Proof of Proposition 1.2. Recall that $G$ is a finite group in which $|x^G|\leq m$ whenever $x$ is an element of prime power order. We wish to show that $G'$ has $m$ -bounded order.

Let $p$ be a prime bigger than $m$ and assume that $p\in \pi (G)$ . Let $P$ be a Sylow $p$ -subgroup of $G$ . Since $|x^G|\leq m$ whenever $x$ is an element of prime power order in $G$ , it follows that $P$ commutes with any element of prime power order and so $P\leq Z(G)$ .

Pick an arbitrary element $g\in G$ and suppose that the order of $g$ is $p_1^{\alpha _1}\ldots p_s^{\alpha _s}$ , where $p_i$ are distinct primes in $\pi (G)$ . Write $g=g_1g_2\ldots g_s$ , where the element $g_i$ is of order $p_i^{\alpha _i}$ . This can be done, for example, choosing $g_i$ as a generator of the Sylow $p_i$ -subgroup of the cyclic group $\langle g\rangle$ . In view of the above, $g_i\in Z(G)$ whenever $p_i\gt m$ . It follows that there are less than $m$ non-central elements in the list $g_1,\ldots ,g_s$ . Each of these non-central elements has centralizer of index at most $m$ . Therefore $|g^G|\leq m^m$ . This holds for an arbitrary element $g\in G$ so the claim follows from Neumann’s theorem.

Proof. We just need to show that $G'\leq [G,T]$ . We can pass to the quotient $G/[G,T]$ and assume that $[G,T]=1$ . Then $T\leq Z(G)$ . Since $G$ is generated by the conjugates of $T$ , we deduce that $G=T$ is abelian. Hence, $G'\leq [G,T]$ and the lemma follows.

Proof. Since $C_G(x)$ has index at most $m$ in $G$ , by Lemma 2.4 we can choose elements $y_1, \ldots , y_m$ such that $l_X(y_i) \le m-1$ and $[G, x]$ is generated by the commutators $[y_i, x]$ . Here, the elements $y_1,\ldots , y_m$ are not necessarily pairwise distinct. For each $i = 1, \ldots , m$ , write $y_i = y_{i1} \ldots y_{i(m-1)}$ where $y_{ij} \in X$ . We can write $[y_i,x]$ as a product of conjugates of $[y_{ij},x]$ . Let $h_1,\ldots ,h_s$ be the conjugates in $G$ of elements from the set $K=\{ x, y_{ij}; 1 \le i \le m,\ 1\le j\le m-1\}$ . Notice that the size of each conjugacy class containing an element of $K$ is less than or equal to $m$ , and $K$ has at most $m(m-1)+1$ elements. Therefore, $s$ is $m$ -bounded. Let $H=\langle h_1,\ldots ,h_s\rangle$ . It is clear that $[G,x]\le H'$ , so it is sufficient to show that $H'$ has finite $m$ -bounded order. Observe that $C_G(h_i)$ has index at most $m$ in $G$ for each $i=1,\ldots ,s$ . It follows that the center $Z(H)$ has index at most $m^s$ in $H$ . Thus, Schur’s theorem [Reference Khukhro9, 2.4.1] tells us that $H'$ has finite $m$ -bounded order.

Proof. Of course, $\overline {X(G)} \subseteq X(\overline {G})$ . We need to establish the inverse containment $X(\overline {G}) \subseteq \overline {X(G)}$ .

Pick $[\overline {x},\overline {y}] \in X(\overline {G})$ . There is a Sylow $p$ -subgroup $\overline {P}$ of $\overline {G}$ such that $[\overline {x}, \overline {y}] \in \overline {P}$ and $\overline {x} \in \overline {P}$ . The full preimage of $\overline {P}$ is $PN$ for some Sylow $p$ -subgroup $P$ of $G$ . Choose $x$ and $y$ such that $\overline {x}=xN$ and $\overline {y}=yN$ . Both elements $x$ and $x^y$ lie in $PN$ .

Write $x = x'n$ , where $x'\in P$ and $n\in N$ . Since $x^y\in PN$ , we also have $x'^y\in PN$ . Thus, there is a Sylow $p$ -subgroup $P_0$ in $PN$ such that $x'^y\in P_0$ . Since $P_0$ is conjugate to $P$ in $PN$ , there is an element $a\in N$ such that $x'^{ya}\in P$ .

So, we have

\begin{equation*}[\overline {x}, \overline {y}] = [x',y]N = [x', ya]N.\end{equation*}

Since $[x', ya] \in P$ and $x' \in P$ , it follows that $[x', ya]\in X(G)$ and, therefore, $[x,y]\in X(G)N$ . This completes the proof.

Proof. For $x\in X(G)$ , let $N_x$ be the normal core of $C_G(x)$ , that is, $N_x = \underset {g \in G}{\bigcap } C_G(x)^g$ . As $[G\, : \,C_G(x)]\le m$ , the index $[G\, : \,N_x]$ divides $m!$ .

Taking into account that $p\gt m$ we conclude that $P\leq N_x$ . Since this happens for all $x \in X(G)$ and since $G'$ is generated by $X(G)$ , it follows that $[G',P]=1$ .

Proof. Let $\pi _1 = \{p_1, \ldots , p_k \}$ the set of prime divisors of $|G|$ which are less than or equal to $m$ and $\pi _2 =\pi (G)\setminus \pi _1$ . Let $H_1$ be a $\pi _1$ -Hall subgroup and $H_2$ a $\pi _2$ -Hall subgroup of $G$ .

