# A group 3-cocycle, trivial on a pair of generating subgroups?

I'm looking for an example of the following situation:

• A group $G$ generated by finite subgroups $H$ and $K$,
• a non-trivial 3-cocycle $\omega \in H^3(G, \mathbb{k}^\times)$

such that

• the restrictions of $\omega$ to a 3-cocycle on each of $H$ or $K$ is a coboundary.

If such an example is possible with at least one of $H^2(H, \mathbb{k}^\times)$ and $H^2(K, \mathbb{k}^\times)$ non-trivial as well that would be even better

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If $\mathbb{k}^\times$ is the multiplicative group of some field, the following works with $\mathbb{k}=\mathbb{F}_3$:
The Quaternion group $Q_8$ is generated by two cyclic subgroups $H,K$ of order 4 and $$H^\ast(H;\mathbb{F}_2)\cong H^\ast(K;\mathbb{F}_2)\cong \mathbb{F}_2[a,b]/(a^2)\; ,\quad|a|=1, |b|=2$$ $$H^\ast(Q_8;\mathbb{F}_2) = \mathbb{F}_2[x,y,z]/(xy,x^3-y^3)\;,\quad |x|=|y|=1,|z|=4$$ Thus $x$ restricts on $H,K$ to some class $c$ having $c^2=0$ and consequently $x^3\neq 0$ restricts to zero on $H,K$.
Thanks Ralph, that's perfect. The same groups also work for $\mathbb k = \mathbb C$, right? – Scott Morrison May 16 '13 at 15:57
No, I'm afraid, but it doesn't: For $\mathbb{k}=\mathbb{C}$ the question is equivalent to finding $\omega \in H^4(G;\mathbb{Z})\cong H^3(G;\mathbb{C}^\times)$ that restricts to zero. Since the cohomology of $Q_8$ is 4-periodic, $H^4(Q_8;\mathbb{Z})$ is generated by a cohomology class z (of order 8) that restricts non-zero on both, H and K. – Ralph May 16 '13 at 22:06
Ah. Do you know an example with $\mathbb k = \mathbb C$? – Scott Morrison May 17 '13 at 2:34
Yes: Let $H$ resp. $K$ be any finite group those integral cohomology has a non-zero class x resp. y of degree 2. Then $xy\in H^4(H\times K;\mathbb{Z})=H^3(H\times K;\mathbb{C}^\times)$ is non-zero and restricts to zero on $H$ and $K$. If you take $H=K=\mathbb{Z}/p\oplus \mathbb{Z}/p$ then you get an example with $H^2(H;\mathbb{C}^\times)\neq 0$. – Ralph May 17 '13 at 7:57