Finite groups having centralizer commutator product property

Abstract

Let $\alpha$ be an automorphism of a finite group $G$ and assume that $G = \big \{\,[g , \alpha ] : g \in G\,\big \}\cdot C_G (\alpha )$. We prove that the order of the subgroup $[G, \alpha ]$ is bounded above by $n^{\log _2 (n+1)}$ where $n$ is the index of $C_G (\alpha )$ in $G$.

1 Introduction

Let $A$ be a finite group that acts on the finite group $G.$ In the case where $(|G|,|A|)=1$, there are several very useful relations between the groups $G$ and $A$, some of which are as follows: (i)  $G=[G,A]\cdot C_G(A)$, (ii) $[G,A,A]=[G,A]$ and (iii) $C_{G/N} (A) = C_G (A)N/N$ for any $A$-invariant normal subgroup $N$ of $G$. Almost all of the research papers studying this kind of action concerned with the situations where the fixed point subgroup $C_G(A)$ has a restricted structure. However, Parker and Quick [1] considered a dual situation by assuming that the index of $C_G(A)$ is bounded. As this assumption clearly gives no restriction to $C_{G}(A)$, they focused their attention on the group $[G,A]$ and proved that $|[G,A]|\leqslant n^{\log _2 (n+1)}$ if $|G:C_G(A)|\leqslant n.$

We consider here a special noncoprime action in view of [1]:

Let $\alpha$ be an automorphism of the finite group $G$ such that for every $x\in G, x=[g,\alpha ]\cdot z$ for some $g\in G$ and $z\in C_G(\alpha )$.

In the literature a finite group $G$ admitting such an automorphism $\alpha$ is called an $\alpha$ -CCP group where the acronym CCP stands for “centralizer commutator product”. Lemma 2.1 below shows that nice relations indicated above which are valid in the case of a coprime action also survive in the setting of $\alpha$-CCP groups. The study of $\alpha$-CCP groups was started by Stein [2] who proved that the subgroup $[G, \alpha ]$ is solvable. The goal of the present paper is to give an upper bound for the order of $[G, \alpha ]$ in terms of the index of $C_G (\alpha )$ in $G$. Namely, we prove the following:

Theorem A

Let $G$ be an $\alpha$-CCP group such that $| G : C_G (\alpha )|\leqslant n$. Then $|[G,\alpha ] | \leqslant n^{\log _2 (n+1)}$.

An internal reformulation of Theorem A can be stated as

Theorem B

Let $H$ be a finite group containing an element x such that $H=\big \{[h,x] : h\in H\big \}\cdot C_H(x)$. If $|H:C_H(x)|\leqslant n$ then $|[H,x]|\leqslant n^{log_{2}(n+1)}$.

Theorem A is the $\alpha$-CCP analogue of [1, TheoremA]. The key lemma in our proof is Lemma 3.1 which we obtain as the $\alpha$-CCP analogue of [1, Lemma 2.1]. The rest of the paper contains the proof of Theorem A and some technical results pertaining to the proof of Theorem A; all of which are proven in a similar fashion as in the proofs of [1, Proposition 2.2], [1, Corollary 2.3] and [1, TheoremA] with obvious changes, namely using Lemma 3.1 instead of [1, Lemma 2.1]. For the sake of completeness we present a proof here for each of them.

In Sect. 2 we state and prove some preliminary facts about $\alpha$-CCP groups. Section 3 is concerned with our key lemma, namely Lemma 3.1, and its consequences. We prove our main result Theorem A and its equivalent Theorem B in Sect. 4.

All groups are assumed to be finite. The notation and terminology are standard.

2 Preliminaries on $\alpha$-CCP groups

Lemma 2.1

The following hold for any $\alpha$-CCP group $G$.

• (i) $G=[G,\alpha ]\cdot C_G(\alpha )$ and $[G,\alpha , \alpha ]=[G,\alpha ]$. Furthermore $G=[G,\alpha ]\times C_G(\alpha )$ whenever $G$ is abelian.

• (ii) Every $\alpha$-invariant subgroup $S$ of $G$ is also an $\alpha$-CCP group and we have $\big \{[x,\alpha ] : x\in S\big \}=\big \{[g,\alpha ] : g\in G\big \}\cap S.$

• (iii) $G/N$ is an $\alpha$-CCP group for any $\alpha$-invariant normal subgroup $N$ of $G$.

