Proof. Assume by contradiction that there exists an element x of G such that ${[A, x]\neq \{1\}}$ and suppose first that $\langle x\rangle \ntrianglelefteq G$ . Let $a\in A$ . From [Reference Robinson7, Part 1, Lemma 2.12],

$$ \begin{align*}[\langle x\rangle , \langle a^n\rangle ]=[x,a^n]^{\langle x\rangle \langle a^n\rangle }.\end{align*} $$

Moreover, since A is an abelian normal subgroup of G,

$$ \begin{align*}[x,a^n]^{\langle x\rangle \langle a^n\rangle }=[x,a^n]^{\langle x\rangle }=([x,a]^n)^{\langle x\rangle }=([x,a]^{\langle x\rangle })^n.\end{align*} $$

It follows that the set $\{([x,a]^{\langle x\rangle })^{r!} \mid r\in \mathbb {N}_0\}$ is finite. Consequently, there exists ${k\in \mathbb {N}_0}$ such that

$$ \begin{align*}([x,a]^{\langle x\rangle})^{k!}=([x,a]^{\langle x\rangle})^{(k+1)!}=\cdots .\end{align*} $$

Thus, the subgroup $([x,a]^{\langle x\rangle })^{k!}$ is divisible. However, $\langle x,a\rangle $ is a finitely generated metabelian group, so that it is residually finite (see [Reference Robinson7, Part 2, Theorem 9.51]) and hence, $[x,a]^{k!}=1$ . Therefore, $[x,a]=1$ and $A \leq C_G(x)$ .

Suppose now that $\langle x\rangle \lhd G$ . Then, there exists an integer n such that $\langle x^n\rangle =[x,A]\leq Z(\langle x,A\rangle )$ . Thus, $[A,x,x]=\{1\}=[[A,x],x]$ . Then, $[A,x^n]=[A,x]^n=\{1\}$ and ${[A,x]=\{1\}}$ . Therefore, $A\leq C_G(x)$ for each $x\in G$ .

Proof. Assume for a contradiction that the statement is false and let T be the largest periodic normal subgroup of G. Clearly, the factor group $G/T$ is also a counterexample, so we may suppose without loss of generality that G has no periodic nontrivial normal subgroups. Among all such counterexamples, choose G in such a way that its soluble radical S has minimal derived length m. Let U be the smallest nontrivial term of the derived series of S and let A be a maximal abelian normal subgroup of G containing U. Clearly, A is torsion-free and so it is contained in $Z(G)$ by Lemma 2.1. If $L/A$ is the largest periodic normal subgroup of $G/A$ , the factor group $L/Z(L)$ is locally finite, so that $L'$ is locally finite by Schur’s theorem (see [Reference Robinson7, Part 1, Corollary to Theorem 4.12]) and hence, $L'=\{1\}$ . Thus, L is abelian, so that $L=A$ , and hence, $G/A$ has no periodic nontrivial normal subgroups. However, the soluble radical of $G/A$ has derived length less then m and so the minimal choice of G implies that $G/A$ is abelian. Therefore, G is nilpotent and hence, $A=C_G(A)=G$ . This contradiction completes the proof of the lemma.

Proof. Let x be an element of G. If $\langle x\rangle \lhd G$ , then x is an ${\textit {FC}}$ -element of G. Otherwise, if $\langle x\rangle \ntrianglelefteq G$ , then there exist $y_1,\ldots ,y_t\in G$ such that

$$ \begin{align*}\{[\langle x\rangle, \langle y\rangle] \mid y\in G\}=\{[\langle x\rangle, \langle y_1\rangle], \ldots, [\langle x\rangle, \langle y_t\rangle]\}.\end{align*} $$

It follows that $[\langle x\rangle ,G]$ is finitely generated and so, it is finite. Therefore, x is an ${\textit {FC}}$ -element.

