Square map is endomorphism iff abelian

Statement
Let $$G$$ be a group and $$\sigma:G \to G$$ be the square map of $$G$$ defined as $$\sigma(x) = x^2$$. Then, $$\sigma$$ is an endomorphism of $$G$$ (i.e., $$\sigma(xy) = \sigma(x)\sigma(y) \ \forall \ x,y \in G$$) if and only if $$G$$ is abelian.

Another way of putting it is that $$G$$ is 2-abelian if and only if it is abelian.

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 and the group is termed a n-abelian group. This statement gives a necessary and sufficient condition for a group where $$n = 2$$ gives an endomorphism. Here are results for other values of $$n$$.

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)

Facts used

 * 1) uses::Associative implies generalized associative: Basically this says that in a group, we can drop and rearrange parentheses at will.
 * 2) uses::Invertible implies cancellative in monoid. Since every element of a group is invertible, cancellation is valid in groups.
 * 3) uses::Abelian implies universal power map is endomorphism

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., $$(xy)^2 = x^2y^2$$ for all $$x,y \in G$$.

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

Proof: We let $$x,y$$ be arbitrary elements of $$G$$.

From abelian to square map being endomorphism
This follows directly from fact (3).