cyclic group - automorphism - $f_a(x) = x^a$

Question

Question:

Let G be a cyclic group of order n. For each $a\in\Bbb{Z}$ define $f:G\to G ,f_a(x) = x^a$.

Prove that $f_a$ is an automorphism of G if and only if $a$ is prime to $n$.

I managed to show that if $a$ is prime to $n$ then $f_a$ is automorphism. When I try to prove the reverse, I need to proof that when $f_a$ is an automorphism, (i.e. the kernel is $e$), $a$ is prime to $n$. I managed to show that only when $m=kn$ will $g^{ma}=e=g^{pn}$ (m,p,k are integers). I felt the two $a$ and $n$ must be relatively prime. But I cannot prove it. Can anyone help?

Answer 1

Consider $kerf$ Let $x\in kerf$ then $x^a=e$.Thus order of $x$ divides $a$.Also order of an element divides order of the group .order of $x$ divides $a$ and $n$ both .But $gcd(a,n)=1$.Thus order of $x =1$ thus $x=e$.And in a finite group injectivity implies surjectivity and showing homomorphism is easy.Conversely let $G=<g>$ let $d=gcd(a,n)>1$.Then $a/d,n/d$ are integers.$f_a(g^{n/d})=g^{na/d}=e.g^{n/d}\in kerf_a$ since $n>n/d>1 g^{n/d}\neq e$ thus contradiction that $d>1$