Proof. Let $\mathcal {L}$ be the set of all normal subgroups of finite index of G. If x is any element of G such that $\langle x\rangle \lhd G$ , then x is an $FC$ -element. Consider $x\in G$ such that $\langle x\rangle \ntrianglelefteq G$ . By hypothesis, there exists a finite subset $\mathcal {F}$ of $\mathcal {L}$ such that

$$ \begin{align*}\{[\langle x\rangle,H]=[x,H]\mid H\in \mathcal{L}\}=\{[x,H]\mid H\in \mathcal{F}\}.\end{align*} $$

Then,

$$ \begin{align*}\bigg[x,\bigcap_{H\in \mathcal{F}}H\bigg]\leq\bigcap_{H\in \mathcal{F}}[x,H]=\bigcap_{H\in \mathcal{L}}[x,H]\leq \bigcap_{H\in \mathcal{L}}H=\{1\}\end{align*} $$

and hence, the subgroup $\bigcap _{H\in \mathcal {F}}H$ is contained in $C_G(x)$ . As the index

$$ \begin{align*}\bigg|G:\bigcap_{H\in \mathcal{F}}H\bigg|\end{align*} $$

is finite, it follows that x has finitely many conjugates in G. Therefore, G is an ${\textit {FC}}$ -group.

Proof. Let a be an element of infinite order of G. Let x be an element of G. If $\langle x\rangle \lhd G$ , then x is an $FC$ -element. Assume that $\langle x\rangle \ntrianglelefteq G$ . Then, there exist $y_1,\ldots ,y_k\in G$ such that

$$ \begin{align*}\{[\langle x\rangle, \langle y,a\rangle] \mid y\in G\}=\{[\langle x\rangle, \langle y_1,a\rangle], \ldots, [\langle x\rangle, \langle y_k,a\rangle]\}.\end{align*} $$

For a fixed i in the set $\{1,\ldots ,k\}$ , the subgroup $\langle x,y_i,a\rangle $ is residually finite and satisfies the maximal condition on normal subgroups. Thus, $\langle x,y_i,a\rangle '$ is finite by Lemma 2.5. Therefore, $[x,G]$ is finite and x is an ${\textit {FC}}$ -element. Let X be a nonnormal subgroup of G. Then, there exist $g_1,\ldots ,g_t\in G$ such that

$$ \begin{align*}\{[X,\langle g,a\rangle]\mid g\in G\}=\{[X,\langle g_1,a\rangle], \ldots, [X,\langle g_t,a\rangle]\},\end{align*} $$

where $[X,\langle g_j,a\rangle ]$ is finitely generated for all $j\in \{1,\ldots ,t\}$ , since G is an ${\textit {FC}}$ -group. Thus, $[X,G]$ is finite. Consider now $X^G=X[X,G]$ . We have

$$ \begin{align*}|X^G:X|=|X[X,G]:X|=|[X,G]:(X\cap [X,G])|<\infty.\end{align*} $$

Thus, each subgroup of G is nearly normal and so $G'$ is finite by Neumann’s theorem [Reference Neumann6].

Proof. Let x be an element of G. If $\langle x\rangle \lhd G$ , then x is an ${\textit {FC}}$ -element of G. Otherwise, if $\langle x\rangle \ntrianglelefteq G$ , then there exist $g_1,\ldots ,g_t\in G$ such that

$$ \begin{align*}\{[\langle x\rangle, \langle g\rangle] \mid g\in G\}=\{[\langle x\rangle, \langle g_1\rangle],\ldots,[\langle x\rangle, \langle g_t\rangle]\}.\end{align*} $$

For a fixed i in the set $\{1,\ldots ,t\}$ , let us consider $E_i=\langle x,g_i\rangle $ . Let X be a nonnormal subgroup of $E_i'$ . Then, there exist $y_1,\ldots ,y_k\in E_i'$ such that

$$ \begin{align*}\{[X,\langle y\rangle] \mid y\in E_i'\}=\{[X,\langle y_1\rangle],\ldots,[X,\langle y_k\rangle]\}.\end{align*} $$

