I have seen some proofs of this fact before: here using orbits, and here using the fact that cosets partition the group. I have found an alternative proof which I am struggling to understand.
Theorem. Let $G,H$ be groups, and $H \leq G$. If $|G:H|=2$ then $H$ is normal in $G$.
Proof. There are just two left cosets, $H=eH$ and $gH$, say. Since distinct left cosets are disjoint, and the union of all left cosets is $G$, $gH =G \setminus H$. But then $gH$ is closed under taking inverses, so $$gH = \{ h^{−1}g^{−1}:h\in H\}=Hg^{−1},$$ so there are also two right cosets $H$ and $Hg^{−1}=gH$, and $xH =Hx = \begin{cases} H & \text{if } x \in H \\ G \setminus H & \text{if } x \notin H \end{cases}$.
[If $G$ is finite, the proof is easier since then $|G:H|=|G|/|H|$, and so doesn’t depend on whether we use left cosets or right cosets. So thereare two left cosets $H$ and $G \setminus H$ and two right cosets $H$ and $G \setminus H$.]
My questions:
- Why is $gH$ closed under taking inverses?
- Why is the proof easier is $G$ is finite? Why can't I just say the two left cosets are $H$ and $G \setminus H$ if $G$ is infinite? It doesn't seem to be an issue in the 2nd post linked.
Thank you so much for your help.
Best Answer