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

Applications

• Exponent two implies abelian: If the exponent of a group is 2 (i.e., the group is nontrivial and every non-identity element has order two) then the group is abelian. The analogous statement is not true for any other prime number, i.e., there can be a non-abelian group of prime exponent. The standard example for an odd prime is prime-cube order group:U(3,p) of order $p^3$.

Majority criterion

• Endomorphism sends more than three-fourths of elements to squares implies abelian

Other $n^{th}$ power maps

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 Lie rings

Here are some related facts for Lie rings:

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

Related facts for other algebraic structures

• Square map is endomorphism not implies abelian for loop. In fact, it is possible to have a noncommutative loop of exponent two.
• Square map is endomorphism not implies abelian for monoid.

Facts used

1. Invertible implies cancellative in monoid
2. Abelian implies universal power map is endomorphism

Proof

From square map being endomorphism to abelian

Given: A group $G$ such that the map $\sigma =x\mapsto x^2$ is an endomorphism, i.e., $(xy{)}^2=x^2{y}^2$ for all $x,y\in G$.

To prove: $xy=yx$ for all $x,y\in G$.

Proof: By definition of endomorphism, we have $(xy{)}^2=x^2{y}^2$, so:

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

Cancelling the leftmost $x$ and the rightmost $y$ (see Fact (1)), we get:

$yx=xy$

and hence $x,y$ commute.

From abelian to square map being endomorphism

This follows directly from fact (2).