I know this means that there are two cosets. I also know that one must be H itself.
This means that the remaining coset(s) must be equal for right and left. Also since there is only one possible coset, this means that for all elements, a,b in G, then $aH = bH$ and $Ha = Hb$, since each element in G acting on the subgroup must produce the same set. This means $aH = Ha$ I believe. From there $H = aHa^{-1}$, which given the information I am not sure if this good or how to use it. Any help would good.
[Math] $H$ is a subgroup of index $2$ in finite group $G$, then every left coset of $H$ is a right coset as well
group-theory
Best Answer
The (left) cosets of $H$ partition the group $G$; since there are two cosets and one of them is $H$, that means that, whatever the other one is (call it $\mathcal{C}$), as a set, you have that $H\cap\mathcal{C}=\emptyset$ and $H\cup \mathcal{C}=G$. These two together tell you that the other one must be equal, as a set, to the complement of $H$ inside of $G$; that is, $\mathcal{C}=G-H$. So the two cosets are $H$ and $G-H$. Of course, $G-H$ can be written as $xH$ for some $x$; in fact, for any $x\notin H$ you have $xH = G-H$.
But the exact same argument holds for right cosets: one of them is $H$, and the other one must be, as a set, the complement of $H$. So the two right cosets are $H$ and $G-H$, and $G-H$ can be written as $Hy$ for some $y$; in fact, for any $y\notin H$ you have $Hy=G-H$.
Now note that in both cases, the cosets are: 1. $H$, the coset of the elements of $H$; and 2. $G-H$, the coset of the elements not in $H$.
So each of the two left cosets is also a right coset and vice-versa.