$\operatorname{ad}(L)$ is semisimple for a Lie algebra $L$ This must be really obvious. But I have $L$ a semisimple Lie algebra $$\begin{array}{rccc}\operatorname{ad}\colon&L&\longrightarrow&GL(L)\\\&X&\mapsto&\left(\begin{array}{rccc}\operatorname{ad}(X)\colon&L&\longrightarrow&L\\\&Y&\mapsto&\operatorname{ad}(X)(Y) = [X,Y]\end{array}\right)\end{array}$$ Why is $\operatorname{ad}(L)$ semisimple?

Because the adjoint representation has kernel $Z(L)$, which is zero for a semisimple Lie algebra $L$, we have $L\cong ad(L)$ in this case.