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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 $[G:N(H)]$. 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: $g_1Hg_1^{-1}=g_2Hg_2^{-1}$ if and only if $g_1g_2^{-1} \in N(H)$
 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.

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