H is a subgroup of index 2 in finite group G, then every left coset of H is a right coset as well

Question

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.

Answer 1

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.

Answer 2

You say "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." but this is not true. Let a be the identity and b not be in H.

But earlier you say "I know this means that there are two cosets. I also know that one must be H itself." And if all cosets are the same size, what can you say about the one that isn't H?

Answer 3

If $|G:H| = 2$ and $1 \neq g \in G$, then $$G = H \; \dot{\cup} \; gH \;\;\; and \;\;\; G = H \; \dot{\cup} \; Hg$$ Therefore $$gH = Hg \;\;\;\; \forall 1 \neq g \in G$$ Of course $$gH = Hg \;\;\;\; for \;\; 1 = g$$ Therefore $$gH = Hg \;\;\;\; \forall g \in G$$