Proof. We first prove that, given $x\in A$ and $y\in B$, the order of $[x,y]$ is $m$-bounded. Let $H=\langle x,y\rangle$. By assumptions, $[A:C_A(y)]\leq m$ and $[B:C_B(x)]\leq m$. Hence there exists an $m$-bounded number $l$ such that $x^{l}$ and $y^{l}$ are contained in $Z(H)$ (e.g. we can take $l=m!$). Let $D=A\cap B \cap H$ and $N=\langle D,x^{l},y^{l}\rangle$. Then $H/N$ is abelian of order at most $l^{2}$. Both $x$ and $y$ have centralizers of index at most $m$ in $N$. Moreover every element of $N$ has centralizer of index at most $m$ in $N$. Indeed $|d^{N}| \le |d^{A}|\le m$ for every $d \in D \le A\cap B$. So, every element of $H$ is a product of at most $l^{2}+1$ elements each of which has centralizer of index at most $m$ in $N$. Therefore each element of $H$ has centralizer of $m$-bounded index in $H$. We conclude that $H$ is a BFC-group in which the sizes of conjugacy classes are $m$-bounded. Hence $|H'|$ is $m$-bounded and so the order of $[x,y]$ is $m$-bounded, too.

Now we claim that for every $x \in A$, the subgroup $[x,B]$ has finite $m$-bounded order. Indeed, $x$ has at most $m$ conjugates $\{x^{b_1}, \dots, x^{b_m} \}$ in $B$, where $b_1, \dots, b_m \in B$, so $[x,B]$ is generated by at most $m$ elements. Let $C$ be a maximal normal subgroup of $B$ contained in $C_B(x)$. Clearly $C$ has $m$-bounded index in $B$ and centralizes $[x,B]$. Thus, the centre of $[x,B]$ has $m$-bounded index in $[x,B]$. It follows from Schur's theorem [Reference Robinson30, 10.1.4] that the derived subgroup of $[x,B]$ has finite $m$-bounded order. Since $[x,B]$ is generated by at most $m$ elements of $m$-bounded order, we deduce that the order of $[x,B]$ is finite and $m$-bounded.

Choose $a\in A$ such that $[B:C_B(a)]=\max _{x \in A} [B:C_B(x)]$ and set $n=[B:C_B(a)]$, where $n \le m$. Let $b_1,\dots, b_n$ be elements of $B$ such that $a^{B}=\{a^{b_1},\dots, a^{b_n}\}$ is the set of (distinct) conjugates of $a$ by elements of $B$. Set $U=C_A(b_1,\dots,b_n)$ and note that $U$ has $m$-bounded index in $A$. Given $u\in U$, the elements $(ua)^{b_1},\dots, (ua)^{b_n}$ form the conjugacy class $(ua)^{B}$ because they are all different and their number is the allowed maximum. So, for an arbitrary element $y\in B$ there exists $i$ such that $(ua)^{y}=(ua)^{b_i}=u a^{b_i}$. It follows that $u^{-1}u^{y}=a^{b_i}a^{-y}$, hence

\[ [u,y]=a^{b_i}a^{{-}y} =[a, b_i^{a^{{-}1}}][ y^{a^{{-}1}},a] \in [a, B]. \]

Therefore $[U,B]\leq [a,B]$. Let $a_1,\dots,a_s$ be coset representatives of $U$ in $A$ and note that $s$ is $m$-bounded. As each $[x,B]$ is normal in $B$ and $[U,B]\leq [a,B]$, we deduce that $[A,B]=[a,B]\prod [a_i,B]$. So $[A,B]$ is a product of $m$-boundedly many subgroups of $m$-bounded order. These subgroups are normal in $B$ and therefore their product has finite $m$-bounded order.

Proof. Without loss of generality we can assume that $G=AB$. Set $L=\langle B^{G}\rangle =\langle B^{A}\rangle$.

Let $x\in A$. There is an $m$-bounded number $l$ such that $x$ centralizes $y^{l}$ for every $y\in B$. Since $L$ is abelian, $[x,y]^{i}=[x,y^{i}]$ for each $i$ and therefore the order of $[x,y]$ is at most $l$. Thus $[x,B]$ is an abelian subgroup generated by at most $m$ elements of $m$-bounded order, whence $[x,B]$ has finite $m$-bounded order.

