p-1}b^jA$, one gets that $M=H(1, \mu)\cdot v={\rm span}\{v, b\cdot v, \cdots, b^{p-1}\cdot
v\}$. We divide the discussion into the following two cases.

For the case: $i=0$. In this case, $g\cdot v=v$, $a\cdot v=0$ and $b^p\cdot v=\mu(1-g^p)\cdot v=0$. Hence there is an integer $m$ with $0\leqslant m\leqslant p-1$ such that $b^m\cdot v\neq0$ but $b^{m+1}\cdot v=0$. If $m=0$, then $g\cdot v=v$, $a\cdot v=0$ and $b\cdot v=0$. Hence $M=kv\cong S_0$ since $M$ is simple. If $m>0$, then by Lemma \[2.2\] it follows that $ab^m\cdot v=0$ and $gb^m\cdot v=b^m\cdot v$. Thus, $k\{b^m\cdot v\}$ is a non-zero $H(1,\mu)$-submodule of $M$, and so $M=k(b^m\cdot v)\cong S_0$ since $M$ is simple. In this case, $v=\gamma b^m\cdot v$ for some $0\neq
\gamma\in k$, which implies that $b\cdot v=0$, and so $m=0$, a contradiction.

For the case: $1\leqslant i\leqslant t-1$. In this case, $a\cdot
v=(1-\xi^i)v\neq0$. Since $M$ is a simple $H(1, \mu)$-module, it follows from Lemma \[3.2\] that $k\{v, b\cdot v, \cdots,
{b^{p-1}\cdot v}\}$ is a basis of $M$. In this case, $M$ is isomorphic to $S_i$. In fact, let $0\neq x\in X_i$. Then there is an $A$-module isomorphism $f: X_i\rightarrow kv$, $f(x)=v$, where $kv$ is obviously an $A$-submodule of $M$. Since $M=H(1, \mu)\cdot v$, we have an $H(1, 