Countable central factors in the derived subgroup of a locally normal group

Problem

If $G$ is locally normal, must $G'$ be an internal central product of countable subgroups?

\begin{gathered} \forall X\subseteq G\text{ finite},\quad\exists K\trianglelefteq G:\ X\subseteq K,\ |K|<\infty,\\ G'\stackrel{?}{=}\langle H_i:i\in I\rangle,\qquad |H_i|\le\aleph_0,\quad[H_i,H_j]=1\quad(i\ne j). \end{gathered}

No bound on $|G|$ is imposed; intersections of the central factors may be nontrivial.

The decomposition theorem

A group $G$ is locally normal if, for every finite subset $X\subseteq G$ , there is a finite normal subgroup $K\trianglelefteq G$ containing $X$ . Write $G'$ for the derived subgroup, $C_G(X)$ for the centralizer of $X$ , and $[a,b]=aba^{-1}b^{-1}$ .

Theorem 1. Let $G$ be locally normal. There is a family $(H_i)_{i\in I}$ of countable subgroups of $G'$ such that

\tag{1} [H_i,H_j]=1\quad(i\ne j),\qquad G'=\langle H_i:i\in I\rangle.

Here countable includes finite. Equation (1) expresses $G'$ as an internal central product: each element belongs to a product of finitely many factors, and distinct factors commute. Their intersections need not be trivial. Theorem 1 answers Problem 6.9 of the Kourovka Notebook (1).

Throughout the proof, $G$ is fixed and locally normal, and $D=G'$ . Normality of a subgroup always means normality in $G$ , unless another ambient group is specified.

Uniform commutator corrections

Definition 2. A normal subgroup $N\trianglelefteq G$ is admissible if

\tag{2} G=N C_G(L) \quad\text{for every finite }L\trianglelefteq G\text{ with }L\le N.

Thus every conjugation induced by an element of $G$ on such an $L$ is induced by an element of $N$ . Both $1$ and $G$ are admissible.

Lemma 3. Let $N\trianglelefteq G$ , let $S$ be finite, and let $P(s,x)$ be a predicate for $s\in S$ and $x\in N$ . Suppose that, for each finite normal subgroup $L\le N$ , some $s\in S$ satisfies $P(s,x)$ for all $x\in L$ . Then some $s\in S$ satisfies $P(s,x)$ for all $x\in N$ .

Proof. Otherwise, choose $x_s\in N$ with $\neg P(s,x_s)$ for each $s\in S$ . Local normality puts the finite set $\{x_s:s\in S\}$ in a finite normal subgroup $K$ of $G$ . The subgroup $L=K\cap N$ is finite and normal and contains every $x_s$ , contradicting the hypothesis for $L$ . This also covers $S=\varnothing$ , since the hypothesis for $L=1$ requires an element of $S$ . ◻

Lemma 4. If $N$ is admissible and $a,b\in G$ , there exists $s\in D\cap N$ such that

\tag{3} [a,b]s^{-1}\in C_G(N).

Proof. For every finite normal $L\le N$ , admissibility gives $n\in N$ with $an^{-1}\in C_G(L)$ . Since $L$ is normal, so is $C_G(L)$ . In the quotient $G/C_G(L)$ , the images of $a$ and $n$ are equal; hence

\tag{4} [a,b][n,b]^{-1}\in C_G(L).

Choose a finite normal subgroup $K$ containing $b$ . Every $[n,b]$ lies in $K$ , so $S=\{[n,b]:n\in N\}$ is finite. Apply Lemma 3 to the condition that $[a,b]s^{-1}$ commute with $x$ , using (4) for its local hypothesis. The resulting $s\in S$ belongs to $D\cap N$ , since $N$ is normal, and satisfies (3). ◻

Proposition 5. For every admissible $N$ ,

\tag{5} D=(D\cap N)C_D(N).

Proof. The two factors on the right commute, and their product is a subgroup of $D$ . Lemma 4 places every commutator in that product. Since commutators generate $D$ , equality follows. ◻

Countable admissible enlargements

If $K$ is finite and normal, conjugation defines a homomorphism $G\to\operatorname{Aut}(K)$ with kernel $C_G(K)$ . In particular,

\tag{6} [G:C_G(K)]<\infty.

For normal subgroups, products below are subgroup products; they agree with the corresponding joins in the subgroup lattice.

Lemma 6. Let $N$ be admissible, let $L\le N$ be finite and normal, and let $K$ be finite and normal. Then

\tag{7} [G:N C_G(LK)]\le [G:C_G(K)]<\infty.

Proof. Put $C=C_G(L)$ , $E=C_G(K)$ and $Q=N(C\cap E)$ . We have $C_G(LK)=C\cap E$ and $NC=G$ by admissibility. Thus $CQ=G$ , and

[G:Q]=[C:C\cap Q] \le [C:C\cap E]\le [G:E].

The final index is finite by (6). ◻

Lemma 7. Let $N$ be admissible and $K$ finite and normal. There is a finite normal subgroup $J\ge K$ such that

\tag{9} G=NJ C_G(LK) \quad\text{for every finite normal }L\le N.

