y}=\a(x,y)\l(x)\l(y)\l(xy)^{-1}\widetilde{xy}\ \ {\mbox for\ \ all }\ \ x,y\in G.$$

Hence, performing a diagonal change of basis $\{\bar g| g\in G\}$ results in replacing $\a$ by a cohomologous cocycle. The advantage of this observation is that in choosing a distinguished cocycle cohomologous to $\a$, we can work exclusively within the algebra $F^{\a}[G]$.

We say that $\a\in Z^2(G,F^*)$ is a [*normal cocycle*]{} if $\a(x,g)=\a(xgx^{-1},x)$ for all $x\in G$ and all $\a$-regular $g\in G$.

Therefore $\a$ is normal if and only if $\bar x\bar g\bar x^{-1}=\overline{xgx^{-1}}$ for all $x\in G$ and all $\a$-regular $g\in G$.

\[a2.1\][\[K\]]{} Let $\{g_1,...,g_r\}$ be a full set of representatives for the $\a$-regular conjugacy classes of $G$ and, for each $i\in \{1,...,r\}$, let $T_i$ be a left transversal for $C_G(g_i)$ in $G$. Put $\widetilde{tg_it^{-1}}=\bar t \overline{g_i}\overline{t}^{-1},1\leq i \leq r$, and $\tilde{g}=\bar g $ if $g$ is not $\a$-regular . Then $\tilde {x}\tilde {g}\tilde {x}^{-1}=\widetilde{xgx^{-1}}$ for all $x\in G$ and all $\a$-regular $g\in G$. Thus any cocycle $\a\in Z^2(G,F^*)$ is cohomologous to a normal cocycle.

\[a2.2\][\[K\]]{} Let $F$ be an arbitrary field, let $\a\in Z^2(G,F^*)$ and let $\{g_1,...,g_r\}$ be a full set of representatives for the $\a$-regular conjugacy classes of $G$. Denote by $T_i$ a left transversal for $C_G(g_i)$ 