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
View other elementary non-basic facts
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this

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., $\sigma (xy)=\sigma (x)\sigma (y)\mathrm{\forall }x,y\in G$) 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.

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 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 $n$.

• n-abelian iff (1-n)-abelian
• The set of $n$ for which $G$ is $n$-abelian is termed the exponent semigroup of $G$. It is a submonoid of the multiplicative monoid of integers.
• abelian implies n-abelian for all n
• n-abelian implies every nth power and (n-1)th power commute
• n-abelian implies n(n-1)-central
• nth power map is endomorphism iff abelian (if order is relatively prime to n(n-1))
• nth power map is surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center
• (n-1)th power map is endomorphism taking values in the center implies nth power map is endomorphism
• Frattini-in-center odd-order p-group implies p-power map is endomorphism
• Frattini-in-center odd-order p-group implies (mp plus 1)-power map is automorphism

Value of $n$ (note that the condition for $n$ is the same as the condition for $1-n$)	Characterization of $n$-abelian groups	Proof	Other related facts
0	all groups	obvious
1	all groups	obvious
2	abelian groups only	2-abelian iff abelian	endomorphism sends more than three-fourths of elements to squares implies abelian
-1	abelian groups only	-1-abelian iff abelian
3	2-Engel group and derived subgroup has exponent dividing three	Levi's characterization of 3-abelian groups	cube map is surjective endomorphism implies abelian, cube map is endomorphism iff abelian (if order is not a multiple of 3)
-2	same as for 3-abelian	(based on n-abelian iff (1-n)-abelian)

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)

Opposite facts for other algebraic structures

Statement	Algebraic structure	What step of the proof fails?	Comment
Square map is endomorphism not implies abelian for loop	loop	The reparenthesization in Step (3) of the proof below, that requires associativity.	In fact, it is possible to have a noncommutative loop of exponent two.
Square map is endomorphism not implies abelian for monoid	monoid	The cancellation in Step (4), which requires that we are working over a cancellative 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 or loop.

From abelian to square map being endomorphism

This follows directly from fact (3).