Proof. Let H be an $(m,k)$ -subnormal subgroup of G, so that there exists a subnormal subgroup X of G of defect at most k such that $\lvert X:H\rvert \leq m$ . If $N=H_{X}$ is the normal core of H in X, then $X/N$ is a finite nilpotent group, so $H/N$ is subnormal in $X/N$ with defect at most m. Hence, H is a subnormal subgroup of G of defect at most $m+k$ .

Proof. The assertion is a consequence of Lemma 2.1 and [Reference De Falco, de Giovanni, Heineken and Musella3, Theorem 2.5].

Proof. Let A be a large abelian subgroup; then A is $(m,k)$ -subnormal in G, so that there exists a subnormal subgroup C of defect at most k such that $\lvert C:A\rvert \leq m$ . It follows that $A_{C}$ , the normal core of A in C, is a large abelian subnormal subgroup of G. Thus, B is a locally nilpotent $\mathfrak {X}(m,k)$ -group containing a large abelian subgroup. The statement is now a consequence of Corollary 2.2.

Proof. Let s be the least positive integer such that $\gamma _{s+1}(G)$ is small and let g be any element of G; if we put

$$ \begin{align*} g^{\gamma_{s}(G)}=\{g^{h} \mid h \in \gamma_{s}(G)\}, \end{align*} $$

the function

$$ \begin{align*} g^{h} \in g^{\gamma_{s}(G)}\longrightarrow [g,h] \in [g,\gamma_{s}(G)]\leq \gamma_{s+1}(G) \end{align*} $$

is a bijection and, therefore, the set $g^{\gamma _{s}(G)}$ has cardinality strictly smaller than $\aleph $ . However,

$$ \begin{align*} \lvert g^{\gamma_{s}(G)}\rvert =\lvert \gamma_{s}(G):C_{\gamma_{s}(G)}(g)\rvert \end{align*} $$

where $C_{\gamma _{s}(G)}(g)$ is the centraliser in $\gamma _{s}(G)$ of g.

Put $F=\langle g_{1}, \ldots , g_{t}\rangle $ ; then $\lvert \gamma _{s}(G):C_{\gamma _{s}(G)}(g_i)\rvert $ is strictly smaller than $\aleph $ , for each $i \in \{1,\ldots ,t\}$ . Moreover,

$$ \begin{align*} C_{G}(F)=\overset{t}{\underset{i=1}{\bigcap}}C_{G}(g_{i}), \end{align*} $$

so that $\lvert \gamma _{s}(G):C_{\gamma _{s}(G)}(F)\rvert $ is strictly smaller than $\aleph $ , and hence $C_G(F)$ is a large subgroup of G.

Proof. It follows from Proposition 2.4 that $C_{G}(F)F$ is an $(m,k)$ -subnormal subgroup of G in which F is normal.

Proof. We can write $G=A\langle x\rangle $ , where A is a normal abelian subgroup of G and x is an element of G. Of course, A is large and it can be regarded as a module over the countable ring $R=\mathbb {Z}\langle x\rangle $ . Moreover, the R-submodules of A are all its G-invariant subgroups. By [Reference De Falco, de Giovanni, Heineken and Musella3, Lemma 2.3], A contains the direct product

$$ \begin{align*} \underset{i \in I}{Dr}\,M_{i}, \end{align*} $$

where I is a set of cardinality $\aleph $ and, for each $i \in I$ , $M_{i}$ is a G-invariant subgroup of A. It follows that there exist two subsets $I_{1}$ and $I_{2}$ of I, each of cardinality $\aleph $ , such that

$$ \begin{align*} I_{1}\cap I_{2}=\emptyset. \end{align*} $$

For $j=1,2$ , put

$$ \begin{align*} B_{j}=\underset{i \in I_{j}}{Dr}\,M_{i}, \end{align*} $$

then $B_{1}$ and $B_{2}$ are G-invariant subgroups of A of cardinality $\aleph $ whose intersection is trivial. Thus, [Reference Lennox9, Theorem A] ensures that, for $j=1,2$ ,

$$ \begin{align*} \lvert \gamma_{\mu (m+k)}(G/B_{j})\rvert \leq m{!}. \end{align*} $$

Therefore, $\lvert \gamma _{\mu (m+k)}(G)\rvert \leq (m{!})^{2}$ and G is finite-by-nilpotent.

