $o\left(a\right)=2^n,o\left(b\right)=2^m$ and $o\left(c\right)=2^t$ with $n\ge m$. Thus $G^{\prime }=\langle c,[c,g]|g\in G\rangle$. By Corollary 6, $\mathsf{\Phi }\left(G\right)$ is abelian. So ${[c,g]}^2=[c^2,g]\in \langle c^2\rangle$, which implies ${\mho }_1\left(G^{\prime }\right)\le \langle c^2\rangle$ and therefore ${\mho }_1\left(G^{\prime }\right)$ is cyclic. Now we consider the following two cases: $c\left(G\right)=2$ and $c\left(G\right)>2$.

Case 1.

$c\left(G\right)=2$.

By Lemma 11, we see $exp\left(G^{\prime }\right)\le 4$. We may assume $exp\left(G^{\prime }\right)=4$. Then $o\left(c\right)=4$ and $[a^2,b]=[a,b^2]=c^2$. Thus $n\ge m\ge 3$ and $c^2\in \langle a\rangle \cap \langle b\rangle$. Without loss of generality, we may assume $\langle a\rangle \cap \langle b\rangle =\langle a^{2^u}\rangle =\langle b^{2^v}\rangle$, $a^{2^u}=b^{2^v}$ and $u\ge v\ge 2$. Let $b_1=a^{-2^{u-v}}b$. Then $[a,b_1^2]=c^2$. If $u>v$ or $v\ge 3$, then $o\left(b_1\right)=2^v$. Thus $\langle c^2\rangle =\langle b_1^{2^{v-1}}\rangle$, which implies $a^{2^{u-1}}\in \langle b\rangle$, a contradiction. So $u=v=2$ and $a^4=b^4$. Noticing that $G=\langle a,b_1\rangle$ and $[a,b_1]=c$, we see $a^4=b_1^4$ by the above. It follows from $o\left(b_1\right)=8$ that $o\left(a\right)=8$. So, we see $G=\langle a,b|a^8=1,a^4=b^4=c^2,[a,b]=c,[c,a]=[c,b]=1\rangle$.

Case 2.

$c\left(G\right)>2$.

In this case, we consider the following two subcases: $G^{\prime }$ is cyclic and $G^{\prime }$ is not cyclic.

Subcase 1.

$G^{\prime }$ is cyclic.

If $o\left(c\right)\le 4$, then $c^2\in Z\left(G\right)$ and $G^{\prime }\cap Z\left(G\right)\lesssim C_2$. So we may assume $t\ge 3$. By Lemma 10, we see $[c,a]=1,c^{-2},c^{-2+2^{t-1}}$ or $c^{2^{t-1}}$. If $\langle [c,a]\rangle =\langle c^2\rangle$, then $exp(G^{\prime }\cap Z\left(G\right))=2$. Thus we may assume $[c,a]=c^{2^{t-1}}$ and $[c,b]=1$. It follows that $[a^2,b]=c^{2+2^{t-1}}$. According to Lemma 10, it is easy to see $\langle c^2\rangle =\langle a^4\rangle$. So $[a^4,b]=1$ and therefore $o\left(c\right)\le 4$, in contradiction to the hypothesis.

Subcase 2.

$G^{\prime }$ is not cyclic.

Since $[a,b]=c$, $[a^2,b]=c^2[c,a]$. By Lemma 10, we see $[c,a]=c^{-2}{a}^{-4},c^{-2}{a}^{-4+2^{n-1}},c^{-2}$ or $c^{-2}{a}^{2^{n-1}}$. Similarly, $[c,b]=c^{-2}{b}^{-4},c^{-2}{b}^{-4+2^{m-1}},c^{-2}$ or $c^{-2}{b}^{2^{m-1}}$. It follows that $G^{\prime }\le \langle c,a^4,b^4\rangle$, $[\langle [c,a]\rangle ,G]\le {\mho }_1\left(\langle [c,a]\rangle \right)$ and $[\langle [c,b]\rangle ,G]\le {\mho }_1\left(\langle [c,b]\rangle \right)$. Then $[G_3,G]\le {\mho }_1\left(G_3\right)\le {\mho }_1\left(G^{\prime }\right)$. So $G_4$ is cyclic.

Now we prove $exp(G^{\prime }\cap Z\left(G\right))=2$. Assume $[c,a]=c^{-2}{a}^{-4}$ or $c^{-2}{a}^{-4+2^{n-1}}$, and $n\ge 4$.

If $[c,b]=c^{-2}$, then $G^{\prime }=\langle c,a^4\rangle$. Since $G^{\prime }$ is not cyclic, we see $[c,a]\ne 1$. Take $g\in G^{\prime }\cap Z\left(G\right)$ and assume $g=c^{2i}{a}^{4j}$. It follows from $[g,b]=1$ that $o\left(g\right)\le 2$. So $exp(G^{\prime }\cap Z\left(G\right))=2$.

If $[c,b]=c^{-2}{b}^{2^{m-1}}$, then $G^{\prime }=\langle c,a^4,b^{2^{m-1}}\rangle$. If $[c,a]=1$, then $a^4\in \langle c\rangle$ and $G^{\prime }=\langle c,b^{2^{m-1}}\rangle$. It is easy to see that $exp(G^{\prime }\cap Z\left(G\right))=2$. Assume $[c,a]\ne 1$. Take $h\in G^{\prime }\cap Z\left(G\right)$ and assume $h=c^{2k}{a}^{4l}$. It follows from $[h,b]=1$ that $o\left(h\right)\le 2$ and so $exp(G^{\prime }\cap Z\left(G\right))=2$.

If $[c,b]=c^{-2}{b}^{-4}$ or $c^{-2}{b}^{-4+2^{m-1}}$, we may assume $m\ge 4$ by the above. It is easy to see that $\langle a^8,b^8\rangle \le \langle c\rangle$. Thus $[b^8,a]=1$, which implies $o\left(b\right)=16$ and $b^8=a^{2^{n-1}}$. On the other hand, we see $[{(a^{2^{n-3}}{b}^2)}^2,a]=b^8$ and therefore $b^8=a^{2^{n-2}}{b}^4$. It follows that $[a,b^4]=1$. However, it is impossible.

Assume $[c,a]=c^{-2}$ or $c^{-2}{a}^{2^{n-1}}$. Without loss of generality, we may assume $[c,b]=c^{-2}$ or $c^{-2}{b}^{2^{m-1}}$. Then $G^{\prime }\le \langle c,a^{2^{n-1}},b^{2^{m-1}}\rangle$. It is clear that $exp(G^{\prime }\cap Z\left(G\right))=2$. □