Let $G$ be an Abelian group with odd order. Show that $\varphi : G \to G$ such that $\varphi(x)=x^2$ is an automorphism

Question

Let $G$ be an Abelian group with odd order. Show that $\varphi : G \to G$ such that $\varphi(x)=x^2$ is an automorphism.

I was able to show that the $\varphi$ function is a homomorphism and one-to-one. But I wasn't able to show that it is onto.

Answer 1

Since $G$ has odd order, that means $G$ is finite. So $\phi$ can be injective if and only if it is surjective. To formally prove surjectivity, argue by contradiction. Suppose there is a $y \in G$ such that $\forall{x} \in G$, $\phi(x) \neq y$. That implies there are two distinct $x_{1}, x_{2}$ such that $\phi(x_{1}) = \phi(x_{2})$ by injectivity.