0
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $n_G \geq m$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.