Order of factor group of a cyclic group Assume that $G= \langle a \rangle$ is a cyclic group, and $H$ is a subgroup of $G$. Assume $H= \langle a^k \rangle$ , where $k$ is the smallest positive integer such that $a^k \in H$. Note that $G/H$ is cyclic, since $G$ is cyclic. Assume that $[G:H]=t$. Prove that $t=k$.
 A: There are a few ways to do this. One way (if the group is finite) is to show directly that $|H|=\frac{n}{k}$ where $n = |G|$ and then use Lagrange's theorem: $|G|=[G:H]|H|$. (You can show that $|H|$ is the smallest positive integer $m$ such that $(a^k)^m=e$ because $H$ is cyclic, then use this to deduce that $|H|=\frac{n}{k}$.)
Edit: possibly an easier way to do this is to use the fact that $G/H$ is cyclic, so it must have a generator. Clearly $aH$ generates as $(aH)^i=a^iH$ so the size of the group is just the order of $aH$. But this is the smallest positive integer $m$ such that $a^mH=H$ i.e. $a^m \in H$. This is, by definition, $k$.
Another way to do it is to use the fact $|G/H|=[G:H]$ as you seem to be suggesting and then explicitly compute what $G/H$ is. In your comment, you successfully identify the quotient group, but if you want to do so formally:
1) Identify the possible elements of $G/H$: Notice that since all elements in $G$ are of the form $a^i$ for some integer $0\leq i\leq n$, all possible cosets are of the form $a^iH$.
2) Show when two of these elements are the same: You can do this by saying that $a^iH=a^jH$ is equivalent to $a^{i-j}\in H$. Thus $a^iH=a^jH$ iff $k | i-j$.
3)Finally, you can now say that any of the (at most $n$) possible cosets from part 1 can be represented by one of the cosets you describe in your comment (by part 2) and that all of these are distinct (by part 2 again) so you have successfully identified $G/H$ which now clearly has $k$ elements.
