Exercise 2.3.35 (If $\text{g.c.d}(k,|G|) = 1$, then $\varphi : G \to G,\ x \mapsto x^k$ is surjective)

Proof. Let $G=⟨x⟩$, where $|x|=n$ and let $k$ be an integer such that $k\wedge n=1$. Consider the map

We show that $\phi$ is surjective. Let $y$ be any element of $G=⟨x⟩$. Then $y=x^a$ for some integer $a$. Since $k\wedge n=1$, there are integers $u,v$ such that $1=\mathit{uk}+\mathit{vn}$, thus $a=\mathit{auk}+\mathit{avn}=\mathit{rk}+\mathit{sn}$, where $r=\mathit{au}$ and $s=\mathit{av}$ are integers. Since $x^n=1$,

This shows that $\phi$ is surjective.

We generalize. Suppose now that $G$ is a finite group of order $n$, and $k\wedge n=1$. We want to show that $\phi :G\to G$ defined by $\phi (x)=x^k$ is surjective (i.e, every element $y\in G$ is a $k^{\text{th}}$ power). Let $y$ be any element of $G$. Consider the subgroup $H=⟨y⟩$. By Lagrange’s Theorem, $|y|=|H|$ divides $n=|G|$. Therefore $y^n=1_G$. As above, there are integers $u,v$ such that $1=\mathit{uk}+\mathit{vn}$. Then

Put $x=y^u$. Then $y=x^k=\phi (x)$ is a $k^{\text{th}}$ power. This shows that $\phi$ is surjective. □