In the book "profinite groups, arithmetic, and geometry" of Shatz, the index $(G:H)$ of a closed subgroup $H$ of a profinite group $G$ is defined to be the supernatural number $lcm\big((G/U):(H/(H\cap U))\big)$ where $U$ runs over the open normal subgroups of $G$. There is an exercise following this definition saying that "$(G:H)=lcm(G:U)$ where $U$ runs over those open normal subgroups of $G$ containing $H$.

If $G$ is a finite group with discrete topology, then the index given is nothing but the number of elements in the coset space $G/H$. However, if we take $G$ to be a finite simple group having a non-trivial proper subgroup $H$, e.g. $Alt_n$ for a suitable $n$, the only normal subgroup contating $H$ is $G$ itself and $\big((G/G):(H/(H\cap G))\big)=1$.

I am not sure if the claim in the exercise is true for infinite profinite groups as they are necessarily non-simple, which means they don't admit trivial counter-examples. But at least the exercise seemed me wrong for finite case. Am I missing something, or this is a well-known misprint which I don't know?


(The first time around I had read your question too quickly and not properly appreciated it. Sorry about that.)

You are right: the exercise on p. 12 of Shatz's book is false, because of the example you suggest. You asked if there were also counterexamples among infinite profinite groups. Certainly: let $n \geq 5$, let $p$ be a prime number greater than $n$, and consider $G = \mathbb{Z}_p \times A_n$. Then the problem persists: take $H = \mathbb{Z}_p \times H'$, where $H'$ is a proper nontrivial subgroup of $A_n$. (Use Goursat's Lemma.)

I checked that this exercise does not appear in Serre's Galois Cohomology. Have you found that it is used at any point of Shatz's book?

It seems plausible to me that you could recover a statement like this by working prime-by-prime with the Sylow subgroups of the groups in question -- certainly there are enough normal subgroups of $p$ groups to detect indices -- but I haven't thought carefully about that.

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    $\begingroup$ Thanks for your answer. You are right, product with any infinite profinite group with the example yields countre-examples to the infinite case. When the normality condition is dropped the exercise becomes correct. So far i did not see any usage of the exercise. $\endgroup$ – safak Nov 19 '10 at 20:31
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    $\begingroup$ infact it is exactly what is written in serre's "it is also the lcm of the indices (G:V) for open V containing H" $\endgroup$ – safak Nov 19 '10 at 21:57

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