Taft-algebra integrals and failure of semisimplicity

The multiplication rule gives $g{\mathrm{\Lambda }}_N={\mathrm{\Lambda }}_N$ by reindexing $i\mapsto i+1(\mathrm{m}\mathrm{o}\mathrm{d}N)$. It also gives $x{g}^i={\zeta }^{-i}{g}^{i}x$, so $x{\mathrm{\Lambda }}_N={\sum }_i{\zeta }^{-i}{g}^i{x}^N=0$. These are the left-integral equations for the generators $g,x$; because they generate $T_N$ and $\epsilon$ is multiplicative, the equations extend to every product and then every linear combination. Thus ${\mathrm{\Lambda }}_N$ is a left integral.

Write any left integral as $t={\sum }_{i,j}{c}_{ij}{g}^i{x}^j$. The equation $gt=t$ forces $c_{ij}=c_{i-1,j}$ for every $i,j$, so for each $j$ the coefficient is independent of $i$. The equation $xt=0$ has coefficient ${\zeta }^{-i}{c}_{ij}$ at each distinct basis vector $g^i{x}^{j+1}$ for $j<N-1$; since ${\zeta }^{-i}\mathrm{\ne }0$, all those $c_{ij}$ vanish. Thus only the $j=N-1$ coefficients can remain, and they are all equal, proving ${\int }_{T_N}^{\mathrm{\ell }}=k{\mathrm{\Lambda }}_N$.

Since $N>1$, each summand of ${\mathrm{\Lambda }}_N$ contains a positive power of $x$, and multiplicativity of the counit gives $\epsilon ({\mathrm{\Lambda }}_N)=0$. Every left integral is a scalar multiple of ${\mathrm{\Lambda }}_N$ by step 2.1, so none has nonzero counit. The Hopf–Maschke criterion [F3] therefore implies that $T_N$ is not semisimple.