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 April 10th, 2012, 02:34 AM #1 Member   Joined: Feb 2012 Posts: 38 Thanks: 0 Normal subgroups Hi everyone. Let $N$ be a normal subgroups of $G$ of finite index $n$. Show that, if $A$ is any subgroup of $G$, then  is finite and $s|n$. Thanks.
 April 11th, 2012, 12:40 AM #2 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Normal subgroups first, let's note that A?N is normal in A: let a be any element of A, and n be any element of A?N. then ana^-1 is in A, since all of a,n and a^-1 are in A, and A is a subgroup, thus closed under multiplication. but since a is also in G (since A is a subgroup of G), and N is normal in G, ana^-1 is in N. thus ana^-1 is in both A, and N, so is in A?N. hence we can form the group A/A?N, which has order [A:A?N]. now, since N is normal, AN is a subgroup of G, and by the second isomorphism theorem: AN/N ? A/A?N. since AN is a subgroup of G containing N, AN/N is a subgroup of G/N, which therefore must be finite (since G/N is finite, of order n), and have order dividing |G/N|. hence s = [A:A?N] = [AN:N] = |AN/N| divides |G/N| = [G:N] = n.

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