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

Proof

From square map being endomorphism to abelian

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

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

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation	What algebraic assumptions does this use?
1	$(xy{)}^2=(x^2)(y^2)$	--	square map is endomorphism	--	--	None, works over any magma
2	$(xy)(xy)=(xx)(yy)$	--		Step (1)	--	None, just using definition of square. Works over any magma.
3	$(x(yx))y=(x(xy))y$	Fact (1)		Step (2)	Reparenthesize	The reparenthesization requires associativity of expressions involving two variables. It works over any semigroup or monoid and even more generally over any diassociative magma.
4	$yx=xy$	Fact (2)		Step (3)	Cancel the right-most $y$ from both sides, then the left-most $x$ from both sides.	The cancellation requires that we are working in a cancellative magma, such as a cancellative monoid or a quasigroup.

From abelian to square map being endomorphism

This follows directly from fact (3).