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 April 20th, 2012, 08:38 AM #1 Member   Joined: Feb 2012 Posts: 38 Thanks: 0 Normal subgroups Hi. Prove that if H is a subgroup of G, then the number of subgroups conjugate with H is equal to . Thanks: PD: N(H):=Normalizer of H July 2nd, 2012, 02:25 AM #2 Newbie   Joined: Jul 2012 Posts: 4 Thanks: 0 Re: Normal subgroups I will provide you with a hint Show that: if and only if July 5th, 2012, 11:09 AM #3 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Normal subgroups suppose xHx^-1 = yHy^-1. then (y^-1x)H(x^-1y) = H. we can re-write this as: (y^-1x)H(y^-1x)^-1 = H, which shows that y^-1x is in N(H). this in turn means that x is in yN(H), that is: xN(H) = yN(H). reversing the argument shows that if xN(H) = yN(H), then x and y give rise to the same conjugate of H. thus the number of conjugates of H, equals the number of (left) cosets of N(H), which is what [G:N(H)] is. 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 1 April 11th, 2012 12:40 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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