Now we choose $a\in A$ such that $[B:C_B(a)]$ is as big as possible. Let $b_1,\dots,b_{m}$ be elements of $B$ such that $a^{B}=\{a^{b_1},\dots,a^{b_{m}}\}$. Set $U=C_A(b_1,\dots,b_m)$ and note that $U$ has $m$-bounded index in $A$. Arguing as in the previous lemma, we see that for arbitrary $u\in U$ and $y\in B$, the conjugate $(ua)^{y}$ belongs to the set $\{(ua)^{b_1},\dots, (ua)^{b_m}\}$. Let $(ua)^{y}=(ua)^{b_i}$. Then $u^{-1}u^{y}=a^{b_i}a^{-y}$ and hence $[u,y]=a^{b_i}a^{-y}\in [a, B]$. Therefore $[U,B]\leq [a,B]$.

Let $V=\cap _{x\in A}U^{x}$ be the maximal normal subgroup of $A$ contained in $U$. We know that $[V,B]$ has $m$-bounded order, since $[V,B]\le [a,B]$. Denote the index $[A:V]$ by $s$. Evidently, $s$ is $m$-bounded. Let $a_1,\dots,a_s$ be a transversal of $V$ in $A$. As $[V,B] \le L=\langle B^{A}\rangle$ is abelian, we have

\[ \langle[V,B]^{G}\rangle=\langle[V,B]^{A}\rangle=\prod_{i=1}^{s}[V,B]^{a_i}. \]

Thus $[V,L]=[V, B^{A}] = \langle [V,B]^{A}\rangle$ is a product of $m$-boundedly many subgroups of $m$-bounded order, and hence it has $m$-bounded order. Write

\[ L=\langle B^{A}\rangle \leq \langle B^{V a_i} \mid i=1, \dots s \rangle \leq [V,L]\prod_{i=1}^{s}B^{a_i}. \]

Thus, it becomes clear that $L$ is a product of $m$-boundedly many conjugates of $B$. Say $L$ is a product of $t$ conjugates of $B$. Then, every $y \in L$ can be written as a product of at most $t$ conjugates of elements of $B$ and consequently $[A: C_A(y)] \le m^{t}.$ Moreover, as $A$ is normal in $G$ and $|a^{B}| \le m$ for every $a\in A$, the conjugacy class $x^{L}$ of an element $x \in A$ has size at most $m^{t}$. Now lemma 2.1 shows that $[A,B] \le [A,L]$ has finite $m$-bounded order.

Proof. In what follows $\bar {G}=G/N$ and $\bar {K}=KN/N$. Write $\bar {K_0}$ for the set of cosets $(N\cap K)h$ with $h\in K$. If $S_0=(N\cap K)h\in \bar {K_0}$, write $S$ for the coset $Nh\in \bar {K}$. Of course, we have a natural one-to-one correspondence between $\bar {K_0}$ and $\bar {K}$.

Write

\begin{align*} |K||G|Pr(K,G)&=\sum_{x\in K}|C_G(x)|=\sum_{S_0\in\bar{K_0}}\sum_{x\in S_0}\frac{|C_G(x)N|}{|N|}|C_N(x)|\\ &\leq\sum_{S_0\in\bar{K_0}}\sum_{x\in S_0}|C_{\bar{G}}(xN)||C_N(x)|=\sum_{S\in\bar{K}}|C_{\bar{G}}(S)|\sum_{x\in S_0}|C_N(x)|\\ &=\sum_{S\in\bar{K}}|C_{\bar{G}}(S)|\sum_{y\in N}|C_{S_0}(y)|.\\ \end{align*}

If $C_{S_0}(y)\neq \emptyset$, then there is $y_0\in C_{S_0}(y)$ and so $S_0=(N\cap K)y_0$. Therefore

\[ C_{S_0}(y)=(N\cap K)y_0\cap C_G(y)=C_{N\cap K}(y)y_0,\quad\text{whence}\ |C_{S_0}(y)|=|C_{N\cap K}(y)|. \]

Conclude that

\[ |K||G|Pr(K,G)\leq \sum_{S\in\bar{K}}|C_{\bar{G}}(S)|\sum_{y\in N}|C_{N\cap K}(y)|. \]

Observe that

\[ \sum_{S\in\bar{K}}|C_{\bar{G}}(S)|=\frac{|K|}{|N\cap K|}\frac{|G|}{|N|}Pr(\bar{K},\bar{G}) \]

and

\[ \sum_{y\in N}|C_{N\cap K}(y)|=|N\cap K||N|Pr(N\cap K,N). \]

It follows that $Pr(K,G)\leq Pr(\bar {K},\bar {G})Pr(N\cap K,N)$, as required.

