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 April 17th, 2012, 09:45 PM #1 Newbie   Joined: Apr 2012 Posts: 16 Thanks: 0 Subgroup proof Let G be a group, a \in G, and let H, K be subgroups of G. Suppose K is normal with H and H is normal with G. Prove that aKa^{-1} is a subgroup of H, and that it is a normal subgroup. Note that aKa^{-1} = {aka^{-1}|k \in K} *Note* must prove subgroup first via two-step check. Checking for 1.) closure, 2.) Inverses. April 18th, 2012, 01:21 PM #2 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Subgroup proof let x,y be in aKa^-1. thus x = aka^-1, for some k in K, and y = ak'a^-1 for some k' in K. then xy^-1 = (aka^-1)(ak'a^-1)^-1 = (aka^-1)(ak'^-1a^-1) = a(kk'^-1)a^-1, and since K is a subgroup kk'^-1 is in K whenever k,k' are, so xy^-1 is in aKa^-1. thus aKa^-1 is certainly a subgroup of G. what we need to do is show that aKa^-1 must be a subset of H. note that K is a subgroup of H, hence aKa^-1 is a subset of aHa^-1. but H is normal in G, so aHa^-1 = H. thus aKa^-1 is contained in H. April 18th, 2012, 10:30 PM #3 Newbie   Joined: Apr 2012 Posts: 16 Thanks: 0 Re: Subgroup proof Does xy^-1 = (aka^-1)(ak'a^-1)^-1 = (aka^-1)(ak'^-1a^-1) = a(kk'^-1)a^-1 prove both closure and inverses? April 19th, 2012, 12:54 PM #4 Senior Member   Joined: Jul 2011 Posts: 227 Thanks: 0 Re: Subgroup proof Yes it does. If you need to check if a is subgroup of then you have to check first what is obvious because is a subgroup of (thus ). Afterwards it satisfies to verify if what Deveno showed you. Tags proof, subgroup 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 Kappie Abstract Algebra 0 April 22nd, 2012 01:52 PM tinynerdi Abstract Algebra 9 February 23rd, 2010 08:30 AM HairOnABiscuit Abstract Algebra 4 October 9th, 2009 11:49 PM envision Abstract Algebra 3 October 4th, 2009 10:37 PM envision Abstract Algebra 1 October 4th, 2009 03:24 AM

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