Proof. For such $L$ , put $Q_L=N C_G(LK)$ . Lemma 6 bounds all indices $[G:Q_L]$ by the same finite number. Choose $L_0$ for which this index is maximal. For any $L$ , the product $L_0L$ is again a finite normal subgroup of $N$ , and

Q_{L_0L}\le Q_{L_0}\cap Q_L.

If $Q_{L_0L}$ were properly contained in $Q_{L_0}$ , its index would be strictly larger. Therefore $Q_{L_0L}=Q_{L_0}$ , and consequently $Q_{L_0}\le Q_L$ for every $L$ .

Choose a finite transversal $T$ for $Q_{L_0}$ in $G$ . Local normality provides a finite normal subgroup $J$ containing $K\cup T$ . It follows that $G=JQ_{L_0}\le JQ_L=NJ C_G(LK)$ , proving (9). ◻

Proposition 8. If $N$ is admissible and $g\in G$ , there is a countable normal subgroup $B$ containing $g$ such that $NB$ is admissible.

Proof. Choose a finite normal subgroup $K_0$ containing $g$ . Repeatedly apply Lemma 7 to obtain finite normal subgroups $K_0\le K_1\le K_2\le\cdots$ such that, for every finite normal $L\le N$ ,

\tag{11} G=NK_{m+1}C_G(LK_m)\qquad(m\ge0).

Then $B=\bigcup_{m\ge0}K_m$ is countable and normal and contains $g$ .

Let $F\le NB$ be finite and normal. For each $x\in F$ , choose $n_x\in N$ and $b_x\in B$ with $x=n_xb_x$ . The finitely many $b_x$ lie in some $K_m$ . The finitely many $n_x$ lie in a finite normal subgroup $L\le N$ : take a finite normal subgroup of $G$ containing them and intersect it with $N$ . Hence $F\le LK_m$ . By (11),

G=NK_{m+1}C_G(LK_m)\le NB C_G(F)\le G.

This proves admissibility of $NB$ . ◻

Adding a countable central factor

Lemma 9. Suppose $N$ is admissible and $B$ is a countable subgroup. Put $M=NB$ . There is a countable subgroup $H\le D\cap M$ such that

\tag{13} [H,N]=1,\qquad D\cap M=(D\cap N)H.

Proof. The image of $D\cap M$ in $G/N$ is countable, since $M/N$ is the image of $B$ . Choose one $d\in D\cap M$ over each element of that image. By Proposition 5, write $d=uc$ with $u\in D\cap N$ and $c\in C_D(N)$ . Here $c=u^{-1}d\in M$ and $cN=dN$ . Let $H$ be generated by the resulting countably many elements $c$ . A subgroup generated by a countable set is countable, because its elements are finite words in that set and its inverses. Thus $H$ is countable, lies in $D\cap M$ , and centralizes $N$ . Its image in $G/N$ equals that of $D\cap M$ , which gives (13). ◻

Lemma 10. A nonempty directed union of admissible subgroups is admissible.

Proof. Let $N=\bigcup_i N_i$ , where the $N_i$ are admissible and directed by inclusion. The union is normal. Every finite normal subgroup $L\le N$ is contained in one $N_i$ . Thus $G=N_iC_G(L)=NC_G(L)$ . ◻

Proof of the decomposition theorem

Proof of Theorem 1. A stage is a pair $(N,\mathcal H)$ in which $N$ is admissible, $\mathcal H$ is a set of countable subgroups of $D\cap N$ , and

\tag{14} [H,K]=1\quad(H,K\in\mathcal H,\ H\ne K),\qquad \langle\mathcal H\rangle=D\cap N.

Order stages by inclusion in both coordinates. The pair $(1,\varnothing)$ is a stage.

Every nonempty chain of stages has an upper bound obtained by taking the union in each coordinate. The subgroup coordinate is admissible by Lemma 10. Any two factors occur together in one stage of the chain and hence commute when distinct. Finally,

D\cap\bigcup_i N_i=\bigcup_i(D\cap N_i) =\left\langle\bigcup_i\mathcal H_i\right\rangle,

so the union is a stage. Zorn’s lemma gives a maximal stage $(N,\mathcal H)$ .

Fix $g\in G$ . Proposition 8 provides a countable normal $B$ with $g\in B$ and $NB$ admissible. Lemma 9 provides a countable $H\le D\cap NB$ centralizing $N$ , with $D\cap NB=(D\cap N)H$ . Every $K\in\mathcal H$ lies in $N$ , so $[H,K]=1$ . Therefore $(NB,\mathcal H\cup\{H\})$ is a stage extending $(N,\mathcal H)$ . Maximality implies $NB=N$ , whence $g\in N$ . Since $g$ was arbitrary, $N=G$ , and (14) is (1). ◻

References

Preprint · Lean (GitHub)

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st ed., 2026, Problem 6.9. arXiv:1401.0300.