Proof. Let x be any element of G and let A be a large abelian normal subgroup of G; the subgroup $\langle A, x \rangle = A\langle x\rangle $ is an abelian-by-cyclic $\mathfrak {X}(m,k)$ -group so, by Lemma 2.6, $\lvert \gamma _{\mu (m+k)}(A\langle x\rangle )\rvert \leq (m{!})^2$ . Therefore, $\langle x \rangle $ has index at most $(m{!})^2$ in $\langle x\rangle \gamma _{\mu (m+k)}(A\langle x\rangle )$ . Since $A\langle x\rangle /\gamma _{\mu (m+k)}(A\langle x\rangle )$ is nilpotent of class at most $\mu (m+k)-1$ , the subgroup $\langle x\rangle \gamma _{\mu (m+k)}(A\langle x\rangle )$ is subnormal in $A\langle x\rangle $ with defect at most $\mu (m+k)-1$ . Finally, $A\langle x\rangle $ is $(m,k)$ -subnormal in G and so it has index at most m in $(A\langle x\rangle )^{G,k}$ which is subnormal in G with defect at most k.

By an application of [Reference Lennox9, Theorem B], G is locally finite-by-nilpotent.

Let X be a cyclic subgroup of G. By the above argument, there exists a chain

$$ \begin{align*}X=X_n \leq \cdots \leq X_0=G\end{align*} $$

such that either $X_{i+1}$ is subnormal in $X_i$ with defect at most $\mu (m+k)-1$ or $X_{i+1}$ has index at most $(m{!})^2$ in $X_i$ ; we shall prove by induction on the length of the chain that for every cyclic subgroup X of G, there is a positive integer s dividing $\beta (m)=((m{!})^2)!$ such that $X^{s}$ is subnormal in G.

We can assume that $X^{s}$ is subnormal in $X_1$ , for some positive integer s dividing $\beta (m)$ . If $X_1$ is subnormal in G, then $X^{s}$ is subnormal in G. So we can assume that the index $\vert G:X_1 \vert \leq (m{!})^2$ ; then the core Y of $X_1$ in G has index at most $((m{!})^2)!= \beta (m)$ in G. Hence, $X^{\beta (m)}$ is a normal subgroup of $Y\cap X^{s}$ , which is subnormal in Y. It follows that $X^{\beta (m)}$ is subnormal in G, and hence $X^{\beta (m)} \leq B$ .

Therefore, $G/B$ has finite exponent.

Proof. Since N is nilpotent of cardinality $\aleph $ , there exists a nonnegative integer r such that $\lvert Z_{r}(N)\rvert < \aleph $ and $\lvert Z_{r+1}(N)\rvert =\aleph $ . So, the section $\overline {A}=Z_{r+1}(N)/Z_{r}(N)$ is an abelian normal subgroup of $\overline {G}=G/Z_{r}(N)$ . Let $\overline {E}$ be any finitely generated subgroup of $\overline {G}$ and put $\overline {L}=\overline {E}\overline {A}$ . Since $\overline {N}\cap \overline {L} \leq C_{\overline {L}}(\overline {A})$ , then $\overline {L}/C_{\overline {L}}(\overline {A})$ is finite, so that Lemma 2.8 ensures that $\overline {A}$ contains a subgroup $\overline {H}$ of the form

$$ \begin{align*} \overline{H} =\underset{i \in I}{Dr}\,\overline{A}_{i} \end{align*} $$

where I is a set of cardinality $\aleph $ and, for each $i \in I$ , $\overline {A}_{i}$ is a nontrivial normal subgroup of $\overline {L}$ . So we can find two $\overline {L}$ -invariant subgroups $\overline {H}_{1}$ and $\overline {H}_{2}$ of $\overline {H}$ , both of cardinality $\aleph $ and such that

$$ \begin{align*} \overline{H}_{1}\cap \overline{H}_{2}=\overline{E}\cap \overline{H}_{1}\overline{H}_{2}=\{1\}. \end{align*} $$

Therefore,

$$ \begin{align*} \overline{E}= \overline{E}\overline{H}_{1}\cap \overline{E}\overline{H}_{2} \end{align*} $$

is the intersection of two $(m,k)$ -subnormal subgroups of $\overline {G}$ and, by [Reference Dixon, Ferrara and Trombetti7, Lemma 2.5], $\overline {E}$ is $(m^{2},k)$ -subnormal. An application of [Reference Lennox9, Theorem A] completes the proof.

