2-cocycle for trivial group action is constant on axes

Statement

Suppose $G$ is a group and $A$ is an abelian group. Suppose $f:G\times G\to A$ is a 2-cocycle for trivial group action of $G$ on $A$. In other words, $f$ satisfying the following condition (that we will refer to as the 2-cocycle identity):

$f(g,hk)+f(h,k)=f(gh,k)+f(g,h)\ \forall \ g,h,k,\in G$

Then, if we denote by $1$ the identity element of $G$, the following holds:

$f(g,1)=f(1,g)=f(1,1)\ \forall \ g\in G$

The phrase "constant on axes" in the name of this page refers to the fact that if we think of $G\times G$ visually as a two-dimensional space, the "axes" are $G\times 1$ and $1\times G$, and the assertion here is that the 2-cocycle function is constant on these axes (and because the axes intersect at $(1,1)$, this constant value is forced to be the same across both axes).

Relation with concept of normalized 2-cocycle

This fact helps set the stage for the definition of normalized 2-cocycle for trivial group action, which is a 2-cocycle for trivial group action where these equal values are all zero (the identity element of $A$).

Related facts

• 2-cocycle for trivial group action is symmetric between element and inverse

Proof

To avoid symbol confusion, we use the letter $x$ for the $g$ in the statement to be proved.

Given: A group $G$ with identity element $1$, an arbitrary element $x\in G$, an abelian group $A$, $f:G\times G\to A$ satisfying the 2-cocycle identity:

$f(g,hk)+f(h,k)=f(gh,k)+f(g,h)\ \forall \ g,h,k,\in G$

First part

To prove: $f(x,1)=f(1,1)$

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Setting $g=x,h=1,k=1$ in the 2-cocycle identity, we get $f(x,1)+f(1,1)=f(x,1)+f(x,1)$.		$f$ satisfies the 2-cocycle identity
2	Cancelling $f(x,1)$ from both sides of Step (1), we get $f(1,1)=f(x,1)$.		$A$ is an abelian group so we can cancel elements	Step (1)
3	Swapping the left and right side, we get $f(x,1)=f(1,1)$ as desired.			Step (2)

Second part

To prove: $f(1,x)=f(1,1)$

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Setting $g=1,h=1,k=x$ in the 2-cocycle identity, we get $f(1,x)+f(1,x)=f(1,x)+f(1,1)$.		$f$ satisfies the 2-cocycle identity
2	Cancelling $f(1,x)$ on the left from both sides, we get $f(1,x)=f(1,1)$ as desired.		$A$ is an abelian group so we can cancel elements	Step (1)

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