Showing $x^m$ is an automorphism

Question

Let $m,n \in \mathbb N$ and they are relatively prime and $G$ is a group with $|G|=n$.

Show that $T:G \to G$ is an automorphism of $G$ where $T(x)=x^m$

First I showed it's a homomorphism. I have to show bijectiveness. I wrote for being onto in brief : $x^m \in G$ and we can find $x \in G$ which satisfies that.

But I'm not sure how should I show being (1-1) and how can I use being relatively prime.

I know it's so basic question but I've stuck.

Thanks in advance for any help! :)

Correction : $G$ is commutative

Answer 1

To show injectiveness, assume that there exists $a$ and $b$ such that $a^m = b^m$, and then try and prove why you should have $a = b$.

Hint : in a group of order $n$, what is the condition you need on $k$ to have $x^k = 1$, with $x \in G$ ?

Answer 2

You can prove that the map is surjective directly.

Let $a,b \in \mathbb Z$ such that $am+bn=1$.

Then $x = x^1 = x^{am+bn} = (x^a)^m (x^n)^b = (x^a)^m = T(x^a)$.

Since $G$ is finite, surjective implies bijective.