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 rewrite 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. 

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