Can we see directly from the cocycle condition that 2-cocycles are symmetric?

Let $A$ be an abelian group and let $C$ be a cyclic group. All central extensions of $C$ by $A$ are abelian because in any such extension

$$ 1\rightarrow A\rightarrow E\rightarrow C\rightarrow 1$$

any lift $x$ of a generator for $C$ to $E$ generates $E$ over $A$. Since $A\subset E$ is central, $x$ commutes with $A$, thus $E=\langle x,A\rangle$ is abelian.

Any 2-cocycle $f\in Z^2(C,A)$ allows the construction of such an $E$, in the usual way: $E:= \{(a,c):a\in A, c\in C\}$, with the group operation given by

$$(a,c)(a',c') = (aa'f(c,c'), cc')$$

since the action of $C$ on $A$ is trivial. Since $E$ is abelian by the above argument, this formula implies that $f$ is symmetric in its inputs!

Since this works for any cocycle, it must be that the symmetry is forced only by the cocycle condition

$$f(x,y)+ f(xy,z) = f(y,z) + f(x,yz)$$

and the fact that $C$ is cyclic.

Is there a direct way (without going through the group extensions) to see that the cocycle condition together with $C$ being cyclic imply that $f(x,y) = f(y,x)$?

My first thought was to let $x=r^a, y=r^b,z=r^c$, where $r\in C$ is a generator, and substitute into the cocycle condition. This basically reduces to the case $C=\mathbb{Z}$, but I don't see what to do next. The argument must use the cyclicness in an essential way, i.e. just abelianness is insufficient, since there do exist nonabelian central extensions of (noncyclic) abelian groups (e.g. $Q_8$ or more generally any extraspecial group).

Solution 1:

(Edit after incorrect first answer)

Let $f$ be a cocycle (central extension with abelian quotient): writing additively, the cocycle condition is $f(a,b)+f(a+b,c)=f(a,b+c)+f(b,c)$. Then $g:(a,b)\mapsto f(a,b)-f(0,0)$ is also a cocycle.

When $c=0$ the cocycle relation yields $g(a,0)=0$ for all $a$, and similarly $g(0,a)=0$ for all $a$.

When $c=a$ the cocycle relation yields $g(a,b)-g(b,a)=g(a,a+b)-g(a+b,a)$. Now assume the quotient is $\mathbf{Z}$ and take $(u,v)$ with $g(u,v)-g(v,u)$ nonzero with $|u|+|v|$ minimal. Then $uv\neq 0$. If $uv<0$, putting $a=u,b=v$ yields a smaller counterexample. If $0<u\le v$ putting $a=u$, $a+b=v$ also yields a smaller counterexample, so we have a contradiction.

When the quotient group is cyclic we obtain a cocycle on $\mathbf{Z}$ by composition, which is then symmetric.