I am reading a proof on p.51 of Robert Steinberg in his book "Endomorphisms of Algebraic Groups" and I am having a bit of difficulty understanding one point in the proof.

The setting is as follows. Consider an algebraic group $G$ over an algebraically closed field of arbitrary characteristic. We say that an automorphism $\sigma:G\to G$ is *semisimple* if there exists an embedding $G\hookrightarrow G'$ such that $\sigma$ is realised by conjugation in $G'$ by a semisimple element. The theorem is:

If $G$ is solvable and $\sigma:G\to G$ is a semisimple automorphism then there exists a maximal torus $T\subseteq G$ such that $\sigma(T) = T$.

I shall paraphrase the proof on pp.51-52 of Steinberg. Assume $G$ is connected. We choose at first an arbitrary maximal torus $T\subseteq G$, and let $U$ be the normal subgroup of $G$ of all unipotents; then we have $G = T\ltimes U$. We embed $G$ in a larger group so that $\sigma:G\to G$ is realised by conjugation by a semisimple element $s$. Then $sTs^{-1}$ and $T$ are maximal tori of $G$, so we can write $T = usTs^{-1}u^{-1}$ where $u\in G$ and by our decomposition, we can assume $u\in U$.

It then suffices to show that $us$ is actually conjugate to $s$: i.e. $us = xsx^{-1}$ for some $x\in G$, for then $\sigma$ will fix $x^{-1}Tx$.

Then Steinberg says: *"and on replacing $us$ by its semisimple part we may assume it to be semisimple."* After this, he proceeds to show that if $us$ is semisimple, then it indeed is conjugate to $s$, which completes the proof.

Question: why can we assume that $us$ is semisimple? This

wouldfollow if the semisimple part could be written as $u's$ where $u' \in U$, but I cannot see why this is true at the moment.

Thanks!