# Power automorphism and abelian groups

First of all I'm not sure if "Power automorphism" is the correct term, so I apologize if it is not.

"Let $G$ be an abelian group of order $n$, and $m$ an integer. $f:G\rightarrow G$ s.t. $f(a)=a^m$. Find a sufficient and necessary condition such that $f$ will be an automorphism."

I know that $f$ will be an automorphism iff $\operatorname{Im}(f)_m=\{a^m|a\in G\}=G$.

Obviously that is correct for $m=1+nt$ for all $t\in\mathbb{Z}$. I also know the inverse map is a bijection so it also holds for $m=-1$.

I think that I'm not being very systematic. Does anyone have an idea regarding how to approach this problem? Thanks in advance =]

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Your "if and only if" is not necessarily true if your group is infinite. (This doesn't matter here, but is good to know!) –  user1729 Aug 27 '13 at 19:11
Is the current answer insufficient? (I ask because you keep bumping the question.) If the question you asked in your comment to the answer is still sticking you, you could either edit your question to say this or start another question referencing this one and asking that point. (Although I have answered this comment now.) –  user1729 Aug 29 '13 at 16:32
Thank you, I will try to do so from now on. –  Py42 Aug 29 '13 at 17:56

Hint 1

This map is always a homomorphism.

Hint 2.1

By the pigeon-hole principle, it is enough to prove it is injective.

Hint 2.1.1

Suppose $x \in \ker(f)$. We know $x^{m} = 1$, but also $x^{n} = 1$, so $x^{\gcd(n, m)} = 1$.

Hint 2.1.2

So if $\gcd(n, m) = 1$ we have...

Hint 2.1.3

Conversely if $k = \gcd(n, m) > 1$, let $p$ be a prime dividing $k$, and let $x$ be an element of order $p$...

Hint 2.2

By the pigeon-hole principle, it is enough to prove it is surjective.

Hint 2.2.1

If $\gcd(n, m) = 1$, use Bézout to show that the map is surjective.

Hint 2.2.2

If $k = \gcd(n, m) > 1$, use Bézout to show that the map is not surjective.

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Thank you, I much enjoyed the hint-structure. Regarding the last one, how can you use Bézout's identity to show that if $gcd(n,m)=k>1$, the map isn't surjective? –  Py42 Aug 27 '13 at 17:17
@Py42 If $k>1$ then $1$ will not be in the image. –  user1729 Aug 29 '13 at 16:33
f(1)=1, isn't it? –  Py42 Aug 29 '13 at 17:34
@Py42 1 isn't the identity here, 0 is. –  user1729 Aug 29 '13 at 18:35
@user1729 Which element did you mean by $1$? For $(m,|G|)=k>1$ the image is $\{g^m|g\in G\}$. I'm not finding a contradiction to $f$ being injective/surjective. –  Py42 Aug 29 '13 at 19:52