a basis of $S_i$.

$(3)$ If $M$ is a simple $H(1, \mu)$-module, then $M$ is isomorphic to some $S_i$.

We have already known that $S_0$ is a simple $H(1, \mu)$-module and dim$S_0=1$. Now let $1\leqslant i\leqslant t-1$ and take $0\neq x\in X_i$. Let $v=1\otimes x\in S_i$. Then $g\cdot v=\xi^i v$ and $a\cdot v=(1-\xi^i)v$. Since $S_i$ is a cyclic $H(1, \mu)$-module generated by $v$, it follows from Lemma \[3.2\] that $S_i$ is a simple $H(1, \mu)$-module with dim$S_i=p$. Moreover, $\{v, b\cdot v, \cdots, b^{p-1}\cdot v\}$ is a basis of $S_i$, and $v$ is the unique common eigenvector of the actions of $g$ and $a$ on $S_i$ up to a non-zero scale multiple. Thus, $S_0, S_1, \cdots, S_{t-1}$ are non-isomorphic simple $H(1, \mu)$-modules. This shows Parts (1) and (2).

Now let $M$ be a simple $H(1,\mu)$-module. Since $k$ is an algebraically closed field and $ga=ag$, there is a non-zero vector $v\in M$ such that $g\cdot v=\alpha v$ and $a\cdot v=\beta v$ for some $\alpha, \beta\in k$. Hence $A\cdot v=kv$. Since $g^n=1$, $\alpha^n=\alpha^{p^st}=(\alpha^t)^{p^s}=1$. Hence $\alpha^t=1$, and consequently $\alpha=\xi^i$ for some $0\leqslant
i\leqslant t-1$. Since $a^p=1-g^p$, we have $\beta^p=1-\xi^{ip}=(1-\xi^i)^p$. It follows that $\beta=1-\xi^i$. Since $M$ is a simple $H(1, \mu)$-module and $H(1,
\mu)=\sum\limits_{0\leqslant j\leqslant 