Suppose that $H$ is a (cyclic) subgroup of order $m$ of a cyclic group
$G$ of order $n$. What is $G/H$?
It's a very simple question but I am still struggling with getting accustomed to the notion of the quotient group.
Now, I know that $m$ divides $n$ and that $G/H$ will be a group of cosets such that $\{gH:g\in G\}$. As a result, there will be $n/m$ of such cosets. I was wondering I can say anything more specific than just that?
Best Answer
The quotient of a cyclic group is again cyclic. A cyclic group is a group which is generated by a single element. If $x$ is a generator of $G$, then $xH$ is a generator of $\frac{G}{H}$. So $\frac {G}{H}$ is a cyclic group of order $\frac {n} {m}$. Can't get much more specific than that!