$t$ is a non-zero element of $H$. Since $g^n=1$, $g(\sum\limits_{0\leqslant i\leqslant n-1}g^i)=\sum\limits_{0\leqslant i\leqslant n-1}g^i$. It follows that $gt=t=\varepsilon(g)t$. By the definition of $H$, we also have $at=(\sum\limits_{0\leqslant i\leqslant n-1}g^i)a^2bab^3=0=\varepsilon(a)t$ and $bt=(\sum\limits_{0\leqslant i\leqslant n-1}bg^i)abab^3=\sum\limits_{0\leqslant i\leqslant n-1}(ig^ia+g^ib)abab^3
=\sum\limits_{0\leqslant i\leqslant n-1}g^ibabab^3=\sum\limits_{0\leqslant i\leqslant n-1}g^iabab^4=0=\varepsilon(b)t$. Since $g, a, b$ are generators of $H$, it follows that $\int_H^l=kt$. On the other hand, we have $a(a+b)=a^2+ab=ab$ and $bg=g(a+b)$. Hence $tg=(\sum\limits_{0\leqslant i\leqslant n-1}g^i)abab^3g=(\sum\limits_{0\leqslant i\leqslant n-1}g^i)ga(a+b)a(a+b)^3
=(\sum\limits_{0\leqslant i\leqslant n-1}g^i)abab^3=\varepsilon(g)t$. We also have $ta=(\sum\limits_{0\leqslant i\leqslant n-1}g^i)abab^3a
=(\sum\limits_{0\leqslant i\leqslant n-1}g^i)abab(ab^2+aba)
=(\sum\limits_{0\leqslant i\leqslant n-1}g^i)baba(ab^2+aba)
=0=\varepsilon(a)t$ and $tb=(\sum\limits_{0\leqslant i\leqslant n-1}g^i)abab^4=0=\varepsilon(b)t$. Thus, $\int_H^r=kt=\int_H^l$, and so $H$ is unimodular. It is easy to check that $S^2(g)=g$, $S^2(a)=g^{-1}ag$ and $S^2(b)=g^{-1}bg$. Hence $S^2$ is inner since $S^2$ is an algebra automorphism. It follows that $H$ is a symmetric Hopf algebra.

In the rest of this section, we assume $p>2$. Let $n=p^st$ with $p\nmid t$ and $s\geqslant 1$. Let $\lambda, \mu\in k$. Now we give some properties of $H(\lambda,\mu)$.

\[2.2\] In $H(\lambda,\mu)$, we have

$(1)$ $bg^i=ig^ia+g^ib$, $ba^j=a^jb+\frac{j}{2}a^{j+1}$ and $bg^ia^j=(i+\frac{j}{2})g^ia^{j+1}+g^ia^jb$ for all $i,\ j\geqslant 0$. In particular, $g^p$ is central in $H(\lambda,\mu)$.

$(2)$ If $1\leqslant m\leqslant p-1$, then $$ab^m =\sum_{0\leqslant i\leqslant m}\alpha_{m,i}b^{m-i}a^{i+1},$$ where $\alpha_{m,i}\in k$ with $\alpha_{m,0}=1$, $\alpha_{m,1}=-\frac{m}{2}$ and $\alpha_{m,2}=\frac{1}{4}m(m-1)$.

$(3)$ If $1\leqslant m\leqslant p-1$, then $$gb^m =\sum_ {0\leqslant 