$H(\lambda, \mu)$ generated by $a$. It follows that $(H(\lambda, \mu)a)^n=H(\lambda, \mu)a^n=0$. Thus, $H(\lambda, \mu)a\subseteq J$, and so $a\in J$. Similarly, we have $b^n=0$. Consider the quotient algebra $H(\lambda, \mu)/\langle a\rangle$ of $H(\lambda, \mu)$ modulo $\langle a\rangle$. Then $H(\lambda, \mu)/\langle a\rangle$ is generated, as an algebra, by $\overline{g}$ and $\overline{b}$. In this case, we have $\overline{g}\overline{b}=\overline{b}\overline{g}$. It follows that the ideal $\langle\overline{b}\rangle$ of $H(\lambda, \mu)/\langle a\rangle$ generated by $\overline{b}$ satisfies $\langle\overline{b}\rangle^n=0$. Therefore, $b\in J$.

\(2) Assume $t>1$ and $\lambda\mu\neq0$. Then $g^{p^m}\neq1$ for all $m\geqslant 0$. Hence $a^{p^m}=\lambda^{p^{m-1}}(1-g^{p^m})\neq0$ for all $m\geqslant 0$. This means that $a$ is not a nilpotent element, and so $a\notin J$. Similarly, $b\notin J$.

\[2.5\] If $\lambda\neq0$, then $H(\lambda,\mu)\cong
H(1,\lambda^{-1}\mu)$.

Assume $\lambda\neq0$. Let $g$, $a$, $b$ and $g_0$, $a_0$, $b_0$ denote the generators of $H(\lambda,\mu)$ and $H(1,\lambda^{-1}\mu)$, respectively. Then in $H(1,\lambda^{-1}\mu)$ we have $g_0^n=1$, $g_0^{-1}(\lambda^{\frac{1}{p}}a_0)g=\lambda^{\frac{1}{p}}a_0$, $g_0^{-1}(\lambda^{\frac{1}{p}}b_0)g_0=\lambda^{\frac{1}{p}}a_0+\lambda^{\frac{1}{p}}b_0$. $(\lambda^{\frac{1}{p}}a_0)^p=\lambda(1-g_0^p)$, $(\lambda^{\frac{1}{p}}b_0)^p=\mu(1-g_0^p)$, and $(\lambda^{\frac{1}{p}}b_0)(\lambda^{\frac{1}{p}}a_0)
=(\lambda^{\frac{1}{p}}a_0)(\lambda^{\frac{1}{p}}b_0)+{\frac{1}{2}}(\lambda^{\frac{1}{p}}a_0)^2$. It follows that there is an algebra map $\varphi:H(\lambda,\mu)\rightarrow H(1,\lambda^{-1}\mu)$ such that $\varphi(g)=g_0$, $\varphi(a)=\lambda^{\frac{1}{p}}a_0$ and $\varphi(b)=\lambda^{\frac{1}{p}}b_0$. It is easy to see that $\varphi$ is a Hopf algebra homomorphism. Similarly, there exists a Hopf algebra homomorphism $\psi: H(1, \lambda^{-1}\mu)\rightarrow H(\lambda, \mu)$ such that $\psi(g_0)=g$, $\psi(a_0)=\lambda^{-\frac{1}{p}}a$ and $\psi(b_0)=\lambda^{-\frac{1}{p}}b$. Obviously, $\varphi\circ\psi={\rm id}$ and $\psi\circ\varphi={\rm id}$, and so $H(\lambda,\mu)\cong
H(1,\lambda^{-1}\mu)$.

Simple 