Normalization of group $2$-cocycles

Question

Let $G$ be a group and $a\colon G\times G\to U(1)$ be a $2$-cocycle on $G$. In other words, $a$ is a function that satisfies

$$a(s,t)a(st,u)=a(s,tu)a(t,u)$$

for all $s,t,u\in G$.

Two such cocycles $a$ and $a'$ are said to be cohomologous if there exists a function $b\colon G\to U(1)$ such that

$$a'(s,t)=a(s,t)b(st)\bar{b}(s)\bar{b}(t).$$

I have seen it claimed that every $2$-cocycle $a$ is cohomologous to a "standard" cocycle $a'$ satisfying

$$a'(s,s^{-1})=1,$$

but I could not verify this. The best I could get is that there exists an $a'$ such that $a'(e,e)=1$, but this appears to be weaker than the above condition.

Question: Given a $2$-cocycle $a$ on $G$, is it always possible to find a cohomologous $a'$ such that $a'(s,s^{-1})=1$ for all $s\in G$?

Answer 1

Yes.

Suppose $a$ is normalized: $a(e,x)=a(x,e)=1$.

First notice that $a(x, x^{-1})=a(x^{-1},x)$. This follows from the cocycle equation. Now choose the coboundary $$b(x)=\left((a(x,x^{-1})\right)^{1/2}=b(x^{-1}),$$ it follows that $$ a'(x,x^{-1}) = \frac{a(x,x^{-1})b(e)}{b(x)b(x^{-1})} = \frac{a(x,x^{-1})}{a(x,x^{-1})} =1 $$