• (iv) If $[g, \alpha ]^f \in C_G (\alpha )$ for some $f$ and $g$ in G, then $g\in C_G (\alpha )$.

• (v) $C_{G/N} (\alpha ) = C_G (\alpha ) N / N$ for any $\alpha$-invariant normal subgroup $N$ of $G$.

• (vi) $\big \{\,[g, \alpha ] : g\in G \,\big \}$ is a transversal to $C_G (\alpha )$. Furthermore $\alpha ^G$ is a transversal to $C_H (\alpha ^{a})$ for any $a\in G$ in the semidirect product $H = G \langle \alpha \rangle$.

Proof

This lemma gives almost the same information as in [2, Proposition 2.2] on an $\alpha$-CCP group $G$. We need only to show that $G=[G,\alpha ]\times C_G(\alpha )$ when $G$ is abelian: Notice that $[G,\alpha ]=\big \{\,[g,\alpha ] : g\in G \,\big \}$ when $G$ is abelian and also observe that for any $[g,\alpha ]\in C_G(\alpha )$, we have $[g,\alpha ]=1$ by Lemma 2.1(iv). $\square$

The following lemma is crucial in proving our key lemma Lemma 3.1.

Lemma 2.2

Let $G$ be an $\alpha$-CCP group and set $H = G \langle \alpha \rangle$. Then

• (i) the map $f_{\alpha ^{a}} : \alpha ^G \longrightarrow \alpha ^G$ defined by $f_{\alpha ^{a}} (\alpha ^g) = (\alpha ^{a})^{\alpha ^g}$ is a bijection for any $a \in G$,

• (ii) for any $X\leqslant H$ with $X \cap \alpha ^G \ne \phi$ and for any $\alpha ^{a} \in X$ we have

  $$\begin{aligned} (\alpha ^{a})^X = (\alpha ^{a})^{X \cap \alpha ^G}= X \cap \alpha ^G. \end{aligned}$$

Proof

$\alpha ^G$ is a transversal to $C_H (\alpha ^{a})$ by Lemma 2.1(vi). If $g$ and $h$ are elements of $G$ such that $(\alpha ^{a})^{\alpha ^g } = (\alpha ^{a})^{\alpha ^h}$, then $\alpha ^g (\alpha ^h)^{-1} \in C_H (\alpha ^{a})$ and so $\alpha ^g = \alpha ^h$. This proves $(i)$ since $\alpha ^G$ is finite.

It is straightforward to verify that $(\alpha ^{a})^{X \cap \alpha ^G} \subseteq (\alpha ^{a})^X \subseteq X \cap \alpha ^G$. If $\alpha ^y \in X \cap \alpha ^G$, then $\alpha ^y = (\alpha ^{a})^{\alpha ^h}$ for some $h \in G$ by part (i). This yields $\alpha ^y \in (\alpha ^{a})^{\alpha ^G}$. Notice that $f_{\alpha ^{a}} (X \cap \alpha ^G) \subseteq X \cap \alpha ^G$ as $\alpha ^{a} \in X$, and so $f_{\alpha ^{a}} (X \cap \alpha ^G) = X \cap \alpha ^G$ since $f_{\alpha ^{a}}$ is a bijection. Then $\alpha ^h \in X$ and hence $X \cap \alpha ^G \subseteq (\alpha ^{a})^{X \cap \alpha ^G}$ which establishes the claim (ii). $\square$

3 Some technical lemmas pertaining to the proof of Theorem A

The following results are modifications of Lemma 2.1, Proposition 2.2 and Corollary 2.3 in [1] for $\alpha$-CCP groups.

Lemma 3.1

Let $G$ be an $\alpha$-CCP group and let ${\mathcal O} = {\alpha }^G$. If $I \subseteq {\mathcal O}$ and $\Theta$ is an orbit of $\langle I \rangle$ on ${\mathcal O}$, then $\langle I \rangle \leqslant \langle \Theta \rangle$. Furthermore if some member of $\Theta$ is not contained in $\langle I \rangle$, then $\langle I \rangle < \langle \Theta \rangle$.

