\mu)$-module epimorphism $$\psi: S_i=H(1, \mu)\otimes_AX_i\xrightarrow{{\rm id}\otimes f}
H(1, \mu)\otimes_A(kv)\xrightarrow{\cdot}M$$ given by $\psi(h\otimes x)=h\cdot f(x)=h\cdot v$, $h\in H(1, \mu)$. Since both $S_i$ and $M$ are simple, $\psi$ must be an isomorphism.

For any integer $i$, let $0\leqslant\overline{i}\leqslant t-1$ with $\overline{i}\equiv i$ (mod $t)$. For any positive integer $m$, let $I_m$ denote the identity $m\times m$-matrix over $k$. For any matrix $X$ over $k$, let $r(X)$ denote the rank of $X$.

For $1\leqslant i, j\leqslant t-1$, 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 basis 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 $S_i\otimes S_j$. For any $0\neq u=\sum
x_{i_1,j_1}b^{i_1}\cdot v \otimes b^{j_1}\cdot w\in S_i\otimes S_j$, let $h(u)={\rm max}\{i_1+j_1|x_{i_1,j_1}\neq0\}$ and let $$u(1)={\rm max}\{i_1|x_{i_1,j_1}\neq0 \mbox{ for some $j_1$}\}\mbox{ and }
u(2)={\rm max}\{j_1|x_{u(1),j_1}\neq0\}.$$ With the above notations, we have the following lemma.

\[3.5\] Let $0\neq u\in S_i\otimes S_j$ with $h(u)=u(1)=l>0$. Assume $v_1=g\cdot u-\xi^{i+j}u\neq 0$. Then

$(1)$ $h(v_1)< l$.

$(2)$ If $v_1(2)=0$, then there is an element $u'\in S_i\otimes S_j$ with $h(u')\leqslant l$ and $u'(1)=$ $v_1(1)$ such that $g\cdot u''-\xi^{i+j}u''=0$, or $(g\cdot u''-\xi^{i+j}u'')(1)<v_1(1)$, where $u''=u+u'$.

$(3)$ If $v_1(2)>0$, then there is an element $u'\in S_i\otimes S_j$ with $h(u')\leqslant 