Twisted group algebra of a factor set

Let $Q$ be a finite group and let $\alpha :Q\times Q\to {\mathbb{C}}^{\times }$ be a normalized two-cocycle in the multiplicative convention of The factor set satisfies the two-cocycle equation, that is $\alpha (1,q)=\alpha (q,1)=1,\alpha (q,r)\alpha (qr,s)=\alpha (r,s)\alpha (q,rs)(q,r,s\in Q)\mathrm{.}$ The twisted group algebra ${\mathbb{C}}^{\alpha }[Q]$ is the complex vector space with basis the symbols $u_q$ indexed by $q\in Q$, equipped with the bilinear product that is defined on basis elements by $u_q{u}_r:=\alpha (q,r)u_{qr}(q,r\in Q)$ and extended to all of ${\mathbb{C}}^{\alpha }[Q]$ by bilinearity. The product is determined by this rule, since the $u_q$ form a basis, and it is a proposed associative unital algebra structure whose properties are verified in Projective representations and twisted algebra modules. The space is finite-dimensional, of dimension $\mathrm{\mid }Q\mathrm{\mid }$ over $\mathbb{C}$.

The element $u_1$ is the proposed unit: by the defining rule and the normalization, $u_1{u}_q=\alpha (1,q)u_q=u_q$ and $u_q{u}_1=\alpha (q,1)u_q=u_q$ for every $q$. Each basis element is a candidate unit of the algebra: from the cocycle equation at $(q,q^{-1},q)$ one gets $\alpha (q,q^{-1})=\alpha (q^{-1},q)$, hence $u_q{u}_{q^{-1}}=\alpha (q,q^{-1})u_1,u_{q^{-1}}{u}_q=\alpha (q^{-1},q)u_1,$ so $u_q$ has inverse $\alpha (q,q^{-1}{)}^{-1}{u}_{q^{-1}}$. For $\alpha \equiv 1$ the construction is the ordinary complex group algebra $\mathbb{C}[Q]$, and the twisted algebra is in general not commutative, because $\alpha (q,r)$ need not equal $\alpha (r,q)$.