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:

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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?

  • $\begingroup$ Hint: every normal subgroup of index $n$ is the kernel of some continuous homomorphism $\hat{F}(X)\to S_n$. $\endgroup$ – YCor Feb 4 '19 at 3:34
  • $\begingroup$ 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... $\endgroup$ – 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 $N$.

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  • $\begingroup$ 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 $\phi_N \neq \phi_M: \hat{F}(X) \to S_n$. $\endgroup$ – KarlPeter Feb 4 '19 at 12:24
  • $\begingroup$ 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? $\endgroup$ – KarlPeter Feb 4 '19 at 12:24
  • $\begingroup$ As I said, $N=\ker(\phi_N)$, so $\phi_N=\phi_M$ implies $N=M$. $\endgroup$ – Eric Wofsey Feb 4 '19 at 16:16
  • $\begingroup$ but how do you construct explicitely the injective morphism $N \to \phi_N$? $\endgroup$ – KarlPeter Feb 4 '19 at 16:47
  • $\begingroup$ ...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? $\endgroup$ – KarlPeter Feb 4 '19 at 16:53

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