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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!

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up vote 4 down vote accepted

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.

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I think that I'm missing something... in $S\wr{}G=S^{|G|}\rtimes{}G$, isn't the subgroup of $S^{|G|}$ formed by vectors with equal components a normal subgroup? In the case of $S$ elementary abelian the normal subgroups are exactly the sub-representations, why in the case that $S$ is non-abelian there should be no invariant subgroup? If $S$ is replaced by $S^{m_n}$, furthermore, all elements of $(S^{m_n})^{|G|}$ tuples with the first component equal (of the $m_n$) appears to be invariant under $G$. –  Maurizio Monge Jan 5 '12 at 19:43
If $S$ is any nonabelian group $S^n$, the diagonal embedding of $S$ is not a normal subgroup! Hint: for $n=2$, compute $(g,1)(h,h)(g,1)^{-1}$... In a finite product $\prod_{i\in I}S_i$ of nonabelian simple groups, the only normal subgroups are the $\prod_{i\in J}S_i$ for $J$ ranging over subsets of $I$. It's an easy verification. –  Yves Cornulier Jan 9 '12 at 0:01
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Yves has answered your first question, so I'll address your more general aim.

One can regard the Frattini subgroup of a profinite group $G$ as the set of 'non-generators': a non-generator is an element $x$ such that for any subset $X$ of $G$ for which $\overline{\langle X,x \rangle}=G$, then $\overline{\langle X \rangle} = G$. It is a general fact about profinite groups that the Frattini subgroup is pronilpotent, so if the groups you are interested in are not virtually pronilpotent, there is no easy reduction of generation problems to a fixed finite quotient. So the situation in pro-$p$ groups really is very different to the general case.

One can also define the Mel'nikov subgroup (it goes by other names) $M(G)$ to be the intersection of all normal subgroups $N$ such that $G/N$ is simple. This plays a similar role to the Frattini subgroup, but you have to consider normal closures: we have $x \in M(G)$ if and only if, whenever the subset $X$ of $G$ is contained in a proper closed normal subgroup, then $X \cup \{x\}$ is also contained in a proper closed normal subgroup. A consequence is that if any subset $X$ of $G$ satisfies $XM(G) = G$, then $G$ is the closure of the group generated by the conjugates of $X$.

The natural question at this point is: when is $G/M(G)$ finite? In the pro-$p$ case, this condition precisely corresponds to finite generation (as a topological group); indeed $M(G) = \Phi(G)$ if $G$ is pronilpotent. In the general case, the property '$G/M(G)$ is finite' is not comparable with finite generation, and is not even stable under passing to subgroups of finite index. However, if you additionally assume $G$ is a pro-$\pi$ group, where $\pi$ is a finite set of primes, then $G$ finitely generated implies $G/M(G)$ is finite, although this appears to need the fact that there are only finitely many finite simple $\pi$-groups up to isomorphism (this follows from the classification of finite simple groups - I don't know any elementary proof). In the other direction, $G$ may be infinitely generated even if $G/M(G)$ is finite and $G$ is prosoluble, involving only two primes - there is an example in a recent paper of J. Wilson, for instance: my internet is bad here but I think the paper is called 'Large hereditarily just infinite groups'.

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