# For a finite abelian $G$, $f: G\to G$ defined by $f(g)=g^2$ is an isomorphism iff $|G|$ is odd [duplicate]

Let $$G$$ be an abelian group of finite order, and define $$f: G\to G$$ by $$f(g)=g^2$$. I would like to prove that $$f$$ is an isomporphism if and only if $$G$$ has an odd order.

I am able to prove that $$f$$ (defined above) is a homomorphism if and only if $$G$$ is abelian fairly simply. If $$G$$ is abelian, then for all $$a, b\in G$$, $$f(ab)=(ab)^2=abab=a^2b^2=f(a)f(b)$$ So $$f$$ is indeed a homomorphism. And if $$f$$ is a homomorphism, then for all $$a, b\in G$$, $$f(ab)=f(a)f(b) \implies (ab)^2=a^2b^2 \implies a^{-1}ababb^{-1}=a^{-1}aabbb^{-1}\implies ba=ab$$ Meaning $$G$$ is abelian.

This proof holds for any (not necessarily finite) group.

Applying this to the original problem, we are now left with proving that $$f$$ is bijective if and only if $$G$$'s order is odd. Since $$f$$ is from a finite $$G$$ to itself, it will suffice to show that:

$$|G| \space \text{is odd} \iff f \space \text{is injective (or surjective)}$$ But here I am stuck. I believe there is a simple way to finish off the proof, but I can't think of it, and I couldn't find this question asked here before. I appreciate any help.

• Think about the kernel. Can there be a non-identity element in the kernel? Feb 3 at 14:14
• I think the kernel is the set of $g\in G$ that are the inverse of themselves. I also know that the kernel of a homomorphism is a normal subgroup of the domain... Am I on the right track..? Feb 3 at 14:20
• When $n=2m+1$, we have $1=g^n=(g^m)^2g$ and so $g=(g^{-m})^2$. Thus, the map $g \mapsto g^2$ is surjective.
– lhf
Feb 3 at 14:20

Suppose $$|G|$$ is odd. Let $$g\in{ker(f)}$$, meaning that $$g^2=1$$. Thus $$g=1$$ by Lagrange, otherwise $$g$$ is an element of order 2 in a group of odd order.

Now, suppose $$|G|$$ is even. Then, $$G$$ admits an element of order 2, let's say $$h\in{G}$$. Thus $$ker(f)$$ isn't trivial, because $$h\in{ker(f)}$$, meaning that $$f$$ is not an isomorphism.

• Nice. What about surjectivity? Feb 3 at 14:27
• @FShrike An injective function between sets of the same cardinality...
– Pedro
Feb 3 at 14:30
• As @HappyDay stated, since $f$ is a function between finite sets of the same cardinality, you have $f$ is bijective iff $f$ is injective. Feb 3 at 14:31
• Can someone quickly explain why $G$ necessarily has an element of order $2$ in the second case? Feb 3 at 14:36
• @HappyDay In general, if p is a prime dividing the order of the group, then G admits an element of order p (that's called Cauchy's lemma). But in this case, simply, G is abelian, then it admits a subgroup of order $d$ for every divisor of $|G|$, in particular admits a cyclic subgroup of order $d$ for every divisor of $|G|$. You take the generator as your $h$. Feb 3 at 14:41

Your analysis is good. In summary, if $$G$$ is a (multiplicative) group, then the map $$f\colon G\to G$$, $$f(x)=x^2$$ is

1. a homomorphism if and only if $$G$$ is abelian;

2. surjective if and only if it is injective, when $$G$$ is finite.

Now, suppose the map is not injective. Then there exists an element $$x\in\ker f$$, $$x\ne1$$, meaning that $$x^2=1$$ and so $$\langle x\rangle=\{1,x\}$$ is a subgroup of order two. By Lagrange's theorem, the order of $$G$$ is even.

Conversely, if the order of $$G$$ is even, then there is a subgroup of $$G$$ having order two, by Cauchy's lemma, which implies that there exists $$x\in G$$, with $$x\ne1$$ such that $$x^2=1$$, and therefore $$f$$ is not injective.

Actually, Cauchy's lemma is not needed. Suppose $$G$$ (abelian or not) is a finite group of even order. Then you can consider the equivalence relation $$\sim$$ on $$G$$ defined by $$\text{x\sim y if and only if either y=x or y=x^{-1}}$$ (the proof this is an equivalence relation is easy). The equivalence classes have cardinality $$1$$ or $$2$$. If you remove the equivalence classes of cardinality $$2$$, you remain with an even number of classes of cardinality $$1$$; since the identity element lies in such an equivalence class, there must be another one, so another element $$x\ne1$$ that equals its reciprocal; but $$x=x^{-1}$$ is the same as $$x^2=1$$.

$$f$$ is injective if and only if $$kerf=\{x\in G;g^{2}=0\}=0$$. It is clear that $$kerf=0$$ if and only if $$G$$ has no element of even order (Because suppose that $$g\in G$$ such that $$g^{m}=0$$ where $$m$$ is even. Then, $$m=2n$$ for some $$n$$. Hence, $$(g^{n})^{2}=0$$. This shows that $$g^{n}\in kerf$$ which is a contraction).