Let $G$ be a finite group of order $d$ and $n$ be an integer with $\gcd(n, d)=1$. Prove the mapping $f:G\to G$, $f(x)=x^n$ is bijective.

Question

I have trouble with an exercise in group theory.

Let $G$ be a finite group of order $d$ and $n$ be an integer with $\gcd(n, d)=1$. Prove the map $f:G\to G$, $f(x)=x^n$ is bijective.

Since the group may not be abelian, the mapping may not be a homorphism, so I can't use notions related to that like kernel. I don't know how to make use of the order condition.

Is this a classical exercise or theorem?

Any hint is welcome, thanks!

Answer 1

Since it is a map between finite sets it is enough to show that it is injective. There exists $u,v$ with $un+vd=1$. Lagrange implies that $x^d=1$ for every $x\in G$, we deduce that $x=x^{un+vd}=(x^n)^u$ thus $x^n=y^n$ implies that $(x^n)^u=(y^n)^u$ and $x=y$ thus $f$ is injective and surjective.

Answer 2

It is perhaps fun to note that the converse also holds, namely

Let $G$ be a finite group of order $d$ and $n$ be an integer. If the map $f:G\to G$, $f(x)=x^n$ is bijective, then $\gcd(n, d)=1$.

Proof: assume that $\gcd(n,d) \neq 1$, say the prime $p$ divides both $n$ and $d$. Cauchy's Theorem asserts the existence of an $x \in G$, with $order(x)=p$, $x \neq e$, the identity element of $G$. However, $f(x)=x^n=(x^p)^{\frac{n}{p}}=e^{\frac{n}{p}}=e$. Since $f$ is injective this would imply $x=e$ a contradiction.