Then we have $gt=t=\varepsilon(g)t$, $at=a^p(\sum\limits_{0\leqslant i\leqslant
n-1}g^i)b^{p-1}=\lambda(1-g^p)(\sum\limits_{0\leqslant i\leqslant
n-1}g^i)b^{p-1}=0=\varepsilon(a)t$ and $$\begin{split}
bt&=(\sum_{0\leqslant i\leqslant
n-1}bg^ia^{p-1})b^{p-1}\\
&=[\sum_{0\leqslant i\leqslant
n-1}(i+\frac{p-1}{2})g^ia^p+g^ia^{p-1}b]b^{p-1}\\
&=a^p(\sum_{0\leqslant i\leqslant
n-1}ig^i)b^{p-1}+\frac{p-1}{2}a^p(\sum_{0\leqslant i\leqslant
n-1}g^i)b^{p-1}+b^p(\sum_{0\leqslant i\leqslant n-1}g^i)a^{p-1}\\
 &=0=\varepsilon(b)t.
\end{split}$$ Since $g,a,b$ are generators of $H(\lambda,\mu)$, one gets that $\int_H^l=kt$. On the other hand, since $ba=a(b+\frac{1}{2}a)$, we have $(a+b)^{p-1}=b^{p-1}+a\sum\limits_{0\leqslant j\leqslant
p-2}\alpha_ja^jb^{p-2-j}$ for some $\alpha_j\in k$. Hence $$\begin{split}
tg&=(\sum_{0\leqslant i\leqslant
n-1}g^i)a^{p-1}b^{p-1}g\\
&=(\sum_{0\leqslant i\leqslant
n-1}g^i)a^{p-1}g(a+b)^{p-1}\ \ ({\rm by}\ bg=g(a+b))\\
&=(\sum_{0\leqslant i\leqslant
n-1}g^i)a^{p-1}[b^{p-1}+a\sum_{0\leqslant j\leqslant p-2}\alpha_ja^jb^{p-2-j}]\\
&=t+a^p(\sum_{0\leqslant i\leqslant
n-1}g^i)(\sum_{0\leqslant j\leqslant p-2}\alpha_ja^jb^{p-2-j})\\
&=t=\varepsilon(g)t,\\
\end{split}$$ $$\begin{split}
ta&=(\sum_{0\leqslant i\leqslant
n-1}g^i)a^{p-1}b^{p-1}a\\
&=(\sum_{0\leqslant i\leqslant
n-1}g^i)a^{p-1}a(b+\frac{1}{2}a)^{p-1}\ ({\rm by}\  ba=a(b+\frac{1}{2}a))\\
&=a^p(\sum_{0\leqslant i\leqslant n-1}g^i)(b+\frac{1}{2}a)^{p-1}\\
&=0=\varepsilon(a)t\\
\end{split}$$ and $$\begin{split}
tb&=b^p(\sum_{0\leqslant i\leqslant
n-1}g^i)a^{p-1}=0=\varepsilon(b)t,
\end{split}$$ where we use the facts that $a^p=\lambda(1-g^p)$ and $b^p=\mu(1-g^p)$ are central elements in $H(\lambda, \mu)$. Thus, $\int_H^r=kt=\int_H^l$, and so $H$ is unimodular. It is easy to check that $S^2$ is inner. It follows that $H$ is a symmetric Hopf algebra.

\[2.4\] Let $J$ be the Jacobson radical of $H(\lambda,\mu)$. Then

$(1)$ If $t=1$, then $a, b\in J$.

$(2)$ If $t>1$ and $\lambda\mu\neq0$, then $a,b \not\in J$.

\(1) Assume $t=1$. Then $n=p^s$. Since char$k$=$p$ and $a^p=\lambda(1-g^p)$, we have $a^n=a^{p^s}=[\lambda(1-g^p)]^{p^{s-1}}=\lambda ^{p^{s-1}}(1-g^{p^s})=
\lambda^{p^{s-1}}(1-g^n)=0$. On the other hand, we have $ag=ga$ and $ab=ba-\frac{1}{2}a^2=(b-\frac{1}{2}a)a$. Hence $aH(\lambda, \mu)=H(\lambda,\mu)a$, and consequently $H(\lambda, \mu)a$ is equal to the ideal $\langle a\rangle$ of 