Proof

To ease the notation set $K = \langle I \rangle$ and let $\Theta = ({\alpha }^x)^K$. It should be noted that $K \langle \Theta \rangle$ is a subgroup of $G$ because $K$ normalizes $\langle \Theta \rangle$. Set now $L = K \langle \Theta \rangle$. Since $L \cap {\alpha }^G \ne \phi$ and ${\alpha }^x \in L$, we have

$$\begin{aligned} ({\alpha }^x)^L = (\alpha ^x)^{L \cap \alpha ^G} = L \cap \alpha ^G \end{aligned}$$

by Lemma 2.2(ii). Then, for any generator $\alpha ^y$ of $K$, we have

$$\begin{aligned} \alpha ^y \in L \cap \alpha ^G = (\alpha ^x)^{K \langle \Theta \rangle } \subseteq \langle (\alpha ^x)^K \rangle = \langle \Theta \rangle . \end{aligned}$$

This completes the proof. $\square$

Lemma 3.2

When $G$ is an $\alpha$-CCP group the group $\langle {\alpha }^G\rangle$ can be generated by $\log _2 \Big ( \frac{2 (n+p-1)}{p}\Big )$ conjugates of $\alpha$ where $p$ is the smallest positive divisor of the order of $\alpha$ and $|G:C_G(\alpha )|\leqslant n$.

Proof

We let ${\mathcal O} = {\alpha }^G$ and consider the action of $\langle {\alpha }\rangle$ on ${\mathcal O}$ by conjugation. Suppose first that $\langle \alpha \rangle$ has a fixed point ${\alpha }^x$ which is different from $\alpha$. Then $[\alpha , x] \in C_G (\alpha )$ and hence $[\alpha , x]= 1$ by Lemma 2.1(iv). This contradiction shows that $\alpha$ is the only fixed point of $\langle \alpha \rangle$ in its action on ${\mathcal O}$.

Define $K_0 = 1$, $K_1 = \langle \alpha \rangle$ and for $j > 1$, $K_j = \langle K_{j-1}, {\alpha }_j \rangle$ where at each stage ${\alpha }_j \in {\mathcal O}$ is chosen to maximize the order of $K_j$. Since $G$ is finite, there exists $k$ such that $K_k = \langle {\alpha }^G\rangle$ and $K_{k-1} \ne \langle {\alpha }^G\rangle$. Now $\langle {\alpha }^G\rangle = \langle \alpha _1, \alpha _2, \ldots , \alpha _k\rangle$ where $\alpha _1=\alpha$. Fix $j \in \{1,\ldots ,k\}$ and let $I=\{\alpha _1, \alpha _2, \ldots , \alpha _j\}$. Now $K_j=\langle I\rangle .$ Choose an orbit $\Theta$ of $K_j$ with representative $B$ where $B\nleq K_j$. Then $K_j< \langle \Theta \rangle$ by Lemma 3.1. If $\Theta$ were also an orbit of $K_{j-1}$, then we would have

$$\begin{aligned} K_j< \langle B^{K_j}\rangle =\langle B^{K_{j-1}}\rangle \leqslant \langle B,K_{j-1}\rangle \end{aligned}$$

contradicting the choice of $\alpha _j.$ Therefore $\Theta$ is a union of at least two orbits of $K_{j-1}$ on $\mathcal {O}.$ Notice also that $B\nleq K_i$ for each $i=1, \ldots ,j$. Thus $\Theta$ is a union of at least $2^{j-1}$ orbits of $\langle \alpha \rangle$ on $\mathcal {O}$, each of which has length at least $p.$ Since $\alpha _j\leqslant K_j$ for $i\leqslant j$ we see that the set $\Omega = \{\alpha _1\}\cup \{{\alpha _2}^{K_1}\}\cup \ldots \cup \{{\alpha _i}^{K_{i-1}}\}$ is contained in $K_j.$ Therefore $\Omega \cap {\alpha _{i+1}}^{K_i}$ is empty as $\alpha _{i+1}\nleq K_i.$ Then $\mathcal {O}\supseteq \{\alpha _1\}\cup \{{\alpha _2}^{K_1}\}\cup \ldots \cup \{{\alpha _k}^{K_{k-1}}\}$ and the right hand side is a disjoint union. So

$$\begin{aligned} n\geqslant |\mathcal {O}|\geqslant 1+p(1+2+\cdots +2^{k-2})=1+p(2^{k-1}-1). \end{aligned}$$

Consequently we have $k-1 \leqslant \log _2 \big ( \frac{n+p-1}{p}\big )$ as claimed. $\square$

Lemma 3.3

Let $G$ be an $\alpha$-CCP group. Suppose that $G$ is a $p$-group for some prime $p$ with $|G : C_G (\alpha )|\leqslant p^m$. Then $|[G, \alpha ]| \leqslant p^{\frac{m^2 + m}{2}}$.