By Corollary 2.4, $\langle y_j\rangle ^{E_i'}$ is finite for each $j\in \{1,\ldots ,k\}$ . It follows that $[X,E_i']$ is finitely generated and so it is finite. Observe that

$$ \begin{align*}|X^{E_i'}:X|=|X[X,E_i']:X|=|[X,E_i']:(X\cap [X,E_i'])|<\infty.\end{align*} $$

Therefore, each subgroup of $E_i'$ is nearly normal and so $E_i"$ is finite [Reference Neumann6]. For a fixed i, since $E_i/E_i"$ is finitely generated, it is residually finite and satisfies the maximal condition on normal subgroups. Thus, $E_i/E_i"$ is an ${\textit {FC}}$ -group by Lemma 2.5 and, by Dietzmann’s lemma (see [Reference Robinson7, Part 1, Corollary 2 to Lemma 2.14]), $E_i'/E_i"$ is finite; it follows that $E_i'$ is finite. Therefore, $[x,G]$ is finite and G is an ${\textit {FC}}$ -group.

Proof of Theorem 1.1.

Assume X is not a normal subgroup of G. Then, there exist $g_1,\ldots ,g_t\in G$ such that

$$ \begin{align*}\{[X,\langle g\rangle] \mid g\in G\}=\{[X,\langle g_1\rangle],\ldots,[X,\langle g_t\rangle]\}.\end{align*} $$

Since G is an ${\textit {FC}}$ -group, $[X,G]$ is finitely generated and thus it is finite. Consider now $X^G=X[X,G]$ . We have

$$ \begin{align*}|X^G:X|=|X[X,G]:X|=|[X,G]:(X\cap [X,G])|<\infty.\end{align*} $$

Thus, each subgroup of G is nearly normal and so $G'$ is finite by Neumann’s theorem [Reference Neumann6].

Proof. Let X be a nonnormal subgroup of G, then $[X,H]=[X,HZ(G)]$ for every subgroup H of G and hence, the set $\{[X,H]\mid H\leq G\}=\{[X,HZ(G)]\mid H\leq G\}$ is finite.

Proof. The result is an immediate consequence of Lemma 3.1 and Theorem 1.1.

Proof. Consider $\bar {G}=G/N$ . Then, for each nonnormal subgroup $\bar {X}$ of $\bar {G}$ , the set $\{[\bar {X},\bar {H}] \mid \bar {H}\leq \bar {G}\}$ is finite and hence, by Theorem 1.1, $\bar {G}'$ is finite.

Proof. Let i be the smallest positive integer such that the ith term $Z_i(G)$ of the upper central series of G is infinite. Then, $G/Z_{i-1}(G)$ satisfies the hypothesis and has an infinite centre. It follows from Corollary 3.2 that ${G'Z_{i-1}(G)}/{Z_{i-1}(G)}$ is finite, so $G'$ is finite since $Z_{i-1}(G)$ is finite.

Proof. To prove that $G'$ is finite, first suppose that G is a periodic ${\textit {FC}}$ -group. Let X be a nonnormal subgroup of G. Let us consider $\bar {G}=G/X_G$ and $\bar {X}=X/X_G$ . If $\bar {X}$ is finite, then it is nearly normal in $\bar {G}$ . It follows that X is nearly normal in G since

$$ \begin{align*}|\bar{X}^{\bar{G}}:\bar{X}|=|X^G/X_G:X/X_G|=|X^G:X|<\infty.\end{align*} $$

Suppose that $\bar {X}$ is infinite. Then, the set $\{[\bar {H},\bar {K}] \mid \bar {K}\leq \bar {X}, \bar {K} \text { is infinite}\}$ is finite for each subgroup $\bar {H}$ of $\bar {X}$ . Since $\bar {X}$ is an ${\textit {FC}}$ -group, $|\bar {X}'|$ is finite (see [Reference Cascella1, Theorem B]). Since $\bar {X}\ntrianglelefteq \bar {G}$ , by hypothesis, there exist $\bar {g}_1,\ldots ,\bar {g}_t\in \bar {G}$ such that

$$ \begin{align*}\{[\bar{X},\langle \bar{X},\bar{g}\rangle] \mid \bar{g}\in \bar{G}\}=\{[\bar{X},\langle \bar{X},\bar{g}_1\rangle],\ldots,[\bar{X},\langle \bar{X},\bar{g}_t\rangle]\}.\end{align*} $$

