$\rho(g)^n=\rho(g^n)=\rho(1)=1$, $\rho(g)=\xi^i$ for some $0\leqslant i\leqslant t-1$. Since $b^p=\mu(1-g^p)$, $\rho(b)^p=\mu(1-\rho(g)^p)=\mu(1-\xi^{ip})=(\mu^{\frac{1}{p}}(1-\xi^i))^p$, and so $\rho(b)=\mu^{\frac{1}{p}}(1-\xi^i)$. Thus, $\rho=\rho_i$. This completes the proof.

For the case of $\lambda\neq 0$, by Lemma \[2.5\], we may assume $\lambda=1$. Let $S_0$ be the trivial $H(1, \mu)$-module given by the counit $\varepsilon: H(1, \mu)\rightarrow k$. Then dim$S_0=1$, and $$g\cdot v=v,\ a\cdot v=0,\ b\cdot v=0,\ v\in S_0.$$

Now let $A$ be the subalgebra of $H(1, \mu)$ generated by $g$ and $a$. Then $A$ is a Hopf subalgebra of $H(1, \mu)$. Hence $H(1, \mu)$ is a free right (left) $A$-module [@Mon93]. Note that $A$ is a commutative algebra. For $1\leqslant i\leqslant t-1$, there is an algebra map $\rho_i: A\rightarrow k$ defined by $\rho_i(g)=\xi^i$ and $\rho_i(a)=1-\xi^i$. Let $X_i$ denote the corresponding left $A$-module. Then dim$X_i=1$, $g\cdot x=\xi^ix$ and $a\cdot x=(1-\xi^i)x$ for all $x\in X_i$. Let $S_i=H(1, \mu)\otimes_AX_i$. Then $S_i$ is a non-zero left cyclic $H(1, \mu)$-module generated by $1\otimes x$, where $0\neq x\in X_i$.

\[3.4\] Let $0\leqslant i\leqslant t-1$. Then we have

$(1)$ $S_0, S_1, \cdots, S_{t-1}$ are non-isomorphic simple $H(1, \mu)$-modules.

$(2)$ If $i\neq 0$, ${\rm dim}S_i=p$ and there is a $0\neq v\in S_i$ such that $g\cdot v=\xi^i v$ and $a\cdot v=(1-\xi^i)v$. Moreover, $\{v, b\cdot v, \cdots, b^{p-1}\cdot v\}$ is 