Proof

Firstly we handle the case where $G$ is of class at most two by induction on the order of $G.$ By Lemma 2.1(i) we have $[G,\alpha ]=[G,\alpha ,\alpha ]$ and $G/G'=[G/G',\alpha ]\times C_{G/G'}(\alpha )$. Then $G=[G,\alpha ]$ by induction and hence $C_{G/G'}(\alpha )=1$, that is $C_G(\alpha )\leqslant G'$. Thus $|G:G'|\leqslant p^m.$ In this case the proof is in a similar fashion as in the proof of [1, Proposition 3.1]. For the sake of completeness we present it here. Let the abelian group $\bar{G}=G/G'$ be the direct product of nontrivial cyclic subgroups $\langle \bar{x_i}\rangle$ for $i=1,\ldots ,d$ where $|\bar{x_i}|=p^{m_i}.$ We have $G=\langle x_1,\ldots ,x_d\rangle$ since $G'\leqslant \Phi (G).$ It is straightforward now to verify that $G'=\langle [x_j,x_i] : 1\leqslant i<j\leqslant d \rangle$ since $G'\leqslant Z(G).$ Set $H_i=\langle x_{i+1},\ldots ,x_d, G'\rangle$. Then $G'=\prod _{i=1}^{d-1}[H_i,x_i]$ for $i=1,\ldots ,d-1.$ We have $|[H_i,x_i]|\leqslant |H_i/G'|=p^{m_{i+1}+\cdots +m_d}$ due to the fact that $h\longmapsto [h,x_i]$ defines a homomorphism from $H_i/G'$ onto $[H_i,x_i]$. Thus $|G|\leqslant \prod _{i=1}^{d}p^{m_i}\prod _{i=1}^{d-1}|[H_i,x_i]|\leqslant p^M$ where $M=\sum _{i=1}^{d}im_i.$ It can be proven by induction on $d$ that $M\leqslant (m^2+m)/2$. This completes the proof when $G$ is of class at most two.

Suppose now that $G$ has class $c$ with $c\geqslant 3.$ Again assume $|G|$ minimal, therefore $G=[G,\alpha ].$ The proof in this case is in a very similar fashion as in the proof of [1, Theorem B]. Note that $\gamma _{c-1}(G)$ is abelian. We also observe that $[\gamma _{c-1}(G),\alpha ]\ne 1$, because otherwise $[\gamma _{c-1}(G),\alpha , G]=1=[G,\gamma _{c-1}(G),\alpha ]$ and hence $\gamma _{c-1}(G)\leqslant Z(G)$ by the Three Subgroup Lemma. Let now $N$ be of minimal order among all normal $\alpha$-invariant subgroups of $G$ contained in $\gamma _{c-1}(G)$ and are not centralized by $\alpha$. Let $|G/N:C_{G/N}(\alpha )|=p^r$. As $C_{G/N}(\alpha )=C_G(\alpha )N/N$ by Lemma 2.1(v) we have $|G:C_G(\alpha )N|=p^r$. Note that $G/N$ and $N$ are both $\alpha$-CCP groups by Lemma 2.1(i). It follows by induction that $|[G/N,\alpha ]|\leqslant p^{\frac{r^2 + r}{2}}.$ As $[G/N,\alpha ]=[G,\alpha ]N/N=G/N$ we have $|G/N|\leqslant p^{\frac{r^2 + r}{2}}$. Let now $|N:C_N(\alpha )|=p^s.$ Since $N$ is abelian we have $N=[N,\alpha ]\times C_N(\alpha )$ and so $|[N,\alpha ]|=p^s$. It remains to bound $|N/[N,\alpha ]|$ suitably. As $N$ is contained in $\gamma _{c-1}(G)$ we have $[N,G]\leqslant \gamma _{c}(G)\leqslant Z(G).$ Hence for $g\in G$ the map $x\mapsto [x,g]$ for $x\in N$, is a homomorphism with kernel $C_N(g)$, in particular $[N,G]$ lies in the kernel and $|N:C_N(g)|=|[N,g]|$. Set now $H=[N,\alpha ][N,G]$. Observe that $1\ne [N,\alpha ]=[N,\alpha ,\alpha ]\leqslant [H,\alpha ]$ by Lemma 2.1(i). It follows by minimality of $N$ that $H=N$. Thus

$$\begin{aligned} |[N,g]|=|N:C_N(g)|\leqslant |N:[N,G]|=|[N,\alpha ][N,G]:[N,G]|\leqslant |[N,\alpha ]|. \end{aligned}$$

