How can a subgroup have multiple cosets? I am currently reading An Introduction To The Theory Of Groups, by Joseph Rotman, and in a section describing cosets, there is an exercise question as follows;
Let $H$ be and subgroup of $G$ having exactly two right cosets (right coset in the definition below). Show $g^2 \in H$ for every $g \in G$.
Firstly, using the usual definition of a coset;
$St = \{st:s\in S\}$ taking $t$ as the representative of {$t$}, where $S$ and $t$ are subsets of $G$
My question following this is, if the representative of the subset, $T = \{t\}$, is $t$ as stated, how can a subgroup have multiple cosets if the subsets $S$, $t$ and the binary operation remains constant? The way I am approaching it in my head is, as the choice of representative of $T$ is arbitrary, thus cosets would all be equal no matter what choice for $t$ to be.  
I am more than likely far off approaching this problem in the correct manner, but I am currently really struggling with this section in the book (along with the written structure) so any advise, or clearer reading material suggestions would also be much appreciated.  
Thank you for your time.
 A: Maybe an example is best. You know the cyclic group of order $2$, right? Denote it $G=\{1,x\}$. Well $\{1\}$ is a subgroup of $G$ (why?). Now observe that both $1\{1\}$ and $x\{1\}$ are cosets of $\{1\}$. So $\{1\}$ and $\{x\}$ are cosets of $\{1\}$. So $\{1\}$ has precisely two distinct cosets in $G$. Make sense?
A: Perhaps start by considering the two options: either $g \in H$ or $g \not \in H$. In the first case, it is clear that $g^2 \in H$. For the latter, it is not so obvious.
Suppose that $g \in G \backslash H$.
If $Hg = H$ then $g$ must be in $H$ by the closure of $H$. This is a contradiction, so we know that $Hg \not = H$. In fact, we cannot have that $h_1 g = h_2$ for any $h_1, h_2 \in H$ so the coset $Hg$ must contain nothing in $H$. 
Let this second coset of $H$ be denoted $J$. We have that $Hg = J$ and $J \cap H = \{0\}$.
Now suppose that $Hg^2 = J$. This would mean that $Jg = J \Leftrightarrow J = Jg^{-1}$. However, we know that $J = Hg$ so $Jg^{-1} = H$. Thus $Hg^2 \neq J$. Therefore $Hg^2 = H$.
I would say that the key point of this is that if you take some subgroup $H$ of $G$, and you repeatedly take right cosets with increasing powers of $g$ ($H,Hg,Hg^2,\dots$) you can't have that $Hg^n = Hg^{n+1}$ unless $g \in H$. In this case, we restrict ourselves to having just two cosets, so the only thing that $Hg^2$ can be is $H$.