For a fixed $\bar {g}\in \{\bar {g}_1,\ldots ,\bar {g}_t\}$ , the subgroup $[\bar {X},\langle \bar {X},\bar {g}\rangle ]=\langle \bar {X}',[\bar {X},\bar {g}]\rangle ^{\langle \bar {X},\bar {g}\rangle }$ is finite by Dietzmann’s lemma (see [Reference Robinson7, Part 1, Corollary 2 to Lemma 2.14]). Thus, $[\bar {X},\bar {G}]$ is finite. Let us consider $\bar {X}^{\bar {G}}=\bar {X}[\bar {X},\bar {G}]$ . Then,

$$ \begin{align*}|\bar{X}^{\bar{G}}:\bar{X}|=|\bar{X}[\bar{X},\bar{G}]:\bar{X}|=|[\bar{X},\bar{G}]:[\bar{X},\bar{G}]\cap \bar{X}|<\infty\end{align*} $$

and again, X is nearly normal in G. Therefore, each subgroup of G is nearly normal and so $G'$ is finite by Neumann’s theorem [Reference Neumann6].

Suppose now that G is a nonperiodic ${\textit {FC}}$ -group. Then, by Lemma 2.2, $G'$ is a periodic ${\textit {FC}}$ -group and $G"$ is finite and, by applying Corollary 2.6 to $G/G"$ , it follows that $G'$ is finite.

Proof. Assume for a contradiction that the statement is false, so there exists an element g of G with prime-power order $q^n$ such that the subgroup $[g,A]$ is infinite. Observe that $\langle g\rangle \ntrianglelefteq G$ . If $q=p$ , the subgroup $\langle g,A\rangle $ is nilpotent (see [Reference Robinson7, Part 2, Lemma 6.34]), so that, by Corollary 3.4, $\langle A,g\rangle '$ is finite and hence, $[g,A]$ is finite. This contradiction shows that $q\neq p$ . Thus, A is a completely reducible $\langle g\rangle $ -module and so

$$ \begin{align*}A=\underset{i \in I}{\mathrm{Dr}} \, A_i,\end{align*} $$

where each $A_i$ is a finite $\langle g\rangle $ -invariant subgroup of A on which g acts irreducibly. Let $I'$ be the subset of I consisting of all elements i such that $[\langle g\rangle ,A_i]=[g,A_i]\neq \{1\}$ . Then, $I'$ is infinite and $[\langle g\rangle ,A_i]=[g,A_i]=A_i$ for each $i\in I'$ . It follows that

$$ \begin{align*}[\langle g\rangle, \langle A_j\mid j\in J\rangle]=\underset{j \in J}{\mathrm{Dr}} \, A_j\end{align*} $$

for every subset J of $I'$ . However, $I'$ contains infinitely many infinite subsets, giving a contradiction to the condition that for each nonnormal subgroup X of G, the set $\{[X,H] \mid H\leq G, H \text { is infinite}\}$ is finite.

Proof. By Lemma 3.5, it can be assumed that the subgroup A is infinite. Let x be any element of G. If $\langle x\rangle \lhd G$ , then x is an ${\textit {FC}}$ -element. Suppose that $\langle x \rangle \ntrianglelefteq G$ . Since $G/A$ is an ${\textit {FC}}$ -group, there exist finitely many elements $y_1,\ldots ,y_k$ of G such that ${\langle x\rangle ^GA=E_xA}$ , where $E_x=\langle x^{y_1},\ldots ,x^{y_k}\rangle $ . For each $i=1,\ldots ,k$ , the subgroup $[x^{y_i},A]$ is finite and hence, the centraliser $C_A(x^{y_i})$ has finite index in A. Thus, the index $|A:C_A(E_x)|$ is also finite and so the normal subgroup $C_A(E_x)=C_A(E_xA)$ of G is infinite. By hypothesis, the set

