Help understanding proof of "a subgroup of index $2$ is normal" 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.
 A: *

*Suppose otherwise, if an element of $a \in gH$ had an inverse in $H$ ($a^{-1}\in H$), then that would imply $a \in H$, contradicting the fact that cosets are distinct.

*I think you’re right about this one, the proof seemingly over complicated proving what the two cosets are. Cosets partition a group regardless of whether or not a group is infinite, so if there are two cosets only, then it’s $H$ and $G-H$.

A: About (2):
This is because the index of a subgroup $H$ in $G$ is commonly defined as the number of left cosets. Thus there might be more than two right cosets. If $G$ is finite, then we can immediately see that this can't be the case: $G$ partitions into the left cosets $H$, $aH$, and into the right cosets $H$, $Hb_1 $, $Hb_2 $, ..., $Hb_n $. As all these cosets have $|H|$-many elements in them, they can only add up to having a total of $|G|$-many elements if $n=1$.
If $G$ is infinite, then we can't use this easy cardinality argument, and so the author has to argue that there can only be the right coset $Ha^{-1}$.