We also observe that $[N,G']=1$ by the three subgroup Lemma as $[N,G,G]=1=[G,N,G]$. This gives that $NC_G(\alpha )\leqslant G'\leqslant C_G(N)$. As $N\leqslant \gamma _{c-1}(G)\leqslant G'$ we get $NC_G(\alpha )\leqslant G'\leqslant C_G(N).$ Therefore $|G:C_G(N)|\leqslant p^r.$ Let $Y$ be a minimal generating set for $G$ modulo $C_G(N).$ Then $|Y|\leqslant r.$ Since $[N,G]\leqslant Z(G)$ we also see that $[N,G]=\prod _{y\in Y}[N,y]$. Thus $|[N,G]|\leqslant |[N,\alpha ]|^{|Y|}\leqslant p^{sr}$. So $|N|=|[N,G][N,\alpha ]|\leqslant p^{s(r+1)}$ whence $|G|=|G/N|\cdot |N|\leqslant p^{(r^2+r)+s(r+1)}$. This establishes the claim as

$$\begin{aligned} 1/2\big ((r^2+r)+s(r+1)\big )\leqslant 1/2(r^2+r)+1/2(s^2+s)+sr \nonumber \\ \quad \leqslant 1/2\left( (r+s)^2+r+s\right) \leqslant 1/2(m^2+m). \end{aligned}$$

$\square$

4 Proof of the main results

In this section we present a proof of Theorem A and deduce Theorem B.

Proof of Theorem A

Let $G$ be a minimal counterexample to the theorem. Then $G = [G,\alpha ]$ by induction as $[G,\alpha ] = [G,\alpha ,\alpha ]$ by Lemma 2.1(i). As a consequence $C_G (\alpha ) \leqslant G'$, and $G$ is nonabelian. The main result of [2] gives that the group $G$ is solvable and hence $F (G)\ne 1$. If $\left[ F(G),\alpha \right] = 1$, then $G \leqslant C_G \left( F (G)\right) = Z\left( F (G)\right)$  by the Three Subgroup Lemma, which is a contradiction as $G$ is nonabelian. Thus $\left[ F (G),\alpha \right] \ne 1$ and hence there is a prime $p$ dividing $|F(G)|$ such that $\left[ O_p(G),\alpha \right] \ne 1.$ Notice that if $\left[ Z_2(O_p(G)),\alpha \right] =1$, then $Z_2\left( O_p(G)\right) \leqslant Z(G)$ by the Three Subgroup Lemma as $[G,\alpha ]=G$. This forces $O_p(G)=Z_2(O_p(G))=Z(O_p(G))$ which contradicts the fact that $\left[ O_p(G),\alpha \right] \ne 1.$ Let $Q$ be minimal element of the set $\{S : S$ is a normal $\alpha$-invariant subgroup of $G$ which is contained in $Z_2\left( O_p(G)\right)$ such that $[S,\alpha ]\ne 1\}$. Clearly $[Q',\alpha ]=1$ by the minimality of $Q$ and so $Q'\leqslant Z(G)$ by the Three Subgroup Lemma. Set now $Q_0= \langle [Q,\alpha ]^G \rangle$. Note that both $Q$ and $|G/Q|$ are $\alpha$-CCP groups. So we have $[Q,\alpha ]=[Q,\alpha ,\alpha ]$ by Lemma 2.1 $(i)$. Thus $1\ne \big [[Q,\alpha ]\big ]\leqslant [Q_0,\alpha ]$ and hence $Q=Q_0$ by the minimality of $Q$. Now $|QC_G(\alpha ):C_G(\alpha )|=|Q:C_Q(\alpha )|=p^m$ for some $m.$ Let $|G:QC_G(\alpha )|=r$. Then $r\leqslant {\frac{n}{p^m}}.$ We observe by Lemma 3.3 that $|[Q, \alpha ]|\leqslant p^{\frac{m^2 + m}{2}}.$ Set $R=C_{[Q,\alpha ]}(\alpha )$. Then $R\leqslant [Q,\alpha ]'\leqslant Q'$ and hence $R\leqslant Z(G)$. Now

$$\begin{aligned} |[Q,\alpha ]/R|=|[Q,\alpha ]C_Q(\alpha ):C_Q(\alpha )|=|Q:C_Q(\alpha )|=p^m. \end{aligned}$$