Proof of proposition 1.2. Set

\[ X=\{x\in K \mid |x^{G}|\leq 2/\epsilon\}\quad \text{and}\quad B=\langle X\rangle. \]

Note that $K \setminus X=\{ x\in K \mid |C_G(x)|\leq (\epsilon /2) |G| \}$, whence

\begin{align*} \epsilon |K||G| &\le | \{ (x,y) \in K\times G \mid xy=yx \} |=\sum_{x\in K}|C_G(x)|\\ &\le \sum_{x \in X} |G| + \sum_{x \in K \setminus X} \frac{\epsilon}{2} |G|\\ &\le |X| |G| +(|K| - |X|)\frac{\epsilon}{2}|G|. \end{align*}

Therefore $\epsilon |K| \le |X|+ ({\epsilon }/{2}) (|K| - |X|)$, whence $({\epsilon }/{2}) |K| < |X|$. Clearly, $|B| \ge |X| > ({\epsilon }/{2}) |K|$ and so the index of $B$ in $K$ is at most $2/\epsilon$. As $X$ is symmetric and $(2/\epsilon ) |X| > |K|$, it follows from lemma 2.5 that every element of $B$ is a product of at most $6/\epsilon$ elements of $X$. Therefore $|b^{G}| \le (2/\epsilon )^{6/\epsilon }$ for every $b \in B$.

Let $L=\langle B^{G}\rangle$. Theorem 2.4 tells us that the commutator subgroup $[L,L]$ has $\epsilon$-bounded order. Let us use the bar notation for the images of the subgroups of $G$ in $G/[L,L]$. By lemma 2.3,

\[ Pr(\bar K, \bar G) \ge Pr(K,G)\geq\epsilon. \]

Moreover, $[\bar K: \bar B] \le [K:B] \le {\epsilon }/{2}$ and $|\bar b^{ \bar G}| \le |b^{G}| \le (2/\epsilon )^{6/\epsilon }$. Thus we can pass to the quotient over $[L,L]$ and assume that $L$ is abelian.

Now we set

\[ Y= \{ y \in G \mid |y^{K}| \le 2/\epsilon\} = \{ y \in G \mid |C_K(y)| \ge ({\epsilon}/{2}) |K|\}. \]

Note that

\begin{align*} \epsilon |K||G| & \le | \{ (x,y) \in K \times G \mid xy=yx \} |\\ &\le \sum_{y \in Y} |K| + \sum_{y \in G \setminus Y} \frac{\epsilon}{2} |K|\\ &\le |Y| |K| +(|G| - |Y|)\frac{\epsilon}{2}|K| \le |Y| |K|+\frac{\epsilon}{2}|G| |K|. \end{align*}

Therefore $({\epsilon }/{2}) |G| < |Y|.$

Set $E= \langle Y \rangle$. Thus $|E| \ge |Y| > ({\epsilon }/{2}) |G|$, and so the index of $E$ in $G$ is at most $2/\epsilon$. As $Y$ is symmetric and $(2/\epsilon ) |Y| > |G|$, it follows from lemma 2.5 that every element of $E$ is a product of at most $6/\epsilon$ elements of $Y$. Since $|y^{K}|\le 2/\epsilon$ for every $y\in Y$, we conclude that $|e^{K}|\le (2/\epsilon )^{6/\epsilon }$ for every $e\in E$. Let $T$ be the maximal normal subgroup of $G$ contained in $E$. Clearly, the index $[G:T]$ is $\epsilon$-bounded.

So, now $|b^{G}| \le (2/\epsilon )^{6/\epsilon }$ for every $b \in B$ and $|e^{B}| \le (2/\epsilon )^{6/\epsilon }$ for every $e\in T$. As $L$ is abelian, we can apply lemma 2.2 to conclude that $[T,B]$ has $\epsilon$-bounded order and the result follows.

