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?
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?
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.