I was wondering if my proof of the fact that the kernel of a group homomorphism $$ C_m\to C_m,\ x\to x^r$$ has order $(m,r)$ is correct?
If $x^k\mapsto x^{kr}=0$, then $kr=m\alpha,\ \alpha \in\mathbb Z$. Let $d=(r,m)$. Then $$\frac{r}{d}k=\frac{m}{d}\alpha.$$ Since $m/d$ divides $rk/d$ and $(r/d,m/d)=1,\ m/d$ divides $k: k=\beta\frac{m}{d}$. For $\beta=0,1,\dots,d-1$ this gives $d$ distinct non-identity elements $x^k$, which lie in the kernel. $\square$
One particular thing I'm hessitant about: is it correct that $x^k$ for the $k$'s we found above is a nontrivial element because $\beta/d$ is not an integer and hence the expression $(x^m)^\beta/d$ (which may potentially be the trivial element) does not make sense?