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I am trying to understand a result on algebraic groups, namely that if $\sigma:G\to G$ is an isogeny of a connected linear algebraic group over an algebraically closed field, then $\sigma$ stabilizes some Borel subgroup $B$ of $G$, i.e. $\sigma(B)\subset B$. I think that this was originally proved by Steinberg in his book 'endomorphisms of algebraic groups'. In this book the result is given in Theorem 7.2 after which it is stated that the theorem follows from Lemma 7.3. In the header of Lemma 7.3 it then says: '... and let $B$ be a Borel subgroup of $G$ fixed by $\sigma$...' This does not really seem like a good way to prove Theorem 7.2. Note that the assumption that $\sigma$ stabilizes $B$ is in fact used in the proof of Lemma 7.3: '...We can choose $b\in B$ such that $i_b\sigma$ stabilizes a maximal torus $T\subset B$...'

So my question would be if there is an alternative way to prove that this well-known result is true? And am I missing something or is the proof in this book just wrong?

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The paragraph after Lemma 7.3 explains how to use the case of an endomorphism known to stabilise a Borel, together with $G(k)$-conjugacy of Borels, to prove that every surjective endomorphism stabilises a Borel. Note that Steinberg identifies an algebraic group with its group of $k$-rational points, where $k$ is the algebraically closed ground field, so, e.g., “$y \in G$” means “$y \in G(k)$”.

Assuming the lemma for the moment, let $G$ and $\sigma$ be [a linear algebraic group and a surjective endomorphism of it] and let $B$ be a Borel subgroup of $G$. Then $\sigma B$ is also a Borel subgroup … so that (*) $y \sigma B y^{-1} = B$ for some $y \in G$ …. Applying 7.3 to $i_y \sigma$ we get $y^{-1} = x b y\cdot\sigma x^{-1}\cdot y^{-1}$ for some $x \in G$ and $b \in B$, whence $y = b^{-1} x^{-1}\sigma x$. Substituting into (*), we see that $\sigma$ fixes the Borel subgroup $x B x^{-1}$ ….

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  • $\begingroup$ The problem for me is that in order to apply Lemma 7.3 we have to know that there is a $\sigma$ stable Borel subgroup of $G$ (this is in the header of the lemma) so one would need Theorem 7.2 and thus one goes in circles. $\endgroup$ Commented May 23, 2023 at 17:44
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    $\begingroup$ @VictordeVries, re, we do not apply Lemma 7.3 to $\sigma$, which is not known to stabilise a Borel. We apply it to $i_y\sigma$ (i.e., $g \mapsto y\sigma(g)y^{-1}$), which stabilises a Borel because that's how $y$ was chosen. $\endgroup$
    – LSpice
    Commented May 23, 2023 at 17:55
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    $\begingroup$ Thanks so much, I was extremely confused by this, but now I get it. $\endgroup$ Commented May 23, 2023 at 17:57
  • $\begingroup$ @VictordeVries, re, my pleasure! If your journey through algebraic groups is anything like mine, you won't appreciate how much power is buried in those rational conjugacy results until you've worked with them for some time. (I deleted my previous comment, which accidentally omitted the "until …" part and so read as rather pessimistic.) $\endgroup$
    – LSpice
    Commented May 23, 2023 at 18:33

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