$$ \begin{align*}\{[\langle x\rangle,H]\mid H\leq G\}=\{[\langle x\rangle,HC_A(E_x)]\mid H\leq G\}\end{align*} $$

is finite. Let g be an element of $G'$ . If $\langle g\rangle \lhd G$ , then g is an ${\textit {FC}}$ -element. Otherwise, there exist $l_1,\ldots ,l_s\in G'$ such that

$$ \begin{align*} \{[\langle g\rangle, \langle l\rangle]\mid l\in G'\} =\{[\langle g\rangle, \langle l\rangle C_A(E_g)]\mid l\in G'\} =\{[\langle g\rangle, \langle l_1\rangle],\ldots,[\langle g\rangle, \langle l_s\rangle]\}. \end{align*} $$

It follows that $[\langle g\rangle ,G']$ is finitely generated and, thus, it is finite. Therefore, $G'$ is an ${\textit {FC}}$ -group and, by Lemma 3.5, $G"$ is finite, so that, replacing G by $G/G"$ , it is possible to suppose that G is metabelian. Let $\bar {x}$ be an element of G. If $\langle \bar {x}\rangle \lhd G$ , $\bar {x}$ is an ${\textit {FC}}$ -element of G. Otherwise, if $\langle \bar {x}\rangle \ntrianglelefteq G$ , then there exist $h_1,\ldots ,h_r\in G$ such that

$$ \begin{align*} \{[\langle \bar{x}\rangle, \langle h\rangle]\mid h\in G\} =\{[\langle \bar{x}\rangle, \langle h\rangle C_A(E_{\bar{x}})]\mid h\in G\} =\{[\langle \bar{x}\rangle, \langle h_1\rangle],\ldots,[\langle \bar{x}\rangle, \langle h_r\rangle]\}. \end{align*} $$

For a fixed j in the set $\{1,\ldots ,r\}$ , each $\langle \bar {x},h_j\rangle $ is residually finite and so, by Lemma 2.5, $[\bar {x},G]$ is finite. Thus, G is an ${\textit {FC}}$ -group and, from Lemma 3.5, $G'$ is finite.

Proof. Suppose that G contains a periodic abelian normal subgroup A which does not satisfy the minimal condition on subgroups, and assume for a contradiction that $G'$ is infinite. In particular, by Lemma 3.5, G is not an ${\textit {FC}}$ -group. Then, it follows from Lemma 3.3 that G does not contain two infinite normal subgroups with trivial intersection. In particular, the set of primes $\pi (A)$ is finite and so there exists a prime number p such that the socle B of the p-component $A_p$ of A is infinite. As $G/B$ is an ${\textit {FC}}$ -group, Lemma 3.7 implies that the subgroup $[g,B]$ is infinite for some $g\in G$ , and such an element g must have infinite order by Lemma 3.6. Then, $\langle g,B\rangle $ is not an ${\textit {FC}}$ -group, contradicting Corollary 2.6.

Proof of Theorem 1.2.

