# Does the following “symmetric” 2nd cohomology group of a finite group with coefficients in $Z_2$ always vanish?

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.

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groupprops.subwiki.org/wiki/… "The cohomology group is isomorphic to elementary abelian group:E8", I am not great expert, but there are at least 4 different central extentsions of Klein group Z/2xZ/2 wiht Z/2 - trivial, Dihedral, quaternion, and Z/4xZ/2 - so it seems it cannot be H^* = Z/2 – Alexander Chervov Sep 30 '12 at 16:24
Thanks. Yes I'm afraid I had built my whole argument around the fact that the "2nd cohomology" was Z_2, which I got from a quick lookup, but in fact what I had looked up was that H^2(G, U(1)) = Z_2, whereas I needed H^2(G ,Z_2) = (Z_2)^3. – Bruce Bartlett Sep 30 '12 at 18:16
Actually, my statement that the "symmetric" cohomology is zero in this example is still true - checked by a direct calculation. Also, of the 4 extensions of $Z_2 x Z_2$ by $Z_2$, only the product extension and $Z_2 \times Z_4$ satisfy the second symmetry above, so the quaternion and the dihedral group don't come into it. There are only 2 cocycles satisfying both symmetries above, and they are both coboundaries. – Bruce Bartlett Sep 30 '12 at 21:32
I have corrected the error Alexander Cherkov pointed out above. But my question still stands. Any takers? – Bruce Bartlett Oct 1 '12 at 8:10

## 1 Answer

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 $H^2 (\mathbb{Z}_4 \times \mathbb{Z}_2, \mathbb{Z}_2) = (\mathbb{Z}_2)^3$.

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