If H is a subgroup of a finite group G of index [G: H] two, then H is normal in G.

I need to show that the left cosets equals to the right cosets, but what are the cosets here?

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- Nov 8th 2009, 05:11 PMapple2009normal subgroup and cosets
If H is a subgroup of a finite group G of index [G: H] two, then H is normal in G.

I need to show that the left cosets equals to the right cosets, but what are the cosets here? - Nov 8th 2009, 11:10 PMNonCommAlg
there are more than one way to prove this. one way is to write $\displaystyle G=H \cup xH,$ where $\displaystyle x \notin H.$ let $\displaystyle g \in G.$ if $\displaystyle g \in H,$ then clearly $\displaystyle g H g^{-1} = H.$ so we may assume that $\displaystyle g = xh,$ for some $\displaystyle h \in H.$

then $\displaystyle gHg^{-1}=xhHh^{-1}x^{-1}=xHx^{-1}.$ now if $\displaystyle xh_1x^{-1} \in xH,$ for some $\displaystyle h_1 \in H,$ then $\displaystyle x \in H,$ which is a false result. so $\displaystyle xHx^{-1} \subseteq H$ and we're done.

by the way, i just realized that you've already posted this question in here. next time don't post your question twice.