I need help with the following abstract algebra problem:
Given $C_n$ the cyclic group of order $n$, we define the map \begin{align*} \alpha_m : C_n &\to C_n\\ x &\mapsto x^m \end{align*} Prove that $\alpha_m$ is automorfism if and only if $gcd(m,n)=1$.
The first part is proving that $\alpha_m$ is homomorfism, which is easy seen by doing this: $$\alpha_m(xy)=(xy)^m=x^my^m=\alpha_m(x)\alpha_m(y).$$ Now I need to prove it's injective and surjective. For injectivity, I assume there are some $x$ and $y$ that verify $x^m=y^m$. And here's where I'm stucked. Maybe considering $m=kn+r$? Any help will be appreciated.
Hint : $\gcd(n,m)=1$ implies that there exists $u,v$ such that $un+vm=1$. Thus
$$x=x^{1}=x^{un+vm} = (x^n)^u (x^m)^v = (x^m)^v = (y^m)^v = (y^m)^v (y^n)^u = y^{mv+un} = y$$
Notice also that proving injectivity is enough, since the map is a morphism from a finite group into itself.
Edit : For the converse, you can prove that if $\alpha_m$ is surjective, then $\gcd(n,m)=1$. Let's write $(C_n, \times)$ as $(\mathbb{Z}/n\mathbb{Z}, +)$. Then if $\alpha_m$ is surjective, there exists $x \in \mathbb{Z}/n\mathbb{Z}$ such that $$mx=1 \in \mathbb{Z}/n\mathbb{Z}$$ which means that there exists $k$ such that $mx + kn = 1$ in $\mathbb{Z}$, so $\gcd(n,m)=1$.