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

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:

$f(g,g^{-1})=f(g^{-1},g)\ \forall \ g\in G$

Relation with concept of IIP 2-cocycle

This fact helps set the stage for the definition of IIP 2-cocycle for trivial group action.

Facts used

1. 2-cocycle for trivial group action is constant on axes: This says that $f(g,1)=f(1,g)=f(1,1)\ \forall \ g\in G$

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:

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

To prove: $f(x,x^{-1})=f(x^{-1},x)$

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Setting $g=x,h=x^{-1},k=x$ in the 2-cocycle identity, we get $f(x,1)+f(x^{-1},x)=f(1,1)+f(x,x^{-1})$.		$f$ satisfies the 2-cocycle identity
2	We have $f(x,1)=f(1,1)$.	Fact (1)	$f$ is a 2-cocycle
3	We get $f(1,1)+f(x^{-1},x)+f(1,1)+f(x,x^{-1})$.			Steps (1), (2)	Using Step (2), we replace the first term $f(x,1)$ in Step (1) by $f(1,1)$.
4	Cancelling $f(1,1)$ from both sides of Step (3), we get $f(x^{-1},x)=f(x,x^{-1})$		$A$ is an abelian group, so we can cancel elements	Step (3)
5	Swapping the sides of Step (4), we get $f(x,x^{-1})=f(x^{-1},x)$ as desired.			Step (4)

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