Assume $G$ is a profinite group such that the Jordan-Hölder factors appearing in the finite quotients vary in a finite number of isomorphism classes of simple groups. Assume also $G$ to have a finite number of subgroups whose corresponding quotient is simple. Does this imply that $G$ is (topologically) finitely generated?


I'm asking here after some attempt to make work a modification of the principle the for a $p$-group $P$ each set of elements generating $P/\Phi(P)$ is a generating set for $P$. For $P$ groups the question is clearly much simpler, and i have been thinking that elements generating each simple quotient had to be be enough (this is not true, as shown by the simple example $Sym\_n$. But the different symmetric groups have bigger and bigger Jordan-Holder factors). However the issue is not totally trivial, because maximal (non-normal) subgroups are in generally not contained in a proper normal subgroup, so it is not possible to replicate a similar proof smoothly. Note that the hypothesis of having a finite number of factors rules out silly couterexamples like $\prod\_{i=4}^\infty Alt\_i$.


Is anything know about this question? Thanks for the attention!



Is a profinite group with a finite number of simple quotients and Jordan-Hölder factors finitely generated?
I think the answer is no. Fix a nonabelian finite simple group $S$ and a sequence $(m\_n)$ of integers at least 2. Define inductively $G\_1=S$ and $G\_{n}=S^{m\_n}\wr G\_{n-1}$. This group admits only $S=G\_1$ as simple quotient and only $S$ as Jordan-Hölder factor. I claim that, provided $(m\_n)$ grows fast enough, the rank (minimal number of generators) of $G\_n$ is unbounded, so that the inverse limit of the $G\_n$ is not 
finitely generated, although it satisfies your assumptions. 


The claim that $G\_n$ has a unique maximal normal subgroup $W\_n$ is obtained by an easy induction. This uses only that $G\_n=V\_n\rtimes G\_{n-1}$, where the only subgroups of $V\_n$ that are normal in $G\_n$ are $V\_n$ and {1} (we need $S$ be nonabelian here) and $V\_n$ has trivial centralizer in $G\_n$. 


On the other hand, we use that the rank $r(G)$ of a wreath product $G=A\wr B$ is at least $r(A)/|B|$. Indeed, Suppose that $G$ is generated by $r$ elements. Then it's generated by $B$ and $r$ elements of $A^B$, $a\_i=(a\_{ij})\_{1\le j\le |B|}, i=1\dots r$. So $A$ is generated by the $a\_{ij}$. Thus $r(A)\le |B|r(G)$. 


Since the rank of $S^n$ tends to infinity when $n$ tends to infinity, we can define inductively the sequence $(m\_n)$ so that the rank of $G\_n$ is at least $n$. 


Edit: as pointed out by Maurizio in an email, in the example it's not true that the only subgroups of $V\_n=S^{m\_n}$ that are normal in $G\_n$, are $V\_n$ and the trivial group. Actually there are a little more: $T^{m\_n}$, where $T$ is a direct factor in $S^{m\_n}$. This is however enough to run an induction that $G\_n$ has a unique maximal normal subgroup.


