Bound on the Number of Normal Subgroups of Index $n$

Question

I'm reading Tamas Szamuely's "Galois Groups and Fundamental Groups" and have a question about an argument used in lemma 3.4.11 on page 83:

Here $\hat{F}(X)$ is a free profinite group of finite rank $r$ (so $\vert X \vert = r$). Denote by $Q_n(X)$ the set of all open normal subgroups of index $n$ in $\hat{F}(X)$.

Why is the cardinality of $Q_n(X)$ bounded by $(n!)^r$?

My considerations: It boils down to find an injection $i: Q_n(X) \hookrightarrow (Sym(n))^r$.

Another attempt would be to let $X^r$ to act on $Q_n(X)$ transitively but I can't find a concrete argument.

Could anybody help?

Answer 1

An index $n$ normal open subgroup $N$ of $\hat{F}(X)$ is the kernel of a continuous homomorphism to some quotient group $\hat{F}(X)/N$ of order $n$. By Cayley's theoorem, this quotient is isomorphic to a subgroup of $S_n$. So, every such $N$ is the kernel of a continuous homomorphism $\hat{F}(X)\to S_n$. Such continuous homomorphisms are in bijection with maps $X\to S_n$ and there are $(n!)^r$ such maps, so there are at most $(n!)^r$ such subgroups $N$.