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

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.

Elsewhere, user Hagen Von Eitzen has given a nice proof of injectivity; see here. We can prove surjectivity in a very similar way:

Let $G$ be a group of order $2k-1$, and let $g\in G$. Then $g^k\in G$, and $$ (g^k)^2 = g^{2k} = g(g^{2k-1}) = g, $$ so the squaring map is surjective. $\Box$

Note: we didn't actually use the abelian-ness of $G$ here, just the odd-order condition. So it turns out that the squaring map is surjective (and injective) for every group of odd order, abelian or not.

However, in order for the squaring map to be a homomorphism, you do need $G$ to be abelian.

I will show onto by modifying the proof of Nicky Hekster in Prove that $\varphi$ is automorphism which is for the more general $\varphi(x)=x^m$ where $\gcd(m,n)=1$.

Let $y \in G$. We must find $x \in G$ s.t. $x^2=y$. By Bézout's, $\exists k,l \in \mathbb Z$ s.t. $2k+ln=1$. Then $y=y^1=y^{2k+ln} = y^{2k}y^{ln} = y^{2k}(1) = (y^k)^2 \to x=y^k$.

This proof

• makes use of the finiteness of the group having order $n$
• makes use of that $\gcd(2,n)=1$ because $n$ is odd
• does not use that maps $f: S \to S$ where $S$ is finite are one-to-one iff onto
• does not make use that $\varphi$ is one-to-one.

We have shown that $\varphi$ is onto even though it might not be one-to-one. We can actually prove one-to-one from this.