Let $G$ be a finite group. Usually, a 2-cocycle on $G$ with values in $\mathbb{Z}\\_2 = \{+1, -1\}$ is a collection of signs $\epsilon\_{g,h} \in \{+1, -1\}$, $g,h \in G$, satisfying the cocycle equation (written multiplicatively)
$$
\epsilon\_{g,h} \epsilon\_{gh, k} = \epsilon\_{h,k} \epsilon\_{g,hk}
$$
And a 2-coboundary is a 2-cocycle with $\epsilon\_{g,h} = \frac{t\_g t\_g}{t\_{gh}}$ for all $g,h \in G$, with $t : G \rightarrow \{+1, -1\}$ an arbitrary map. Then the second cohomology group is $H^2(G, \mathbb{Z}\\_2)$ = {2-cocycles} / {2-coboundaries}. 


But suppose we demand that our 2-cocyles satisfy the 2-cocycle equation above together with the "conjugate-cyclic" symmetry 
$$
 \epsilon\_{g,h} = \epsilon\_{h^{-1} g^{-1}, g}
$$
as well as the "conjugate symmetric" symmetry,
$$
 \epsilon\_{g,h} = \epsilon\_{h^{-1}, g^{-1}}.
$$
These symmetries make sense from a TQFT perspective if you draw the 2-cocycle as a bunch of trivalent vertices, when the first symmetry corresponds to counterclockwise rotation of the diagram and the second to a kind of vertical flip.


And suppose now that the coboundaries given by $\{t\_g\}$ satisfy $t\_1 = 1$ and $t\_g t\_{g^{-1}} = 1$. This ensures that the "symmetric" 2nd cohomology group $H^2\_{sym} (G) := ${ "symmetric" 2-cocycles} / {"symmetric" 2-coboundaries} makes sense. 


Question: Does this symmetric 2nd cohomology group always vanish?


I've only checked one example, namely $G = \mathbb{Z}\\_2 \times \mathbb{Z}\\_2 = \langle a,b : a^2 = b^2 =(ab)^2 = 1 \rangle$. In normal cohomology, we have $H^2(G, \mathbb{Z}\\_2)$ = $(\mathbb{Z}\\_2)^3$ but according to my calculations, none of the non-trivial 2-cocycles respect the above symmetries, so 
$$
 H^2\_{sym} (G, \mathbb{Z}\_2) = 0.
$$
Is this perhaps always true for arbitrary $G$? Or perhaps I have made a silly mistake. 



Does the following "symmetric" 2nd cohomology group of a finite group with coefficients in $Z_2$ always vanish?
I'm pretty sure now that $H^2\_{sym} (\mathbb{Z}\_4 \times \mathbb{Z}\_2, \mathbb{Z}\_2) = \mathbb{Z}\_2 \times \mathbb{Z}\_2$, so it can be nonzero. For comparison, in ordinary group cohomology $H^2 (\mathbb{Z}\_4 \times \mathbb{Z}\_2, \mathbb{Z}\_2) = (\mathbb{Z}\_2)^3$.


