# Uniqueness of composition series for profinite groups

(It is possible that an answer to this question can be found in the literature, but I couldn't find anything after searching for about an hour.)

Let $G$ be a compact, totally disconnected, second countable group. (Equivalently, a profinite group: it is the inverse limit of finite groups.) It makes sense to talk about normal series: $\{e\} \lhd \cdots\cdots G_2\lhd G_1\lhd G$ where $G_n$ is a closed subrgoup of $G$, $G_n$ is normal in $G_{n-1}$ and $G_{n-1}/G_n$ is finite and $\bigcap_{n\in\mathbb{N}}G_n=\{e\}$. This is the same as having an inverse limit expresion $$G\to \cdots \to H_2\to H_1\to \{e\}$$ where each $H_n$ is finite.

Call the normal series a composition series if each $G_{n-1}/G_n$ is simple. (In the inverse limit picture, this is the same as the kernel of each map $H_n\to H_{n-1}$ being simple.)

Are the simple factors appearing in a composition series unique up to permutation? The result is false without compactness, but I do not know if it is true with it.

• It's true if $G$ is (topologically) finitely generated. In this case, $G$ is determined uniquely by its collection of finite quotients, and the composition factors of the finite quotients determine those of $G$. – Ian Agol Jan 29 '14 at 21:41
• @Eusebio: do you assume that the $G_n$ are closed subgroups? – YCor Jan 29 '14 at 23:20
• @Yves: yes, the subgroups $G_n$ should be closed; I forgot to mention that. I'll edit the question. – Eusebio Gardella Jan 30 '14 at 1:33
• @Ian: I would like a result that can handle infinite products of finite groups too; in particular, groups for which every finitely generated subgroup is finite. – Eusebio Gardella Jan 30 '14 at 1:36
• Did you check whether the proof of the (Zassenhaus) Butterfly Lemma applies in your situation? It doesn't use much --- only the Noether isomorphism theorems and one of Dedekind's modular laws. That would imply that any two such sequences have a common refinement. – abz Jan 30 '14 at 18:20

If $G$ is second countable, it has a composition series: one can obtain this by starting with some countable basis of identity neighbourhoods (which may be taken to consist of open normal subgroups of $G$, since every open neighbourhood of the identity contains an open normal subgroup), taking intersections to form a descending chain of open normal subgroups with trivial intersection, and then refining the series. So let us assume that we have a composition series $(G_n)$. The equivalence class of $(G_n)$ is defined by the number of times, say $f(S)$, each isomorphism type $S$ of finite simple group appears as a factor $G_{n-1}/G_n$.
Let $S$ be a finite simple group. Then there are two possibilities: either $f(S)$ is a non-negative integer, or $f(S) = \aleph_0$. If $f(S)$ is finite, then there is a finite discrete quotient $G/G_n$ of $G$ such that some (hence any) composition series for $G/H$ has $f(S)$ simple factors isomorphic to $S$. Moreover, every finite discrete quotient of $G$ has at most $f(S)$ copies of $S$ in any composition series: This is clear for $G/G_n$ for all $n$ assuming Jordan-Hölder for finite groups, and then we use compactness to show that any finite discrete quotient $G/H$ of $G$ is a quotient of $G/G_n$. If $f(S)$ is infinite, then $G$ has finite quotients with composition series that have arbitrarily many factors isomorphic to $S$. Thus $f(S)$ is an invariant of $G$ as a topological group, independent of the choice of composition series $(G_n)$, so all composition series are equivalent.