Prove if $G$ is an abelian group, then $q(x) = x^2$ defines a homomorphism from $G$ into $G$. Is $q$ ever an isomorphism?

Question

Prove if $G$ is an abelian group, then $q(x) = x^2$ defines a homomorphism from $G$ into $G$. Is $q$ ever an isomorphism?

The first proof was no problem, I'm having trouble with the isomorphism piece though. My book has the following explanation that I don't quite understand:

"In order for $q$ to be an isomorphism, it must be the case that no element other than the identity is its own inverse. $x \in $Ker$(q) \iff q(x) = e \iff x*x = e \iff x^{-1} = x$"

It also has a hint to think about it another way, considering $\Bbb Z_n$ for small values of $n$, which I also can't quite wrap my brain around...

Answer 1

If there was an element other than the identity that is its own inverse, then the homomorphism would send it to the identity (because taking it squared would give you the identity).

Since a homomorphism from a group to itself is an isomorphism only if the identity is the only element in the group that's sent to the identity by the homomorphism, you can't have an element that is its own inverse.

A good example of a group that satisfies this is $\mathbb Z_3$. Generally, notice that the order of the group can't be even (can you see why?)

Answer 2

$q$ is injective iff $G$ has no element of order $2$.

If $G$ is finite, this happens iff the order of $G$ is odd. In this case, $q$ is an isomorphism.

If $G$ is infinite, $q$ might not be surjective even if it is injective. An example is $G=\mathbb Z$.