such that $g\cdot\underline{u}=\xi^{i+j}\underline{u}$, where $\underline{u}=u+\overline{u}$.

Let $u_1=u$, $v_1=g\cdot u_1-\xi^{i+j}u_1\neq 0$, $m_1=v_1(1)$ and $s_1=v_1(2)$. Then it follows from Lemma \[3.5\] that $m_1<l$ and there is an elements $u_1'\in S_i\otimes S_j$ with $h(u_1')\leqslant l$ and $u_1'(1)=m_1<l$ such that $g\cdot u_2=\xi^{i+j}u_2$, or $(g\cdot u_2-\xi^{i+j}u_2)(1)<m_1$, or $(g\cdot u_2-\xi^{i+j}u_2)(1)=m_1$ and $(g\cdot u_2-\xi^{i+j}u_2)(2)<s_1$, where $u_2=u_1+u_1'$. If $g\cdot u_2=\xi^{i+j}u_2$, then the theorem follows. Otherwise, let $v_2=g\cdot u_2-\xi^{i+j}u_2\neq0$, $v_2(1)=m_2$ and $v_2(2)=s_2$. Since $u_1(1)=l$ and $u_1'(1)=m_1<l$, $u_2(1)=l$, and so $h(u_2)=l$. By replacing $u_1$ with $u_2$, it follows from Lemma \[3.5\] that there is an $u_2'\in S_i\otimes S_j$ with $h(u_2')\leqslant l$ and $u_2'(1)=m_2<l$ such that $g\cdot u_3=\xi^{i+j}u_3$, or $(g\cdot u_3-\xi^{i+j}u_3)(1)<m_2$, or $(g\cdot u_3-\xi^{i+j}u_3)(1)=m_2$ and $(g\cdot u_3-\xi^{i+j}u_3)(2)<s_2$, where $u_3=u_2+u_2'$. Since $h(u_1')\leqslant l$ and $h(u_2')\leqslant l$, $h(u_1'+u_2')\leqslant l$. Furthermore, we have $u_2'(1)=m_2<m_1=u_1'(1)$, or $u_2'(1)=m_2=m_1=u_1'(1)$ and $u_2'(2)=s_2<s_1$. It follows that $(u_1'+u_2')(1)\leqslant m_1<l$. We also have $u_3=u_2+u_2'=u_1+u_1'+u_2'$. If $g\cdot u_3=\xi^{i+j}u_3$, then the theorem follows. Otherwise, let $v_3=g\cdot u_3-\xi^{i+j}u_3\neq0$, $v_3(1)=m_3$ and $v_3(2)=s_2$. Since $u_2(1)=l$ and $u_2'(1)=m_2<l$, $u_3(1)=l$, and so $h(u_3)=l$. Then we may repeat the above procedure by replacing $u_2$ with $u_3$, and continue. Thus one may get a series of elements $u_1', u_2', u_3', \cdots$ in $S_i\otimes S_j$ with $h(u_q')\leqslant l$ and $u_q'(1)=m_q<l$ such that $g\cdot u_{q+1}=\xi^{i+j}u_{q+1}$, or $m_{q+1}:=(g\cdot u_{q+1}-\xi^{i+j}u_{q+1})(1)<m_q$, or $m_{q+1}:=(g\cdot u_{q+1}-\xi^{i+j}u_{q+1})(1)=m_q$ and 