November 8th, 2011, 01:18 AM  #1 
Newbie Joined: Oct 2011 Posts: 14 Thanks: 0  subgroup
Let H be a subgroup of G and . Then prove: . Does anyone know how to solve or can give any clue? 
November 13th, 2011, 10:55 AM  #2 
Newbie Joined: Jan 2011 Posts: 2 Thanks: 0  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}.

November 14th, 2011, 05:08 AM  #3 
Member Joined: Jun 2010 Posts: 64 Thanks: 0  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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