I was reading the solution to a problem I had been stuck on and one of the lines was this:
If $p$ is prime, then $| C_{p^r}^{(p^s)}| = p^{ \min \{r,s \}}$, where $C_{p^r}^{(p^s)}$ is the $p^s$-torsion subgroup of the cyclic group $C_{p^r}$.
I don't understand how this line follows. Of course if $s\geq r$, then the result is clear, but I don't know how this was obtained in the case that $s<r$
Every subgroup of a cyclic group is cyclic. So the $p^s$-torsion subgroup $H$ must be cyclic too. The generator $a$ of that subgroup has order $p^k$, $k\le s$ (assuming $s\le r$). On the other hand for every number $m$ dividing the order of an abelian group there is a subgroup of order $m$. Hence there is a subgroup of order $p^s$. All its elements are $p^s$-torsion by Lagrange. Hence this subgroup should be inside $H$. Hence it is equal to $ H$ and $k=s$.