Square map is endomorphism iff abelian

This article describes an easy-to-prove fact about basic notions in group theory, that is not very well-known or important in itself
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Statement

Verbal statement

The square map on a group, viz the map sending each element to its square, is an endomorphism if and only if the group is abelian.

Statement with symbols

Let $G$ be a group and $\sigma :G\to G$ be the map defined as $\sigma (x)=x^2$. Then, $\sigma$ is an endomorphism if and only if $G$ is Abelian.

Related facts

The $n^{th}$ power map for a fixed integer $n$ is termed a universal power map, and if it is also an endomorphism, it is termed a universal power endomorphism. This statement gives a necessary and sufficient condition for a group where $n=2$ gives an endomorphism. Here are results for other values of $n$:

• Inverse map is automorphism iff abelian
• Cube map is endomorphism iff abelian (if order is not a multiple of 3)
• Cube map is automorphism implies abelian
• nth power map is endomorphism iff abelian (if order is relatively prime to n(n-1))
• Frattini-in-center odd-order p-group implies p-power map is endomorphism

Related facts for other algebraic structures

Here are some related facts for Lie rings:

• Multiplication by n map is derivation iff derived subring has exponent dividing n
• Multiplication by n map is endomorphism iff derived subring has exponent dividing n(n-1)

Proof

From endomorphism to Abelian

Suppose $\sigma =x\mapsto x^2$ is an endomorphism of the group $G$. Then for any $x,y\in G$ we want to show that $x$ and $y$ commute. This can be proved as follows:

$\sigma (xy)=\sigma (x)\sigma (y)$ becaus $\sigma$ is an endomorphism

Thus:

$xyxy=x^2{y}^2$

Cancelling the leftmost $x$ and the rightmost $y$, we get:

$yx=xy$

and hence $x,y$ commute.