# Difference between $G$-rank and the maximal $G$-power quotient

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?

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 $n_G \geq m$.