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November 8th, 2011, 01:18 AM   #1
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subgroup

Let H be a subgroup of G and . Then prove:

.

Does anyone know how to solve or can give any clue?
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November 13th, 2011, 10:55 AM   #2
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Re: subgroup

An index of 2 tells us that the subgroup is normal in G. So we can construct a homomorphism from G to it's quotient group G/H = {H, gH}.
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November 14th, 2011, 05:08 AM   #3
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Re: subgroup

[G:H]=2 => G/H={H,xH} for x not in H.

imedently say G/H is a group. so (xH)*(xH)=H , so
let take a g from G, we have 2 cases
case 1. g is at H , so g*g is at H
case 2. g is not at H, so exist a h from H such g=xh,
g*g=(xh)*(xh) involve at (xH)*(xH)=h,
so for every g we have g*g is at H.


The next question that you say that for every g1 , g2 from G we should
have g1*g2 is at H ,THIS IS NOT TRUE I HAVE EXPLAINED IT AT PREVIOUS POST.
because if we have g1 at H and g2=x, g1*g2 is not at H.
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