Let $G$ be a finite simple group, and $F$ a profinite group (I'm really interested in the case where $F$ is free of finite rank, in particular rank 2).


In Ribes-Zalesskii, they define the $G$-rank of $F$ to be the integer $n\_G$ such that if $K\_G$ is the intersection of all open normal subgroups of $F$ whose quotient is isomorphic to $G$, then $F/K\_G \cong G^{n\_G}$.


At first, I thought this is the same as the largest integer $m\_G$ such that $G^{m\_G}$ is a quotient of $F$, but I've realized that this may not be so. 


I believe (by a Goursat's Lemma argument) that $m\_G$ is the cardinality of the set $Surj(F,G)/Aut(G)$.


Certainly, by taking kernels, every equivalence class in $Surj(F,G)/Aut(G)$ determines an open normal $U\le F$ with $F/U = G$, and hence $m\_G\ge n\_G$, but it's unclear to me if you can have two distinct classes in $Surj(F,G)/Aut(G)$ which have the same kernels.


If this is possible, has there been any work regarding when $n\_G = m\_G$, and if they're different, then how large the gap between $n\_G$ and $m\_G$ can be?



Difference between $G$-rank and the maximal $G$-power quotient
Your first thought was correct: the number $n\_G$ is $m\_G$.


More precisely, if $F$ is any finitely generated (profinite) group and $G$ is a finite simple group. Let $T\_G(F)$ be the set of all normal subgroups $N$ in $F$ such that $F/N$ is isomorphic to $G$. Write $K\_G = \bigcap\_{N \in T\_G(F)} N$. As you mentioned $F/K\_G \cong G^{n\_G}$ for some integer $n\_G$.


**Claim:** $n\_G$ is $m\_G = \max \{m \:|\: F \text{ projects onto } G^m \}$.


Clearly, $n\_G \leq m\_G$ as $F$ projects onto $G^{n\_G}$.
Conversely, let $f: F \to G^m$ be a surjective homomorphism. For each $i\leq m$ there is a projection $p\_i$ onto the i-th factor and $ker(p\_i \circ f) \in T\_G(F)$. But hence $\ker(f) = \bigcap\_{i=1}^m \ker(p\_i\circ f)$ contains the group $K\_G$. This means that $G^{n\_G} = F/K\_G$ projects onto $G^m$. This is only possible if $n\_G \geq m$.


