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In the title referenced above a proof of the conjugacy of Borel subalgebras is given on page 84: We assume $L$ semisimple and let $B$ be a standard Borel subalgebra and $B'$ any other Borel subalgebra. We set $N'$ equal to the set of all nilpotent elements of $B \cap B'$ and assume $N'\neq 0$ for case 1. (Note that N' is an ideal of $B \cap B'$). If $x \in N'$ then $ad x$ acts nilpotently on the vector space $B/B \cap B'$ whence there exists nonzero $y$ in this vector space that is annihilated by all of $N'$ (by a linear algebra thm). In other words, there is a $y$ in $B$ outside of $B'$ that is sent into $B \cap B'$ by any $ad x$ for $x \in N'$. But since $[x,y] \in [B,B]$ then $ad [x,y]$ is nilpotent on $L$ (since $B$ is standard) and hence in $N'$. Therefore there is a $y$ in $B$ outside of $B'$ that is in the normalizer, $K$, of $N'$. Humphreys then wants to make a symmetric conclusion: That there is a $y'$ in $B'$ outside of $B$ that is in the normalizer, $K$, of $N'$. But how is this possible? Since $[x,y'] \in [B',B']$ we may only conclude $ad [x,y']$ is nilpotent on $B'$ and not on all of $L$ since we do not know that $B'$ is standard. Thus we can't know for sure if $[x,y] \in N'$ and can't conclude $y'\in K$. Thoughts?

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The situation can be resolved as follows:

As mentioned in the question, we have some $y'\in B'$ outside of $B \cap B'$ such that for all $x \in N'$ we have $[x,y']$ in $B \cap B'$. If all such $[x,y']$ lie in $N'$ there is no problem. So assume there is some $x' \in N'$ with $[x'y']:=z \notin N'$. Then we may Jordan-decompose within $B \cap B'$ as $z=z_s+z_n$ with $z_s \neq 0$, so by virtue of $z$ and $z_n$ commuting and being nilpotent on $B'$ we see $z_s$ is also nilpotent on $B'$. Since $z_s$ is semisimple this implies that on $B'$ we have $ad (z_s) = 0$. So the centralizer of $z_s$ in $L$, say $C$, contains $B'$. Now find a CSA for $L$, say $H'$, including $z_s$. Clearly $H'$ must also must be contained in $C$. Since $z_s \in B' \subseteq C$ we see that $C$ is an ideal in $L$ and therefore semisimple. As a CSA of $L$, $H'$ is also a CSA of $C$ and so there is a BSA $B'' \subseteq C$ containing $H'$. Now $C$ cannot be all of $L$ so by the inductive hypothesis we have $f \in \varepsilon(C)$ sending $B''$ to $B'$ and so also an $F \in \varepsilon(L)$ sending $B''$ to $B'$ . Hence $B'$ contains a conjugate of the CSA $H'$ of $L$, which is of course also a CSA of $L$. But it is not difficult to show that any BSA of $L$ which contains a CSA of $L$ is standard. Thus $B'$ must be standard, whence $[B',B']$ is nilpotent on all of $L$, but of course this contradicts $z \notin N'$.

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