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

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.

• Hint: "odd order" could imply that the group is finite. Apr 3, 2014 at 17:44
• – lhf
Apr 3, 2014 at 18:00

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.

• thank you.that helps. but why G is necessary finite? Apr 3, 2014 at 17:46
• Well, the number of elements is assumed odd, so it certainly must be finite. For an infinite Abelian group without elements of order $2,$ the map $x \to x^{2}$ is an injective homomorphism, but need not be surjective ( eg consider $G = (\mathbb{Z},+), z\phi = 2z.$ Apr 3, 2014 at 17:51

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.