By Lemma 3.2, $H_2$ centralizes $G'$ . Therefore, $G'=K_1\times K_2$ , where $K_1=G'\cap H_1$ and $K_2=G'\cap H_2$ . Obviously, $G$ $''$ has trivial intersection with $K_2$ , so we pass to the quotient $G/K_2$ and simply assume that $G'\leq H_1$ . In particular, $H_1$ is normal and $[H_1,H_2,H_2]=1$ . It follows that $G=H_1\times H_2$ . Therefore, $G''={H_1}''$ and so without loss of generality, we can assume that $G=H_1$ .

Let $\{ P_1, \ldots , P_k \}$ be a Sylow basis of $G$ and $T = \bigcap _{i=1}^k N_G(P_i)$ the system normalizer. Since $G = P_1P_2 \ldots P_k$ , any element $x\in G$ can be written as a product $x = x_1x_2 \ldots x_k$ , where $x_i\in P_i$ . For $a\in G$ , write

\begin{equation*} [x, a] = [x_1x_2 \ldots x_k, a] = [x_1, a]^{x_2 \ldots x_k}[x_2, a]^{x_3 \ldots x_k} \ldots [x_k, a].\end{equation*}

If $a\in T$ , we have $[x_i, a]\in P_i$ and, therefore, $[x,a]$ is a product of at most $k$ elements from $X(G)$ . So, the index $[G\, : \, C_G([x,a])]$ is at most $m^k$ . Obviously, $k$ here is less than $m$ . Thus, for any $x\in G$ , $a\in T$ , we have $|[x,a]^G|\leq m^k$ , which is $m$ -bounded. The combination of Lemma 2.2 and Theorem 2.3 now implies that $[G, T]' = G''$ has $m$ -bounded order.

Proof. Write $G=S_1\times \ldots \times S_k$ , where $S_i$ are the simple factors. Each of these factors contain a subgroup of index at most $m$ whence, because of the simplicity, it follows that $|S_i|\leq m!$ . Therefore, we only need to show that $k$ is $m$ -bounded.

Recall that simple groups have even order [Reference Feit and Thompson6]. For every $i=1,\ldots ,k$ , choose a Sylow 2-subgroup $P_i$ of $S_i$ and elements $x_i\in P_i$ and $y_i\in S_i$ such that $1\neq [x_i,y_i]\in Z(P_i)$ . Indeed, if the subgroup $P_i$ is non-abelian, both elements $x_i$ and $y_i$ can be chosen inside $P_i$ . If $P_i$ is abelian, then the existence of $x_i$ and $y_i$ is immediate from the theorem of Frobenius [Reference Gorenstein7, Theorem 7.4.5]. For every $i=1,\ldots ,k$ set $g_i=[x_i,y_i]$ . Since the index of the centralizer of an element in a simple group is not a prime power [Reference Gorenstein7, Lemma 4.3.2], it follows that $|g_i^{S_i}|\geq 15$ . Let $g=\prod _ig_i$ . Observe that $g=[\prod x_i,\prod y_i]\in \prod P_i$ . We deduce that $g\in X(G)$ and $|g^{G}|\geq 15^k$ . Hence, $15^k\leq m$ and so $k\leq \log _{15}(m)$ .

Proof. Without loss of generality, we may assume that $R=1$ . Then, $F^*(G)$ is semisimple and by Lemma 3.4 the order of $F^*(G)$ is $m$ -bounded. Since $|G|\leq |F^*(G)|!$ , we immediately deduce that the order of $G$ is $m$ -bounded. This proves the lemma.

Proof of Theorem 1.3. Recall that $G$ is a finite group in which $|x^G|\leq m$ whenever $x$ is a commutator of prime power order. We wish to prove that $G''$ has $m$ -bounded order.

We will prove the stronger result that if $G$ is a finite group in which $|x^G|\leq m$ whenever $x\in X(G)$ , then $G''$ has $m$ -bounded order. We observe that the condition is inherited by subgroups and quotients of $G$ .

Let $H=G^{(\infty )}$ be the last term of the derived series of $G$ . Of course, $H=H'$ . Note that $G/H$ is soluble and so by Lemma 3.3, the second commutator subgroup of $G/H$ has $m$ -bounded order. Therefore, it is sufficient to show that $H$ has $m$ -bounded order. Hence, without loss of generality, we assume that $G=H$ , that is, $G=G'$ .

Let $R=R(G)$ be the soluble radical of $G$ . Lemma 3.5 tells us that $R$ has $m$ -bounded index in $G$ . Therefore, we can use induction on the index $[G\, : \,R]$ .

If $[G\, : \,R]=1$ , then $G$ is soluble. Since $G=G'$ , it follows that $G=1$ and we are done. So we assume that $[G\, : \,R]\geq 2$ . Suppose that $G$ contains a normal subgroup $M$ such that $R\leq M\lt G$ . It is clear that $[M\, : \,R(M)]\lt [G\, : \,R]$ , so by induction $|M''|$ is $m$ -bounded. Certainly, $M/M''\leq R(G/M'')$ . Therefore, the index of the soluble radical in $G/M''$ is strictly less than $[G\, : \,R]$ and the result follows by induction.

Hence, we can assume that such subgroups $M$ do not exist and so $G/R$ is simple. Set $X=X(G)$ and observe that, since $G=G'$ , it follows that $G$ is generated by $X$ . By Lemma 2.5, the order of the subgroup $[G,x]$ is $m$ -bounded whenever $x\in X$ .

Taking into account that $G/R$ is a nontrivial simple group, we conclude that there is $x_0\in X$ such that $[G,x_0]R=G$ . Observe that the quotient $G/[G,x_0]$ is soluble. Since $G=G'$ , we deduce that $G=[G,x_0]$ . Hence, the order of $G$ is $m$ -bounded, as required.