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Let $G$ be a finite $p$-group and $(\mathbb{F},\mathcal{O},\mathbb{K})$ a sufficiently large $p$-modular system. Furthermore let $\mu_{p^\infty}\leq \mathcal{O}^\times$ the group of roots of unity of $p$-power order.

If $\alpha$ is a 2-cocyle with values in $\mu_{p^\infty}$ does there always exist a one-dimensional projective representation $G\to\mathcal{O}^\times$ with cocylce $\alpha$ ? In other words: Does the twisted group algebra $\mathcal{O}_\alpha[G]$ always have a one-dimensional representation $\mathcal{O}_\alpha[G]\to\mathcal{O}$?

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    $\begingroup$ I might be confused, but isn't a one-dimensional representation equivalent to a trivialization of $\alpha$ (i.e. a realization of $\alpha$ as a coboundary)? Without thinking more carefully, I guess there could be kernel in $H^2(G; \mu_{p^\infty}) \to H^2(G; \mathcal{O}^\times)$. But then I think you are asking if this is the zero map, which seems unlikely IMHO. $\endgroup$ Jan 28, 2020 at 21:58
  • $\begingroup$ I don't think so. For example cyclic groups fit into a non-splitting extension $1\to C_p \to C_{p^2} \to C_p \to 1$ which defines a non-trivial cocycle $\alpha$, but there is a morphism $\mathcal{O}_\alpha[C_p] \to \mathcal{O}$ which maps a generator of $C_p$ to a primitive $p^2$th root. $\endgroup$ Jan 28, 2020 at 22:42
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    $\begingroup$ And if $\mathbb{K}$ is algebraically closed for example, then $H^2(G,\mu_{p^\infty}) \to H^2(G,\mathcal{O}^\times)$ is even an isomorphism. First $H^\ast(G,\mu) \to H^\ast(G,\mathcal{O}^\times)$ ($\mu$ being the subgroup of all roots of unity) follows from the fact that $\mathcal{O}^\times / \mu$ is uniquely divisible, $\hat{H}^\ast(G,uniquely divisible)=0$ and the long exact sequence. And $H^2(G,\mu) = H^2(G,\mu_{p^\infty})$ follows because $G$ is a $p$-group. $\endgroup$ Jan 28, 2020 at 22:45

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Ah, dammit. The answer is no and the first group I would have looked at next is the counterexample... the extra special group $G=\langle x,y,z \mid x^p=y^p=[x,y]=z, z^p=1\rangle$ fits into an non-trivial extension $1\to C_p \to G \to C_p\times C_p\to 1$ and the corresponding 2-cocycle $\alpha$ gives us the counterexample, because $\mathcal{O}_\alpha[C_p\times C_p] = \mathcal{O}\langle X,Y \mid X^p=Y^p=XYX^{-1}Y^{-1}=\zeta_p\rangle$ does not have one-dimensional representations into something of characteristic zero. If $\rho$ was such a thing, $\rho(X)$ and $\rho(Y)$ would commute so that $\rho(X)\rho(Y)\rho(X)^{-1}\rho(Y)^{-1}$ would have to be both equal to $1$ and to $\zeta_p$.

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    $\begingroup$ While I sympathise with your frustration, you could probably delete the first sentence, and maybe even the second sentence up to the ellipsis. $\endgroup$
    – LSpice
    Jan 29, 2020 at 2:22
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    $\begingroup$ By the way, for people like me who aren't finite-group theorists, it may be helpful to remark that this extra-special group is perhaps also known as the 3-dimensional Heisenberg group over $\mathbb F_p$. $\endgroup$
    – LSpice
    Jan 30, 2020 at 14:40

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