$G$ 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!

  • $\begingroup$ You should also prove that it is actually a homomorphism (this is where you need the group to be abelian). $\endgroup$ Dec 9, 2013 at 20:03
  • $\begingroup$ @TobiasKildetoft - How I prove it? where I should start? $\endgroup$
    – CS1
    Dec 9, 2013 at 20:04
  • 1
    $\begingroup$ Yes, precisely. $\endgroup$ Dec 9, 2013 at 20:10
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    $\begingroup$ 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. $\endgroup$ Dec 9, 2013 at 21:09
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    $\begingroup$ 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. $\endgroup$ Dec 9, 2013 at 21:35

1 Answer 1


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.)


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