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

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?

• Hint: every normal subgroup of index $n$ is the kernel of some continuous homomorphism $\hat{F}(X)\to S_n$. – YCor Feb 4 '19 at 3:34
• Ok, it seems that you have the identification $\{1, 2, ..., n\}$ with $\{gN \vert g \in G\}$ for $N=ker \phi$ for some $\phi:\hat{F}(X) \to S_n$ in mind. So I guess the argument would be that the map $Hom(\hat{F}(X), S_n), \phi \mapsto ker \phi$ provides a surjection... – KarlPeter Feb 4 '19 at 3:48

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 $$N$$.
• Hi thank you for the answer but one aspect seems unclear to me. Essentially we want to show there exist at least as much continuous morphisms $\hat{F}(X) \to S_n$ as open normal subgroups $N$ of index $n$. But here occurs following problem to me: To use this argument we need to show that different ons $N \neq M$ of index $n$ induce different morphisms $\phi_N \neq \phi_M: \hat{F}(X) \to S_n$. – KarlPeter Feb 4 '19 at 12:24
• The first step is to quotient out these normal subgroups so we two quotient groups $\hat{F}(X)/N$ and $\hat{F}(X)/M$ of order $n$. How can we embed them into $S_n$ to induce different morphisms to $S_n$. The problem is that for every such finite group $\hat{F}(X)/N$ we have to take a separate choice for an embedding to "distinguish" all $N$ resp $\hat{F}(X)/N$. Or do I have overseen an argument in your answer which provides a machinery to construct different morphisms $\phi_N$ avoiding this problem? – KarlPeter Feb 4 '19 at 12:24
• As I said, $N=\ker(\phi_N)$, so $\phi_N=\phi_M$ implies $N=M$. – Eric Wofsey Feb 4 '19 at 16:16
• but how do you construct explicitely the injective morphism $N \to \phi_N$? – KarlPeter Feb 4 '19 at 16:47
• ...I guess that firstly you fix a concrete normal subgroup $B$ of order $n$ in $S_n$. And for each $N$ you choose an isomorphism $i_N: \hat{F}(X)/N \cong B$. Then $\phi_N = i_N \circ pr_N$ where $pr_N$ is the canonical projection. Do you mean this construction? – KarlPeter Feb 4 '19 at 16:53