Let $G=\langle a\rangle$ and $H=\langle a^2\rangle$. Find all the right cosets of $H$ in $G$.
Additional info: I understand that a right coset of $H$ in $G$ is of the form $Ha=\{ha:h \in H\}$. But I am not sure if the cyclic groups are finite or infinite, and I don't understand how to find right cosets. Any type of help would be appreciated.
On
It might be helpful to recall that the cosets of a group partition it.
Since $H:= \{ a^{2k} \mid k \in \mathbb Z \}$ contains the group identity element, we see $Ha = \{a^{2k + 1} \mid k \in \mathbb Z \}$. But then $H \cup Ha = \{ a^k \mid k \in \mathbb Z \} = G$, so $H$ and $Ha$ are the only possible cosets.
On
Consider the map $\sigma\colon G\to G$ defined by $\sigma(x)=x^2$. This is a homomorphism, because $G$ is abelian, being cyclic. The kernel of $\sigma$ is $$ \ker\sigma=\{x\in G:x^2=1\} $$ while its image is $H$.
If the order of $G$ is odd, no element $x\ne1$ can have the property that $x^2=1$, so $\sigma$ is injective, hence surjective. So, in this case, $H=G$ and, of course, there's only one coset.
If the order $n$ of $G$ is even, there is exactly one element $x\in G$ such that $x^2=1$ and $x\ne 1$, precisely $a^{n/2}$. (Why?) So the kernel of $\sigma$ has two elements and $|H|=|G|/|\ker\sigma|=n/2$. Therefore $|G|/|H|=2$ and so there are exactly two cosets; since $a\notin H$ (why?), the cosets are $$ H\quad\text{and}\quad Ha. $$
If $G$ is infinite, then $G\cong\mathbb{Z}$ and the image of $H$ under this homomorphism is $2\mathbb{Z}$. Again the cosets are $$ H\quad\text{and}\quad Ha. $$
Hint :
As written in the question, cosets are of the form $Ha$ where $a\in G$. Clearly, $He=H$ is a coset of H in G.
Now use the fact that $aH=bH$ iff $ab^{-1} \in H$.
Note : i don't know if this is the correct way to say it, but in $G/H$, elements of $H$ are considered as identity elements.