index of a closed subgroup of a profinite group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:24:24Z http://mathoverflow.net/feeds/question/46643 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46643/index-of-a-closed-subgroup-of-a-profinite-group index of a closed subgroup of a profinite group safak 2010-11-19T16:15:58Z 2010-11-19T17:22:59Z <p>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 $H$.</p> <p>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 $\big((G/G):(H/(H\cap G))\big)=1$.</p> <p>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?</p> http://mathoverflow.net/questions/46643/index-of-a-closed-subgroup-of-a-profinite-group/46644#46644 Answer by Pete L. Clark for index of a closed subgroup of a profinite group Pete L. Clark 2010-11-19T16:20:04Z 2010-11-19T17:22:59Z <p>(The first time around I had read your question too quickly and not properly appreciated it. Sorry about that.)</p> <p>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 <a href="http://en.wikipedia.org/wiki/Goursat_lemma" rel="nofollow">Goursat's Lemma</a>.)</p> <p>I checked that this exercise does not appear in Serre's <em>Galois Cohomology</em>. Have you found that it is used at any point of Shatz's book? </p> <p>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. </p>