u-(1-\xi^{i+j})u\neq0$, and hence $u$ is not an eigenvector of the action of $a$. It follows that $u_0$ is the unique common eigenvector of the action of $g$ and $a$ up to a non-zero scale multiple. It follows from Theorem \[3.4\] that soc$(S_i\otimes S_j)\cong S_{\overline{i+j}}$.

For case 2: $i+j=t$. In this case, $1$ is the unique eigenvalue of the action of $g$. It follows from Theorem \[3.4\] that any simple submodule of $S_i\otimes S_j$ is isomorphic to $S_0$, and is spanned by a non-zero vector $v'$ with $g\cdot v'=v'$, $a\cdot v'=0$ and $b\cdot v'=0$. Now we have $g\cdot u_0=u_0$ and $a\cdot u_0=0$. By Lemma \[2.2\](2), it follows that $g\cdot(b^l\cdot u_0)=b^l\cdot
u_0$ and $a\cdot(b^l\cdot u_0)=0$ for all $1\leqslant l\leqslant
p-1$. Since $\Delta(b)={b\otimes 1}+{g\otimes b}$, one can see that $(b^l\cdot u_0)(1)=l$, $(b^l\cdot u_0)(2)=0$. It follows that $\{u_0, b\cdot u_0, \cdots, b^{p-1}\cdot u_0 \}$ are linearly independent and contained in $V_{\xi^{i+j}}=V_1$. Furthermore, $b\cdot(b^{p-1}\cdot u_0)=b^p\cdot u_0=0$. Thus, soc$(S_i\otimes
S_j)=k(b^{p-1}\cdot u_0)\cong S_0$.

This completes the proof.

Now we are going to investigate the indecomposable projective modules over $H(\lambda, \mu)$.

Let $e_i=\frac{1}{t}\sum_{j=0}^{t-1}(\xi^{-ip^s}g^{p^s})^j$. Then $\{e_0, e_1, \cdots, e_{t-1}\}$ is a set of primitive orthogonal idempotents in $kG$ since $\xi^{p^s}$ is also a $t$-$th$ primitive root of unity. Now 