Prove that if $H$ is a subgroup of index $2$ in a finite group $G$, then $gH=Hg$ for all $g\in G$

Question

Prove that if $H$ is a subgroup of index $2$ in a finite group $G$, then $gH=Hg$ for all $g\in G$

I do know that index, or $\frac{|G|}{|H|}=2$ implies that $H$ is practically half of $G$, loosely speaking.

Moreover, I do know that every cyclic group of prime order is abelian, or $gH=Hg.$

Lastly, I do know that to show $gH=Hg,$ it is suffice to show that $gH \subseteq Hg $ and $Hg \subseteq gH $.

But I could not tie those clues together and come up with an argument.

Could you please point me in the right direction?

Thanks.

Answer 1

Let $xH$ be a coset of $H$ in $G$. Since cosets partition $G$, either $xH=H$ or it is the other coset $G-H$ (the other coset is made of the leftovers, so it's the set complement of $H$). If $xH=H$, then $x\in H$ so $xH=H=Hx$. Otherwise $x\not\in H$, so $Hx \not= H$. Thus $xH=G-H=Hx$. So all the left and right cosets match.

Answer 2

$G = H \cup gH = H \cup Hg$. Since cosets are disjoint and these are finite sets, it follows that $gH = Hg$.

Answer 3

We know that the cosets partition the group. That is we have that:

$G=H\cup gH$ for some $g\not\in H$ and $H\cap gH=\emptyset$

Now we want to show that $\forall k\in G$ we have $k^{-1}Hk=H$

So we know that either $k^{-1}Hk=H$ or $k^{-1}Hk=gH$

But as $H$ is a subgroup we know that $1\in H$ and so $k^{-1}1k=1\in k^{-1}Hk$.

This shows that $k^{-1}Hk\cap H\neq \emptyset$ and and so $k^{-1}Hk=H$ as required