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 November 8th, 2011, 01:20 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 8th, 2011, 01:24 AM #2 Newbie   Joined: Oct 2011 Posts: 14 Thanks: 0 Re: subgroup There should be: $\forall_{g_1,g_2 \in G\H} g_1g_2 \in H$
 November 10th, 2011, 09:08 AM #3 Member   Joined: Jun 2010 Posts: 64 Thanks: 0 Re: subgroup IS not true that for all g from G g*g is at H, let suppose that G/H is a group and G/H={xH,H } ,since G/H is group and H is zero of G?H we have that x*h is at (xH)(H)=xH so is not at H ,and h at H and we have xh isnot at H ! ,so your question is incorrect.

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