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!
Best Answer
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.