$0\leqslant m< p-1$ such that $\{v, b\cdot v, \cdots, b^m\cdot v\}$ are linearly independent, but $\{v, b\cdot v, \cdots, b^m\cdot v, b^{m+1}\cdot v\}$ are linearly dependent. Hence there are some $\alpha_i\in k$ such that $b^{m+1}\cdot v=\sum\limits_{0\leqslant i\leqslant m}\alpha_ib^i\cdot v$. Thus, $ab^{m+1}\cdot v=\sum\limits_{0\leqslant i\leqslant m}\alpha_iab^i\cdot v$. By Part (1), we have $ab^{m+1} \cdot v= \sum\limits_{0\leqslant j\leqslant m+1}
\alpha_{m+1,j}b^j\cdot v=\beta b^{m+1}\cdot v+ \sum\limits_{0\leqslant j\leqslant m}
\alpha_{m+1,j}b^j\cdot v$ and $$\begin{array}{rcl}
\sum\limits_{0\leqslant i\leqslant m}\alpha_iab^i\cdot v
&=&\sum\limits_{0\leqslant i\leqslant m}
\sum\limits_{0\leqslant j\leqslant i}\alpha_i\alpha_{i, j}b^j\cdot v\\
&=&\alpha_m\beta b^m\cdot v+\sum\limits_{0\leqslant j\leqslant m-1}\gamma_jb^j\cdot v,\\
\end{array}$$ where $\gamma_j\in k$ for $0\leqslant j\leqslant m-1$. Hence we have $$\begin{array}{rcl}
ab^{m+1}\cdot v-\beta b^{m+1}\cdot v&=&\sum\limits_{0\leqslant j\leqslant m}
\alpha_{m+1,j}b^j\cdot v\\
&=&-\frac{m+1}{2}\beta^2b^m\cdot v+\sum\limits_{0\leqslant j\leqslant m-1}
\alpha_{m+1,j}b^j\cdot v\\
\end{array}$$ and $$a(\sum\limits_{0\leqslant i\leqslant m}\alpha_ib^i\cdot v)-
\beta(\sum\limits_{0\leqslant i\leqslant m}\alpha_ib^i\cdot v)
=\sum\limits_{0\leqslant j\leqslant m-1}(\gamma_j-\alpha_j\beta)b^j\cdot v.$$ It follows that $-\frac{m+1}{2}\beta^2b^m\cdot v+\sum\limits_{0\leqslant j\leqslant m-1}
\alpha_{m+1,j}b^j\cdot v=\sum\limits_{0\leqslant j\leqslant m-1}(\gamma_j-\alpha_j\beta)b^j\cdot v.$ Since $-\frac{m+1}{2}\beta^2\neq0$, one gets that $\{v, b\cdot v, \cdots, b^m\cdot v\}$ are linearly dependent, a contradiction.

\(4) It follows from Parts (1) and (3).

\(5) Let $N_0$ be a non-zero submodule of $N$. Then $N_0$ must contain a common eigenvector of $g$ and $a$. Hence $v\in N_0$ by Part (4), and so $N_0=N$. This shows that $N$ is a simple module.

Now we 