Proof. Since $[T,B]$ is normal in $T$, it follows that there are only boundedly many conjugates of $[T,B]$ in $G$ and they normalize each other. Since $N$ is the product of those conjugates, $N$ has $\epsilon$-bounded order.

Proof. According to remark 2.6, $N=\langle [T,B]^{G}\rangle$ has $\epsilon$-bounded order. Let $B_0=\langle B^{G}\rangle$ and note that $B_0 \le K$ and $[T,B_0]\leq N$. Since the index $[K:B_0]$ and the order of $N$ are $\epsilon$-bounded, the stabilizer in $T$ of the series

\[ 1\leq N\leq B_0\leq K, \]

that is, the subgroup

\[ H=\{g\in T\ \vert\ [N,g]=1\ \& \ [K,g]\leq B_0\} \]

has $\epsilon$-bounded index in $G$. Note that $K\cap H\leq Z_3(H)$, whence the result.

Proof of Theorem 1.3. Set $K=F^{*}(G)$. In view of proposition 1.2 there is a normal subgroup $T\leq G$ and a subgroup $B\leq K$ such that the indices $[G:T]$ and $[K:B]$, and the order of the commutator subgroup $[T,B]$ are $\epsilon$-bounded. As $K$ is normal in $G$, according to remark 2.7 the subgroup $T$ can be chosen in such a way that $K\cap T\leq Z_3(T)$. By [Reference Huppert and Blackburn20, corollary X.13.11(c)] we have $K\cap T=F^{*}(T)$. Therefore $F^{*}(T)\leq Z_3(T)$ and in view of [Reference Huppert and Blackburn20, theorem X.13.6] we conclude that $T=F^{*}(T)$ and so $T\leq K$. It follows that the index of $K$ in $G$ is $\epsilon$-bounded. By remark 2.6 the subgroup $N=\langle [T,B]^{G}\rangle$ has $\epsilon$-bounded order. Conclude that $R=\langle B^{G}\rangle \cap C_G(N)$ has $\epsilon$-bounded index in $G$. Moreover $R$ is nilpotent of class at most 2 and $[R,R]$ has $\epsilon$-bounded order. This completes the proof.

Proof of theorem 1.4. Recall that $K$ is a subgroup of the finite group $G$ such that $\gamma _k(G)\leq K$ and $Pr(K,G)\geq \epsilon$. In view of [Reference Erfanian, Rezaei and Lescot10, theorem 3.7] observe that $Pr(\gamma _k(G),G)\geq \epsilon$. Therefore without loss of generality we can assume that $K=\gamma _k(G)$.

Proposition 1.2 tells us that there is a normal subgroup $T\leq G$ and a subgroup $B\leq K$ such that the indices $[G:T]$ and $[K:B]$ and the order of $[T,B]$ are $\epsilon$-bounded. In particular, $|x^{B}|$ is $\epsilon$-bounded for every $x\in T$. Since $B$ has $\epsilon$-bounded index in $K$, we deduce that $|x^{\gamma _k(G)}|$ is $\epsilon$-bounded for every $x\in T$. Now theorem 3.1 implies that $\gamma _{k+1}(T)$ has $\epsilon$-bounded order. Set $R=C_T(\gamma _{k+1}(T))$. It follows that $R$ is as required.

Proof of theorem 1.5. Recall that $w$ is a group-word implying virtual nilpotency while $K$ is a subgroup of a finite group $G$ such that $w(G)\leq K$ and $Pr(K,G)\geq \epsilon$. We need to show that there is an $(\epsilon,w)$-bounded integer $e$ and a $w$-bounded integer $c$ such that $G^{e}$ is nilpotent of class at most $c$.

As in the proof of theorem 1.4 without loss of generality we can assume that $K=w(G)$. Proposition 1.2 tells us that there is a normal subgroup $T\leq G$ and a subgroup $B\leq K$ such that the indices $[G:T]$ and $[K:B]$ and the order of the commutator subgroup $[T,B]$ are $\epsilon$-bounded. According to remark 2.7 the subgroup $T$ can be chosen in such a way that $K\cap T\leq Z_3(T)$. In particular $w(T)\leq Z_3(T)$. Taking into account that the word $w$ implies virtual nilpotency, we deduce from the Burns–Medvedev theorem that there are $w$-bounded numbers $i$ and $c$ such that the subgroup generated by the $i$th powers of elements of $T$ is nilpotent of class at most $c$. Recall that the index of $T$ in $G$ is $\epsilon$-bounded. Hence there is an $\epsilon$-bounded integer $e$ such that every $e$th power in $G$ is an $i$th power of an element of $T$. The result follows.

