$s_{q+1}:=(g\cdot u_{q+1}-\xi^{i+j}u_1)(2)<s_q$, where $u_{q+1}=u_q+u_q'$, $q=1, 2, 3, \cdots$.

We claim that the above procedure will stop. In fact, if $v_q=g\cdot u_q-\xi^{i+j}u_q\neq 0$ for all $q\geqslant 1$, then $m_{q+1}<m_q$, or $m_{q+1}=m_q$ and $s_{q+1}<s_q$ for all $q\geqslant 1$. Since $l>m_1\geqslant m_2\geqslant m_3\geqslant\cdots\geqslant 0$, there is a $q\geqslant 1$ such that $m_q=m_{q+1}=m_{q+2}=\cdots$. Then it follows that $s_q>s_{q+1}>s_{q+2}>\cdots\geqslant 0$. This is impossible. Thus, there exists an integer $m\geqslant 1$ such that $v_q=g\cdot u_q-\xi^{i+j}u_q\neq 0$ for all $1\leqslant q\leqslant m$, but $g\cdot u_{m+1}-\xi^{i+j}u_{m+1}=0$. Then the theorem follows.

\[3.7\] Let $\{S_i\}_{0\leqslant i\leqslant t-1}$ be the complete set of non-isomorphic simple $H(1,\mu)$-modules defined in Theorem \[3.4\]. Then soc$(S_i\bigotimes
S_j)\cong S_{\overline{i+j}}$ and $S_i\bigotimes S_j$ is indecomposable. In particular, $S_0\otimes S_i\cong S_i$ and $S_i\otimes S_0\cong S_i$. Here $0\leqslant i, j\leqslant t-1$.

It is obvious that $S_0\otimes S_i\cong S_i$ and $S_i\otimes S_0\cong S_i$ for all $0\leqslant i\leqslant t-1$. Now let $1\leqslant i, j\leqslant t-1$. Let $\{b^{i_1}\cdot v|0\leqslant i_1\leqslant
p-1\}$ and $\{b^{j_1}\cdot w|0\leqslant j_1\leqslant p-1\}$ be the bases of $S_i$ and $S_j$ as stated in Theorem \[3.4\], respectively. Then $\{b^{i_1}\cdot v\otimes
b^{j_1}\cdot w|0\leqslant i_1, j_1\leqslant p-1\}$ is a basis of $S_i\bigotimes S_j$. By Lemma \[3.2\](1), the matrix of the action of $g$ on $S_i\bigotimes S_j$ with respect to the basis $\{v\otimes 