$\{\bar{g}|g\in
G\}$. Multiplication in $F^{\a}[G]$ is defined distributively by using $$(a \bar{x})(b\bar{y})=ab\a(x,y)\overline{xy}\ \ (a,b\in F,x,y\in G).$$

It is easy to verify that $F^{\a}[G]$ is an $F$-algebra with $\bar{1}$ as the identity element and with $$\bar{g}^{-1}=\a(g^{-1},g)^{-1}\overline{g^{-1}}=\a(g,g^{-1})^{-1}\overline{g^{-1}}\ \ {\mbox for\ \ all}\ \ g\in G.$$

[\[K\]]{} If [*char*]{} $F$ does not divide $|G|$, then $F^{\a}[G]$ is semisimple.

Our aim for this section is to find an explicit basis for the center $Z(F^{\a}[G])$.

An element $g\in G$ is said to be $\a-{\em regular}$ if $$\a(g,x)=\a(x,g)\ \ {\mbox for\ \ all}\ \ x\in C_G(g).$$

Thus $g$ is $\a$-regular if and only if $\bar g \bar x=\bar x\bar g$ for all $x\in C_G(g).$

It is clear that identity element of $G$ is $\a$-regular for any cocycle $\a$.

[\[K\]]{} For any $\a\in Z^2(G,F^*)$, the following properties hold:

i\) An element $g\in G$ is $\a$-regular if and only if it is $\b$-regular for any cocycle $\b$ cohomologous to $\a$.

ii\) If $g\in G$ is $\a$-regular, then so is any conjugate of $g$.

Let $C$ be a conjugacy class of $G$ and let $g\in C$. We say that $C$ is $\a$-[*regular*]{} if $g$ is $\a$-regular.

If we replace $\bar g$ by ${\tilde g}=\l(g)\bar g,\l(g)\in F^*,\l(1)=1$, yields $${\tilde x}{\tilde 