3

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?

Shaun
  • 47,747
İbrahim İpek
  • 1,126

1 Answers1

1

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.

Shaun
  • 47,747