So $|R|={\frac{|[Q,\alpha ]|}{p^m}}\leqslant p^{\frac{m^2 - m}{2}}$. It remains to bound $|G/R|$ suitably.

Set $\bar{G}=G/R.$ The group $\bar{Q}$ is the product of at most $log_2(r+1)$ of the conjugates of $[\bar{Q},\alpha ]$ in $\bar{G}$: To see this let $H=G\rtimes \langle \alpha \rangle .$ Note that $Q\rtimes H$ and $C_G(\alpha )\langle \alpha \rangle Q\leqslant N_H(Q\langle \alpha \rangle )$. Set $\tilde{H}=H/Q$. Now $|\tilde{H}:N_{\tilde{H}}(\langle \tilde{\alpha } \rangle )|\leqslant |H:Q\langle \alpha \rangle C_G(\alpha )|=r.$ By Lemma 3.2 $\langle {\langle \tilde{\alpha } \rangle }^{\tilde{H}}\rangle$ can be generated by at most $k=log_2 (r+1)$ conjugates of $\langle \tilde{\alpha } \rangle$. That is $\langle {(\langle \tilde{\alpha } \rangle )}^{\tilde{H}}\rangle =\langle \tilde{\alpha _1},\ldots ,\tilde{\alpha _k}\rangle$ where each $\alpha _i$ is a conjugate of $\alpha$ and $\alpha _1=\alpha .$ Note that $H=[G,\alpha ]\langle \alpha \rangle ={\langle {\alpha }^H \rangle }C_G(\alpha )=MQC_G(\alpha )$ where $M=\langle \alpha _1,\ldots ,\alpha _k\rangle C_G(\alpha )$. Therefore

$$\begin{aligned} \langle [Q,\alpha ]^G\rangle = \langle [Q,\alpha ]^M\rangle =[Q,\alpha ][Q,\alpha ,M]\leqslant [Q,M]=\prod _{i=1}^k [Q,\alpha _i]. \end{aligned}$$

We are now ready to complete the proof of Theorem A. By the above paragraph we have $|\bar{Q}|=|\langle [\bar{Q},\alpha ]^{\bar{G}}\rangle |\leqslant |[\bar{Q},\alpha ]|^k=p^{mk}$ and so $|Q|\leqslant p^{mk+(\frac{m^2 - m}{2})}$. Notice that $|G/Q|\leqslant r^k$ by induction. Thus

$$\begin{aligned} |G|=|G/Q||Q|\leqslant r^{k}p^{mk+\frac{m^2 - m}{2}}=r^{k}(p^{m})^{k+\frac{m-1}{2}}\leqslant r^k(p^m)^{log_2(n+1)}\leqslant n^{log_2(n+1)}. \end{aligned}$$

This contradiction completes the proof of Theorem A. $\square$

Remark 4.1

As indicated in the introduction one can reformulate Theorem A as Theorem B. Their equivalence can be easily seen as follows:

Suppose that Theorem A is true. Set $H=G\rtimes \langle \alpha \rangle$ and $x=\alpha$ in $H.$ Then $[G,\alpha ]=[H,x]$ and $\big \{[g,\alpha ] : g\in G\big \}=\big \{[h,x] : h\in H\big \}$ and $|G:C_G(\alpha )|=|H:C_H(x)|=n.$ Therefore $|[G,\alpha ]|=|[H,x]\leqslant n^{log_2 (n+1)}$ by Theorem A. Conversely suppose that Theorem B is true and let $H$ be a finite group containing an element x such that $H=\big \{[h,x] : h\in H\big \}C_H(x)$ holds. Set $G=H$ and let $\alpha$ denote the inner automorphism of $G$ induced by $x$. Then by applying Theorem B we have $|[H,x]|\leqslant n^{log_2 (n+1)}$ as desired.