Proposition 26

Suppose that K is of characteristic $p > 0$. For all $a \in L(G)$, $a^p \in L(G)$ and $\operatorname{ad}(a^p) = (\operatorname{ad} a)^p$ (the power $a^p$ being calculated in $U(G)$).

If $a \in L(G)$, $a$ is primitive in $U(G)$, hence $a^p$ is primitive in $U(G)$ (Chapter II, § 1, no. 2, Remark 1) and hence $a^p \in L(G)$. Let $\sigma_a$ (resp. $\tau_a$) be the linear mapping $x \mapsto a * x$ (resp. $x \mapsto x * a$) of $U(G)$ into $U(G)$. For all $x \in U(G)$, $(\operatorname{ad} a)(x) = (\sigma_a - \tau_a)(x)$ and hence $(\operatorname{ad} a)^p = (\sigma_a - \tau_a)^p$. But $\sigma_a$ and $\tau_a$ commute and therefore $(\sigma_a - \tau_a)^p = (\sigma_a)^p - (\tau_a)^p = \tau_a^p - \sigma_a^p$, whence the second assertion.