Let $G$ be a finite abelian group of order $n$. Consider $m$ which is relatively prime to $n$. Prove that every $g\in G $ can be written as g=$x^m$ with $x\in G$.
$\phi:G \to G$ $g,h \in G$ then$\phi (gh)=g^mh^m=\phi(g)\phi(h)$ by abelian group and One one will be sufficient for showing onto.But for one one $\phi(g)=\phi(h)$ that $g^m=h^m$ ,$(gh^{-1})^m$=e but m is relatively prime to n So there exist integer a and b such that ma+bn=1 Now consider $gh^{-1}$=$(gh^{-1})^{ma+bn}$=e therefore g=h Hence one one .whether this my solution is correct ,Or any other way to do this ? I thought in this way we can give automorphism of abelian group.
(Indeed, if $m$ is relatively prime to $r$, then $m$ is an invertible element (where the group action is multiplication) in $\mathbb{Z}/r\mathbb{Z}$.)
So let $g$ be any element in $G$. Then $g^r = 1$ for some $r$ dividing $n$. As $m$ and $n$ are relatively prime, so are $m$ and $r$. So use Fact 1, and let $c$ be the integer such that $cm \equiv 1$ mod $r$. Then $(g^c)^m = g^{cm} = g$ (because $cm \equiv 1$ mod $r$)