Proof. By [Reference Erfanian, Rezaei and Lescot10, theorem 3.7], without loss of generality we can assume that $K=w(G)$. Proposition 1.2 tells us that there is a normal subgroup $T\leq G$ and a subgroup $B\leq w(G)$ such that the indices $[G:T]$ and $[w(G):B]$ and the order of $[T,B]$ are $\epsilon$-bounded. Since $K$ is normal in $G$, according to remark 2.7 the subgroup $T$ can be chosen in such a way that $w(G)\cap T\leq Z_3(T)$. If $w=[x^{n},y_1,\dots,y_k]$, then $[x^{n},y_1,\dots,y_{k+3}]$ is a law in $T$, whence the exponent of $\gamma _{k+4}(T)$ is $w$-bounded. If $w$ is the $k$-Engel word, then the $(k+3)$-Engel word is a law in $T$ and the theorem follows from the Burns–Medvedev theorem [Reference Burns and Medvedev3].

Proof of Theorem 1.6. Proposition 1.2 tells us that there is a normal subgroup $T\leq G$ and a subgroup $B\leq P$ such that the indices $[G:T]$ and $[P:B]$ and the order of the commutator subgroup $[T,B]$ are $\epsilon$-bounded. In view of remark 2.6 the subgroup $N=\langle [T,B]^{G}\rangle$ has $\epsilon$-bounded order. Therefore $C=C_T(N)$ has $\epsilon$-bounded index in $G$. Set $B_0=B\cap C$ and note that $[C,B_0]\leq Z(C)$. It follows that $B_0\leq Z_2(C)$ and we conclude that $B_0\leq O_p(G)$. Let $L=\langle {B_0}^{G}\rangle$. As $B_0 \le L \le O_p(G)$, it is clear that $L$ is contained in $P$ as a subgroup of $\epsilon$-bounded index. Moreover $[L,L]\leq N$ and so the order of $[L,L]$ is $\epsilon$-bounded. Hence the result.

Proof of theorem 1.7. Recall that $G$ is a finite group such that $Pr(P,G) \ge \epsilon$ whenever $P$ is a Sylow subgroup. We wish to show that $G$ has a nilpotent normal subgroup $R$ of nilpotency class at most $2$ such that both the index $[G:R]$ and the order of the commutator subgroup $[R,R]$ are $\epsilon$-bounded.

For each prime $p\in \pi (G)$ choose a Sylow $p$-subgroup $S_p$ in $G$. Theorem 1.6 shows that $G$ has a normal $p$-subgroup $L_p$ of class at most $2$ such that both $[S_p:L_p]$ and $|[L_p,L_p]|$ are $\epsilon$-bounded. Since the bounds on $[S_p:L_p]$ and $|[L_p,L_p]|$ do not depend on $p$, it follows that there is an $\epsilon$-bounded constant $C$ such that $S_p=L_p$ and $[L_p,L_p]=1$ whenever $p\geq C$. Set $R=\prod _{p\in \pi (G)}L_p$. Then all Sylow subgroups of $G/R$ have $\epsilon$-bounded order and therefore the index of $R$ in $G$ is $\epsilon$-bounded. Moreover, $R$ is of class at most $2$ and $|[R,R]|$ is $\epsilon$-bounded, as required.

Proof of theorem 1.8. Recall that $G$ is a finite group admitting a coprime automorphism $\phi$ of prime order $p$ such that $Pr(K,G)\geq \epsilon$, where $K=C_G(\phi )$. We need to show that $G$ has a nilpotent subgroup of $p$-bounded nilpotency class and $(\epsilon,p)$-bounded index.

