# index of a closed subgroup of a profinite group

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?

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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.)
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.