We should first show that $f: G \to G$,
$$f = x^m$$
is indeed a homomorphism: Since $G$ is an abelian group, $f(xy) = (xy)^m = x^my^m = f(x)f(y).$
Next, for showing $f$ is an isomorphism, hence an automorphism, we must show that $f$ is a bijection. For doing so, showing $f^{-1}$ exists is enough. In this case we should make use of Lagrange's theorem and elementary number theory.
$(n, m) = 1$ if and only if there are integer solutions to $mQ + nR = 1$. For such $Q$,
$$mQ \equiv 1 \pmod{n}.$$
Which leaves us with $mQ = kn + 1$. By Lagrange's theorem $a^n = e$ for $\forall a \in G$. Using this fact, we can define $f^{-1}$ to be $x^Q$ because:
$$f^{-1}(x^m) = (x^m)^Q = x^{mQ} = x^1$$
$$x^m \mapsto x.$$
Since $f$ is a bijective homomorphism (i.e. isomorphism) from the group $G$ to itself, it is an automorphism.
QED
Is my proof correct? If it is so, is it rigorous enough?
Best Answer
It seems fine to me. Well done!
There's not much else to say.
It's rigorous enough for me.
One thing I'd like to point out is that you started a sentence with a mathematical symbol. This is usually advised against, for form's sake.
Also, you could say that you get $Q$ and $R$ from Bézout's Identity.