By proposition 1.2 there is a normal subgroup $T\leq G$ and a subgroup $B\leq K$ such that the indices $[G:T]$ and $[K:B]$ and the order of the commutator subgroup $[T,B]$ are $\epsilon$-bounded. Let $T_0$ be the maximal $\phi$-invariant subgroup of $T$. Evidently, $T_0$ is normal and the index $[G:T_0]$ is $(\epsilon,p)$-bounded. Since $\langle [T_0,B]^{G}\rangle \leq \langle [T,B]^{G}\rangle$, remark 2.6 implies that $M=\langle [T_0,B]^{G}\rangle$ has $\epsilon$-bounded order. Moreover, $M$ is $\phi$-invariant. Set $D=C_G(M)\cap T_0$ and $\bar {D}=D/Z_2(D)$, and note that $D$ is $\phi$-invariant.

In a natural way $\phi$ induces an automorphism of $\bar {D}$ which we will denote by the same symbol $\phi$. We note that $C_{\bar {D}}(\phi )=C_D(\phi ) Z_2(D)/Z_2(D)$, so its order is $\epsilon$-bounded because $B\cap D\leq Z_2(D)$. The Khukhro theorem [Reference Khukhro21] now implies that $\bar {D}$ has a nilpotent subgroup of $p$-bounded class and $(\epsilon,p)$-bounded index. Since $\bar {D}=D/Z_2(D)$ and since the index of $D$ in $G$ is $(\epsilon,p)$-bounded, we deduce that $G$ has a nilpotent subgroup of $p$-bounded class and $(\epsilon,p)$-bounded index. The proof is complete.

Proof of theorem 1.9. By hypotheses, $G$ is a finite group admitting an elementary abelian coprime group of automorphisms $A$ of order $p^{2}$ such that $Pr(C_G(\phi ),G)\geq \epsilon$ for each $\phi \in A^{\#}$. We need to show that $G$ has a nilpotent normal subgroup $R$ of nilpotency class at most $2$ such that both the index $[G:R]$ and the order of the commutator subgroup $[R,R]$ are $(\epsilon,p)$-bounded.

Let $A_1,\dots,A_{p+1}$ be the subgroups of order $p$ of $A$ and set $G_i=C_G(A_i)$ for $i=1,\dots,p+1$. According to proposition 1.2 for each $i=1,\dots,p+1$ there is a normal subgroup $T_i\leq G$ and a subgroup $B_i\leq G_i$ such that the indices $[G:T_i]$ and $[G_i:B_i]$ and the order of the commutator subgroup $[T_i,B_i]$ are $\epsilon$-bounded. We let $U_i$ denote the maximal $A$-invariant subgroup of $T_i$ so that each $U_i$ is a normal subgroup of $(\epsilon,p)$-bounded index. The intersection of all $U_i$ will be denoted by $U$. Further, we let $D_i$ denote the maximal $A$-invariant subgroup of $B_i$ so that each $D_i$ has $(\epsilon,p)$-bounded index in $G_i$. Note that a modification of remark 2.6 implies that $N_i=\langle [U_i,D_i]^{G}\rangle$ is $A$-invariant and has $\epsilon$-bounded order. It follows that the order of $N=\prod _iN_i$ is $(\epsilon,p)$-bounded. Let $V$ denote the minimal ($A$-invariant) normal subgroup of $G$ containing all $D_i$ for $i=1,\dots,p+1$. It is easy to see that $[U,V]\leq N$.

Obviously, $U$ has $(\epsilon,p)$-bounded index in $G$. Let us check that this also holds with respect to $V$. Let $\bar {G}=G/V$. Since $V$ contains $D_i$ for each $i=1,\dots,p+1$ and since $D_i$ has $(\epsilon,p)$-bounded index in $G_i$, we conclude that the image of $G_i$ in $\bar {G}$ has $(\epsilon,p)$-bounded order. Now lemma 5.2 tells us that the order of $\bar {G}$ is $(\epsilon,p)$-bounded and we conclude that indeed $V$ has $(\epsilon,p)$-bounded index in $G$. Also note that since $N$ has $(\epsilon,p)$-bounded order, $C_G(N)$ has $(\epsilon,p)$-bounded index in $G$. Let

\[ R=U\cap V\cap C_G(N). \]

Then $R$ is as required since the subgroups $U,V,C_G(N)$ have $(\epsilon,p)$-bounded index in $G$ while $[R,R]\leq N\leq C_G(R)$. The proof is complete.