Let $G$ be abelian. Let $n \in \mathbb{Z}^+$ such that the order of $G$ and $n$ are relatively prime. Show that the function $\varphi : G \rightarrow G$ defined by $\varphi (a) = a^n $ is an automorphism for $a \in G$.
I'm stuck in proving 1-1-ness of the function. Let $a,b$ in $G$ such that $\varphi (x) = \varphi (y)$ then $a^n = b^n$. How can I show that $\varphi$ is 1-1 from here on out?
Best Answer
I assume you have shown that $\phi$ is a homomorphism.
Showing that a homomorphism is injective is equivalent to showing that its kernel is trivial.
Let $a \in \ker \phi$. We want to show that $a=1$. Let $g$ be the order of $G$ and let $k$ be the order of $a$. By Lagrange's theorem, $k \mid g$. Since $a \in \ker \phi$, we know that $a^n = 1$. Hence $k \mid n$. Since $\gcd(n,g) = 1$, there exist $x,y \in \mathbb{Z}$ such that $nx + gy = 1$. Now $k$ divides both $g$ and $n$, so it divides (a multiple of) their sum $nx + gy$. So $k \mid nx + gy = 1$. Therefore $k=1$ (orders of elements are $>0$). So $a$ has order $1$. It follows that $a$ is the neutral element.
$\ker \phi$ is trivial, so $\phi$ is injective.