Proof. Let B be the Baer radical of G; it follows from Corollary 2.3 and Lemma 2.7 that B is nilpotent and $G/B$ is a locally finite group, so that $\gamma _{\mu (m+k)}(G)$ is small by Lemma 2.9. Let F be any finitely generated subgroup of G. Applying Proposition 2.4 to the group $AF$ , we can say that $C_{AF}(F)$ is large, so that also $C_{AF}(F) \cap A$ is a large subgroup. It follows that $Z(AF)$ is large, and hence it contains two large subgroups $U_{1}$ and $U_{2}$ such that

$$ \begin{align*} U_{1}\cap U_{2}=F\cap U_{1}U_{2}=\{1\}. \end{align*} $$

Therefore,

$$ \begin{align*} F=FU_{1}\cap FU_{2} \end{align*} $$

is the intersection of two $(m,k)$ -subnormal subgroups of G and, by [Reference Dixon, Ferrara and Trombetti7, Lemma 2.5], F is $(m^{2},k)$ -subnormal. The statement follows from [Reference Lennox9, Theorem A].

Proof. Let B be the Baer radical of G. By Corollary 2.3, B is a large nilpotent subgroup of G, so that the section $B/B{'}$ has cardinality $\aleph $ . Hence, $G/B{'}$ is an $\mathfrak {X}(m,k)$ -group and contains the large abelian normal subgroup $B/B{'}$ , so that, as a consequence of Theorem 2.10,

$$ \begin{align*} \lvert \gamma_{\mu (m^2+k)}(G/B{'})\rvert \leq (m^{2}){!}. \end{align*} $$

By Hall’s theorem, $\lvert G/B{'}: Z_{2\mu (m^2+k)}(G/B{'})\rvert $ is finite so, with $Z/B' =Z_{2\mu (m^2+k)}(G/B{'})$ , also $\lvert G:ZB\rvert $ is finite. Clearly, $ZB/B{'}$ is nilpotent by Fitting’s theorem, and so the nilpotency of the normal subgroup B of $ZB$ ensures that $ZB$ is a nilpotent normal subgroup of G. It follows that $Z \leq B$ and the statement is proved.

Proof. If B is the Baer radical of G, then B is a large and nilpotent subgroup of G by Corollary 2.3. Since $G/B$ is finite by Lemma 2.11, as a consequence of Lemma 2.9, $\gamma _{\mu (m^2+k)}(G)$ is small.

In addition, Lemma 2.11 ensures the existence of a finitely generated subgroup F such that $G=FB$ . By Corollary 2.5, F is f-subnormal in G, so that there exists an f-series of the form

$$ \begin{align*} F=F_{0}<F_{1}<\cdots<F_{n}=G, \end{align*} $$

where either $\lvert F_{i+1}: F_{i}\rvert $ is finite or $F_{i}$ is normal in $F_{i+1}$ , for each $i \in \{0, \ldots , n-1 \}$ .

We will prove that G is finite-by-nilpotent by induction on the length n of the f-series above. If $n=1$ , since G is an uncountable group, F is normal in G. On the other hand, [Reference Casolo and Mainardis2, Theorem 1.1] tells us that F is also finite-by-nilpotent so we can find a positive integer r such that the G-invariant subgroup $\gamma _{r}(F)$ is finite, and

$$ \begin{align*} G/\gamma_{r}(F)=BF/\gamma_{r}(F)=(F/\gamma_{r}(F))\cdot (B\gamma_{r}(F)/\gamma_{r}(F)) \end{align*} $$

is nilpotent by Fitting’s theorem.

Assume now that $n>1$ . Then $\lvert F_{n-1}{\,:\,}F_{n-1}\cap B\rvert $ is finite and $B\cap F_{n-1}$ is contained in the Baer radical of $F_{n-1}$ , so we can apply the inductive hypothesis to $F_{n-1}$ which has to be a finite-by-nilpotent group. It follows that there exists a finite characteristic subgroup M of $F_{n-1}$ such that $F_{n-1}/M$ is nilpotent (see [Reference Casolo and Mainardis2, Lemma 2.9]). If $F_{n-1}$ is normal in G, also M is G-invariant and

$$ \begin{align*} G/M= (F_{n-1}/M) \cdot (BM/M) \end{align*} $$

is nilpotent by Fitting’s theorem. Assume that $\lvert G:F_{n-1}\rvert $ is finite; then M has finitely many conjugates in G, so that $M^{G}$ is finite by Dietzmann’s lemma. On the other hand,

$$ \begin{align*}G/M^G= (F_{n-1}M^G/M^G) \cdot (BM^G/M^G)\end{align*} $$

is finite-by-(locally nilpotent) (see [Reference Casolo and Mainardis2, Lemma 2.7]), so that G itself is finite-by-(locally nilpotent), and hence actually finite-by-nilpotent by Corollary 2.2.