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 November 8th, 2011, 01:18 AM #1 Newbie   Joined: Oct 2011 Posts: 14 Thanks: 0 subgroup Let H be a subgroup of G and $(G:H)=2$. Then prove: $\forall_{g \in G} g^2 \in H$ $\forall_{g_1,g_2 \in G\H g_1 \dot g_2 \in H}$. 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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