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 April 10th, 2012, 02:34 AM #1 Member   Joined: Feb 2012 Posts: 38 Thanks: 0 Normal subgroups Hi everyone. Let be a normal subgroups of of finite index . Show that, if is any subgroup of , then is finite and . 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. Tags normal, subgroups Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post BrettsGrad Abstract Algebra 2 February 20th, 2013 08:28 AM Fernando Abstract Algebra 2 July 5th, 2012 11:09 AM ejote Abstract Algebra 4 February 1st, 2011 07:30 AM donwu777 Abstract Algebra 1 December 10th, 2008 04:03 PM bjh5138 Abstract Algebra 1 October 7th, 2007 09:20 PM

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