2-central implies 4-abelian

Statement

Suppose $G$ is a 2-central group: its inner automorphism group is a group of exponent two (i.e., an elementary abelian 2-group), or equivalently, every square element of $G$ is in the center.

Then, $G$ is a 4-abelian group: the fourth power map is an endomorphism, and hence a universal power endomorphism, of $G$. In symbols:

$(xy)^{4}=x^{4}y^{4}\ \forall \ x,y\in G$

Related facts

• 3-central implies 9-abelian
• 4-central implies 16-abelian
• 6-central implies 36-abelian

Facts used

1. Exponent two implies abelian
2. Class two implies commutator map is endomorphism
3. Formula for powers of product in group of class two

Proof

Straightforward proof with symbol manipulation

Given: A group $G$ such that every square element is in the center of $G$. Elements $x,y\in G$.

To prove: $(xy)^{4}=x^{4}y^{4}$.

Proof: We start with $(xy)^{4}$ and simplify it in stages

Step no.	New expression for $(xy)^{4}$	Justification of how this is obtained from the previous step
1	$xyxy(xy)^{2}$	expand as $(xy)^{2}(xy)^{2}$ and further expand the first $(xy)^{2}$.
2	$xyx(xy)^{2}y$	use that $(xy)^{2}$, being a square element, is in the center, hence we can commute it past the right most $y$.
3	$xyx^{2}yxy^{2}$	expand the $(xy)^{2}$ from the previous step.
4	$x^{3}y^{2}xy^{2}$	use that $x^{2}$, being a square element, is in the center, so commute it past the $y$ to its left. Now, regroup exponents.
5	$x^{4}y^{4}$	use that $y^{2}$ (the occurrence to the immediate right of $x^{3}$), being a square element, is in the center, so commute it past the $x$ to its right. Now, regroup exponents.

Proof using given facts

Given: A group $G$ such that the inner automorphism group of $G$ has exponent (dividing) two, i.e., every square element is in the center of $G$. Elements $x,y\in G$.

To prove: $(xy)^{4}=x^{4}y^{4}$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	The inner automorphism group of $G$ is abelian, so $G$ is a group of nilpotency class two.	Fact (1)	The inner automorphism group of $G$ has exponent dividing two.		Given-fact direct.
2	$[x,y]^{2}=[x^{2},y]$	Fact (2)		Step (1)	Fact-step combination direct
3	$[x,y]^{2}$ is the identity element of $G$.		The inner automorphism group of $G$ has exponent dividing two. Put another way, any square element is in the center of $G$.	Step (2)	From the given data, $[x^{2},y]$ is the identity element of $G$. Step (2) thus shows that $[x,y]^{2}$ is the identity element of $G$.
4	$x^{4}y^{4}=[x,y]^{6}(xy)^{4}$	Fact (3)		Step (1)	Step-fact combination direct.
5	$x^{4}y^{4}=(xy)^{4}$			Steps (3), (4)	Step (3) tells us that $[x,y]^{2}$ is the identity element, hence we get that $[x,y]^{6}$ is also the identity element. Plug into Step (4) to get the conclusion.