Why is a subgroup with the index of 2 normal to the group? I believe it has something to do with it being half and having nowhere to go.
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This might help http://math.stackexchange.com/questions/84632/subgroup-of-index-2-is-normal – LASV Dec 02 '13 at 19:05
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I do believe this question is sufficiently different to that linked to by Luis in his comment to spare it from closure. – Robert Lewis Dec 02 '13 at 20:23
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@RobertLewis How is it different? – user7530 Dec 02 '13 at 20:51
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Well, you're right; it's about "being half and having nowhere to go."
So let $G$ be a group and let $H$ be a subgroup of $G$ such that $[G:H] = 2$; then $H$ has two cosets of the form $gH$ or $Hg$ in $G$; it has two left cosets, and it has two right cosets. Now for any $g \in G$, $g \notin H$, we have $G = H \cup gH = H \cup Hg$ as the coset partitions of $G$. Since $H \cap gH = H \cap Hg = \varnothing$, we must have $gH = Hg$ since clearly $gH \ne H \ne Hg$; remember $g \notin H$ here. But $gH = Hg$ implies $gHg^{-1} = H$, whence $H$ is normal in $G$.
Hope this helps. Holiday Cheer,
and of course,
Fiat Lux!!!
Robert Lewis
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