TITLE: 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
QUESTION [5 upvotes]: A problem from Artin's Algebra (not the problem I am asking):

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.

I have solved this problem, also we can see here.
Then comes Generalization of the problem:

Let $G$ be an abelian group of finite order. Prove that the map $\varphi : G\rightarrow G$ defined by $\varphi (x)=x^k$ is an automorphism, where $k$ does not have any prime factor same of order of $G$.

Is it ok?

REPLY [6 votes]: Here is a roadmap:

Prove that $\varphi$ is a homomorphism. Use that $G$ is abelian.
Prove that $\varphi$ is injective. Use that the order of $G$ is odd.
Prove that $\varphi$ is surjective. Use that $G$ is finite. 

For the generalization, use that $\gcd(k,n)=1$ implies $ku+nv=1$ for $u,v \in \mathbb Z$ in the second step.