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 would follow 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.


  • $\begingroup$ Here's a more telegraphic version of @JimHumphreys's excellent answer: morphisms of algebraic groups respect the Jordan decomposition, so $\pi((u s)_{\text{semi}}) = \pi(u s)_{\text{semi}} = s_{\text{semi}} = s$, where $\pi$ is the natural projection $G \to G/U \cong T$. $\endgroup$
    – LSpice
    Commented Jan 26, 2018 at 19:01

1 Answer 1


Consider yourself fortunate to be having only "a bit of difficulty" in reading Steinberg's concise proofs. His style typically involves numerous reductions to get to the essential point, along with sentences crammed with dense arguments. Though he is almost always right on target, it can be frustrating to unpack such proofs (as I learned the hard way in graduate school when giving seminar talks on two of his papers from 1963 and 1965).

Here you are looking at his 7.6 in the well-filled AMS memoir published in 1968 but mostly developed earlier. (I'm looking at the typeset version in his collected papers, where the page numbers are different.) It's helpful to add some notation, which he wisely avoids as far as possible. The essential situation involves a connected solvable algebraic group $G= T \ltimes U$ with $T$ any maximal torus and $U$ the unipotent radical. A basic fact is that all maximal tori of $G$ are conjugate under elements of $U$ (a step to the general conjugacy theorem). We are given some embedding of $G$ into a possibly larger algebraic group $G'$ and a "semisimple" automorphism $\sigma$ of $G$ coming from conjugation by a semisimple $s \in G'$. The problem is to show that $\sigma$ stabilizes some maximal torus of $G$.

Now $T$ itself might not be $\sigma$-stable, but $us T s^{-1} u^{-1} = T$ for some $u \in U$. Thus $us \in N_{G'}(T)$, where its Jordan decomposition in $G'$ lives. Note too that $us \in N_{G'}(G)$ since each factor does; thus the Jordan parts of $us$ stabilize $G$ and also $T$. In particular, it is harmless now to replace $us$ by its semisimple part in $G'$ (i.e., assume it is semisimple). This is Steinberg's reduction.

After this reduction, the remaining point is to show that $us$ is actually conjugate in $G'$ to $s$. Moreover, the conjugating element stabilizes $G$ and takes $T$ to a new maximal torus which will be $\sigma$-stable as desired. Here again the argument has to be reduced, now to the case when $U$ is commutative. Etc.

P.S. Steinberg's last steps are also compressed (using the commutativity of $U$ in a devious way), but his reference to page 4-14 in Expose 4 of the 1956-58 Chevalley seminar here is again elementary.

As Steinberg points out, a different proof (which actually uses more of the Chevalley structure theory) was given by David J. Winter in Corollary 4 here.

  • $\begingroup$ Thank you very much for your answer. There is one point here that is still not clear to me; replacing $us$ by its semisimple part is essentially conjugating by the inverse of the unipotent part, so we get (us)_u^(-1)usTs^-1u^-1(us)_u = T. If we let u‘=(us)_u^(-1)us then we have u‘s semisimple, but the next part of the theorem uses that u‘ should be in U. Does this follow? this seems to be the crucial part of the proof. $\endgroup$
    – user1437
    Commented Nov 11, 2012 at 14:55
  • $\begingroup$ @Jason: What you are doing is too complicated. Just forget about $us$, since its semisimple part in $G'$ already takes the maximal torus $T$ of $G$ to itself. If you next show that this new semisimple element is conjugate in $G'$ to $s$, it will follow that $\sigma$ preserves a conjugate of $T$. $\endgroup$ Commented Nov 11, 2012 at 17:01
  • $\begingroup$ P.S. Maybe you are worried about the format of the semisimple part of $us$? It lies in the closed subgroup of $G'$ generated by $s, U$ with $U$ normal, thus looks like $u' s^i$; but the Jordan decomposition $u s = u"\cdot u' s^i$ forces $i=1$. So think of $us$ as $u's$ in the rest of the argument. (This is my guesswork, since Steinberg avoids writing it all down.) $\endgroup$ Commented Nov 11, 2012 at 18:03
  • $\begingroup$ Another P.S. My formulation assumes $s$ has finite order, but in general the idea is the same: $us$ has semisimple part $u's'$ with $s'$ in the closed (diagonalizable) subgroup of $G'$ generated by $s$ while $u'$ lies in the unipotent radical $U$ of the group generated by $s, U$ together. $\endgroup$ Commented Nov 11, 2012 at 23:23
  • $\begingroup$ @Jim: Yes, what I was doing was too complicated indeed. I think my misunderstanding was with format of the semisimple part of $us$. Thank you very much for your answer and comments, again. $\endgroup$
    – user1437
    Commented Nov 12, 2012 at 15:46

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