will compute simple modules over $H(\lambda,\mu)$. Note that $H(\lambda, \mu)=\mathcal{B}(V)\# kG$ if $\lambda=\mu=0$. We first consider the case of $\lambda=0$.

\[3.3\] Let $\mu\in k$. Then there are $t$ non-isomorphic simple modules $T_i$ over $H(0,\mu)$, $0\leqslant i\leqslant t-1$. Each $T_i$ is $1$-dimensional and given by $$g\cdot v=\xi^iv,\  a\cdot v=0,\ b\cdot v=\mu^\frac{1}{p}(1-\xi^i)v,\ v\in T_i.$$

Let $0\leqslant i\leqslant t-1$. Then it is easy to see that there is an algebra map $\rho_i: H(0, \mu)\rightarrow k$ such that $\rho_i(g)=\xi^i$, $\rho_i(a)=0$ and $\rho_i(b)=\mu^\frac{1}{p}(1-\xi^i)$. It follows that $T_0, T_1, \cdots, T_{t-1}$ given in the theorem are non-isomorphic 1-dimensional simple $H(0, \mu)$-modules.

By the proof of Lemma \[2.4\](1), one knows that the ideal $\langle a\rangle$ of $H(0, \mu)$ generated by $a$ is equal to $H(0,
\mu)a=aH(0, \mu)$. Since $a^p=0$, $\langle a\rangle^p=(H(0,
\mu)a)^p=H(0, \mu)a^p=0$. Hence $\langle a\rangle\subseteq J$, the Jacobson radical of $H(0, \mu)$. Thus, any simple $H(0, \mu)$-module is a simple module over the quotient algebra $H(0, \mu)/\langle
a\rangle$. However, $H(0, \mu)/\langle a\rangle$ is a commutative algebra and $k$ is an algebraically closed field. It follows that any simple $H(0, \mu)$-module is 1-dimensional and determined by an algebra map from $H(0, \mu)$ to $k$. Now let $\rho: H(0,
\mu)\rightarrow k$ be an algebra map. Then $\rho(a)=0$. Since 