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$G$ is a multiplicative group, $K$ is a field. Let $\gamma$ and$\tilde{\gamma}$ be two twistings of $K^t[G]$ related by the equation $\tilde{\gamma}(x,y)=\delta(x)\delta(y)\delta(xy)^{-1}\gamma(x,y)$. Is function $\delta:G\rightarrow:K^0$ unique?

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No. You can multiply $\delta$ pointwise by a homomorphism $G \to K^{\times}$.

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  • $\begingroup$ can you explain it in more details please? $\endgroup$ – math15 Jan 10 '16 at 19:15
  • $\begingroup$ @math15: suppose $\delta'(x) = \delta(x) f(x)$ also satisfies the above equation. Then $f(x) f(y) f(xy)^{-1} = 1$; equivalently, $f(xy) = f(x) f(y)$. So $f$ can be a homomorphism $G \to K^{\times}$. $\endgroup$ – Qiaochu Yuan Jan 10 '16 at 19:23

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