Prove that $\varphi$ is automorphism

Question

$G$ is commutative group. $|G|=n$.
$m\in \mathbb{N}$ and $\gcd(m,n)=1$.

I need to prove that $\varphi :G\to G$, $\varphi(x)=x^m$ is automorphism of G.

My try:
I assume that $a\in \ker(G)$, so $a\in G$ and in one hand: $a^m=e$ (because $\varphi(a)=e$), and at the other hand $a^n=e$ because $a\in G \Longrightarrow$ because $\gcd(m,n)=1,\;a$ must be $e$, so $\ker(\varphi)=\left\{e\right\}$.
And that's mean that $\varphi$ is Aut.

I'm right? my proof is OK?

Thank you!

You should also prove that it is actually a homomorphism (this is where you need the group to be abelian). — Tobias Kildetoft, Dec 09 '13 at 20:03
@TobiasKildetoft - You mean: for $a,b\in G$: $\varphi(a\cdot b)= (ab)^m=a^m\cdot b^m=\varphi(a)\cdot \varphi(b)$? — CS1, Dec 09 '13 at 20:09
Yoav, in order to be an automorphism, you have to show that $\varphi$ is injective and surjective. For a finite group injectivity is equivalent to surjectivity. And ker$(\varphi)={e}$, so injectivity is assured. Another way is to use Bézout's Theorem: there are integers $k$ and $l$ such that $km+ln=1$. If $y \in G$ then $y=y^{km+ln}=y^{km}.y^{ln}=(y^k)^m$. So for each $y \in G$ you can find an $x \in G$ namely $x=y^k$, such that $\varphi(x)=y$. So the map is surjective. — Nicky Hekster, Dec 09 '13 at 21:09
YoavFridman, How do you go from $\gcd(n,m) = 1$ (and $a^m = e$ and $a^n = e$) to $a = e$? It seems to me you need an argument like @NickyHekster 's above, using Bezout's Theorem. — Magdiragdag, Dec 09 '13 at 21:35
Possible duplicate of Let $G$ be an abelian group of odd order. Prove that the map $\varphi : G\rightarrow G$ defined by $\varphi (x)=x^2$ is an automorphism — , Oct 12 '18 at 02:21

score 2 · Answer 1 · answered Dec 30 '14 at 01:45

As pointed out in the comments, the proof given above is nearly correct--it just needs to show that $\phi$ is a morphism. (This was done in the comments, but since comments are supposed to be ephemeral, I'll include the outline below.)

If $a, b\in G$, then $\phi(a)\phi(b) = a^mb^m = (ab)^m = \phi(ab)$. Thus, $\phi$ is a homomorphism. (This requires $G$ to be abelian.)

Prove that $\varphi$ is automorphism

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