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April 20th, 2012, 08:38 AM   #1
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Normal subgroups


Prove that if H is a subgroup of G, then the number of subgroups conjugate with H is equal to .


PD: N(H):=Normalizer of H
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July 2nd, 2012, 02:25 AM   #2
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Re: Normal subgroups

I will provide you with a hint
Show that:
if and only if
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July 5th, 2012, 11:09 AM   #3
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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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