The condition of the statement is obviously sufficient. To prove the converse statement, first suppose that the group is soluble and assume for a contradiction that it is neither finite-by-abelian nor a finite extension of a group of type $p^{\infty }$ for some prime number p. Let G be a counterexample with minimal derived length and let A be the smallest nontrivial term of the derived series of G. Then, the factor group $G/A$ is not a counterexample and hence, A must be infinite. Moreover, since A is periodic by Lemma 2.2, it follows from Lemma 3.8 that A satisfies the minimal condition on subgroups. In particular, A is a finite extension of a divisible nontrivial subgroup D, and D is not contained in $Z(G)$ by Corollary 3.2. If g is any element of infinite order of G, the metabelian subgroup $\langle g,D\rangle $ is an ${\textit {FC}}$ -group by Corollary 2.6 and so $[D,g]=\{1\}$ . It follows that G cannot be generated by its element of infinite order, and so it is periodic by Lemma 2.2. Let F be the Fitting subgroup of G and consider an element x of G such that $[D,x]\neq \{1\}$ . Then, the centraliser $C_D(x)$ has infinite index in D and so the index $|F:C_F(x)|$ is also infinite. Let B be a maximal abelian normal subgroup of F containing A. Since $F/A$ is hypercentral, it is easy to see that $C_F(B)=B$ . As G is periodic, the normal closure $K=B^{\langle x,F\rangle }$ is generated by finitely many conjugates of B and so it is nilpotent. Let L be the subgroup generated by all elements of prime order of K. Then, the Sylow subgroups of L have finite exponent and so $L\cap D$ is finite. However, it follows from Lemma 3.3 that $\langle x\rangle K$ cannot contain two infinite normal subgroups with finite intersection, so that L is finite and hence, all abelian subgroups of K have finite socle. Therefore, K satisfies the minimal condition on abelian subgroups and so it is a Černikov group by a result of Šunkov [Reference Šunkov8]. In particular, B is a Černikov group and then F itself is a Černikov group, since periodic groups of automorphisms of Černikov groups are likewise Černikov groups (see [Reference Robinson7, Part 1, Theorem 3.29]). As $C_G(F)\leq F$ , a second application of the same result shows that G is a Černikov group. Since G is a counterexample, its largest divisible subgroup J is not of type $p^{\infty }$ for any prime p and so there exists, in J, a strictly ascending chain of infinite subgroups

$$ \begin{align*}J_1<J_2<\cdots <J_n<J_{n+1}<\cdots .\end{align*} $$

Let g be an element of G. If $\langle g\rangle \lhd G$ , then g is an ${\textit {FC}}$ -element of G. If $\langle g\rangle $ is nonnormal in G and $C_{J}(g)=J$ , then again g is an ${\textit {FC}}$ -element. By Lemma 3.5, G cannot be an ${\textit {FC}}$ -group, so there exists an element x of G such that $\langle x\rangle \ntrianglelefteq G$ and $C_{J}(x)<J$ . It follows from the hypothesis that there exists a positive integer m such that ${[\langle x\rangle ,J_m]=[\langle x\rangle ,J_n]}$ for every positive integer $n\geq m$ . In particular, for a fixed $n\geq m$ , since J is an abelian normal subgroup of G, it follows that $[x,J_n^{\langle x\rangle }]=[x,J_m^{\langle x\rangle }]=\{[x,a] \mid a\in J_m^{\langle x\rangle }\}$ . Then, $J_n\leq \langle J_m,x\rangle C_{\langle x\rangle J}(x)$ for each $n\geq m$ and hence, the subgroup $C_{\langle x\rangle J}(x)$ is infinite. Consider now $C_J(x)$ . Then, $|C_{\langle x\rangle J}(x):C_J(x)|$ is finite and so $C_J(x)$ is infinite. Let H be the subgroup $\langle x,C_G(J)\rangle $ . Thus, $C_J(x)\leq Z(H)$ , so $H'$ is finite by Corollary 3.2 and hence, $[x,J]=\{1\}$ . Then, $J\leq C_G(x)$ and, by Lemma 3.5, $G'$ is finite.

Suppose finally that G is a soluble-by-finite group such that for each nonnormal subgroup X of G, the set $\{[X,H] \mid H\leq G, H \text { is infinite}\}$ is finite and let S be the largest soluble normal subgroup of G. Obviously, it can be assumed that S is not a finite extension of a group of type $p^{\infty }$ for any prime p. If x is any element of G, then the subgroup $\langle x,S\rangle $ is a soluble group such that for each nonnormal subgroup X of G, the set ${\{[X,H] \mid H\leq G, H \text { is infinite}\}}$ is finite and hence, it follows from the previous part of the proof that $\langle x,S\rangle $ is an ${ \textit {FC}}$ -group. Then, x has finitely many conjugates in $\langle x,S\rangle $ and so also in G. Therefore, G is an ${\textit {FC}}$ -group and $G'$ is finite by Lemma 3.5.