TITLE: Subgroup of index $2$ is Normal
QUESTION [56 upvotes]: Please excuse the selfishness of the following question:

Let $G$ be a group and $H \le G$ such that $|G:H|=2$. Show that $H$ is normal.

Proof:
Because $|G:H|=2$, $G = H \cup aH$ for some $a \in G \setminus H$. 
Let $x\in G$. Then $x \in H$ or $x \in aH$.
Suppose $x \in H$. Then $xhx^{-1} \in H$ because $H$ is a group.
Suppose $x \in aH$. Then $ahH(ah)^{-1} = ahHh^{-1}a^{-1} = a(hHh^{-1})a^{-1}$ ***1
***1: Can I say now that $x \in H$, based on the that $hHh^{-1} \in H$ and that $aa^{-1} = e$ ?
Thank you for the time you spent reading my doubts.

REPLY [2 votes]: If $[G:H]=2$, then necessarily $Ha=aH$ for every $a\in G\setminus H$.