$a$ and $b$. Hence the theorem follows.

In the rest of this section, assume $t>1$.

\[3.2\] Let $M$ be an $H(\lambda,\mu)$-module. If there exists an element $0\neq v\in M$ such that $g\cdot v=\alpha v$ and $a\cdot v=\beta v$ for some $\alpha,\beta\in k$ with $\beta\neq 0$, then the following statements holds:

$(1)$ If $1\leqslant m\leqslant p-1$, then $$ab^m \cdot v= \sum_{0\leqslant j\leqslant m}
\alpha_{m,j}b^j\cdot v\ \mbox{ and }\ gb^m\cdot v=\sum_ {0\leqslant j\leqslant m}\beta_{m, j}b^j\cdot v,$$ where $\alpha_{m,j}, \beta_{m, j}\in k$ with $\alpha_{m,m}=\beta$, $\alpha_{m, m-1}=-{\frac{m}{2}}\beta^2$, $\beta_{m, m}=\alpha$, and $\beta_{m, m-1}=-m\alpha\beta$.

$(2)$ $N ={\rm span}\{v, b\cdot v, \cdots, b^{p-1}\cdot v\}$ is an submodule of $M$.

$(3)$ $\{v, b\cdot v, \cdots, b^{p-1}\cdot v\}$ are linearly independent.

$(4)$ Consider the actions of $g$ and $a$ on $N$. Then $\alpha$ and $\beta$ are the only eigenvalues of $g$ and $a$, respectively, with multiplicity $p$. Moreover, $v$ is the unique common eigenvector of $g$ and $a$ up to a non-zero scale multiple.

$(5)$ $N$ is a simple $H(\lambda,\mu)$-module.

\(1) It follows from Parts (2) and (3) of Lemma \[2.2\].

\(2) Since $b^p=\mu(1-g^p)$, it follows from Part (1).

\(3) Suppose that $\{v, b\cdot v, \cdots, b^{p-1}\cdot v\}$ are linearly dependent. Since $v\neq 0$, there exists an $m$ with 