# Showing a group is surjective to show it's an automorphism.

Let $G$ be a group. The set of all automorphisms of $G = \operatorname{Aut}(G$), with $(\operatorname{Aut}(G), \circ)$ also being a group.

Consider $C_n=\langle g:g^n=1\rangle$, the cyclic group of order $n$. For each positive integer $m$ with $1\leq m\leq n$ define a map $f_m:C_n\rightarrow C_n$ as follows: for $r\in \mathbb{Z}, f_m(g^r):=g^{rm}$

$1)$ Suppose that gcd$(m,n)=1$ and deduce that $f_m$ is an automorphism:

My attempt: If gcd$(m,n)=1$, then this implies the kernel is trivial, thus we know it is injective. Furthermore since $G$ is acting on itself, it is an automorphism if it is an ismorphism. Thus we now need to show that it is surjective (to make it bijective) and a homomorphism. I have shown it is a homomorphism myself but don't know how to show it is surjective. How can I show it is surjective?

• Please clean up your question: there are quite a few things that have nothing to do with what you are asking or that simply do not make sense :-( – Mariano Suárez-Álvarez Feb 22 '17 at 20:06

For finite sets $A$ and $B$ of the same cardinality, an injective function $f:A\to B$ is automatically surjective and a surjective function $g:A\to B$ is automatically injective.