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I'm trying to understand the proof that any group of index $2$ is normal.

Let $G$ be a group and $H$ a subgroup of index $2$.

I understand that if $x\in H$ then $xH=H=Hx$ since $H$ is a subgroup. However, I don't understand why if $x\in G\setminus H$ then $xH=G\setminus H.$

I know there are other proofs of this on this site but none of them specifically address this question.

MHW
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    To address your specific question: Since $H$ has index $2$, by definition $G$ is partitioned into two left cosets of $H$, call them $H$ and $gH$. If $x \in G \setminus H$ then $x \not\in H$ and so $x$ must be in $gH$, and this implies that $G\setminus H \subset gH$. Conversely, if $x \in gH$ then $x \not\in H$, so $x \in G \setminus H$. This shows the opposite containment $gH \subset G \setminus H$. Since both containments hold, we have $gH = G\setminus H$. Therefore, if $x \in G \setminus H$ then $x \in gH$, i.e. the coset containing $x$ is $gH$, i.e. $xH = gH = G \setminus H$. –  May 18 '16 at 14:12

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If $ H $ is a subgroup of index 2, then it has two cosets in $ G $, which we may denote by $ H $ and $ G - H $. If $ g \in H $, then clearly we have $ gH = Hg = H $. Otherwise, $ gH, Hg \neq H $ since $ g \notin H $, however any coset of $ H $ is either $ H $ or $ G - H $, so we must have that $ gH = Hg = G - H $. Since the left and right cosets coincide, we conclude that $ H $ is normal.

Ege Erdil
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