# Semisimple or Unipotent Borels are the Whole Group

Posted by Jason Polak on 03. August 2013 · Write a comment · Categories: math

($G$ is an algebraic group over an algebraically closed field $k$ and we identify $G$ with $G(k)$.)

In the previous post, to prove that a two-dimensional connected group $G$ is solvable, we used that if some Borel in it is either semisimple or unipotent, then it is actually all of $G$.

Here is a short proof: We first prove that $B$ is actually nilpotent. For $B = B_u$, its unipotent part, this is true since connected unipotent groups are nilpotent (and thus solvable, but we already knew that). If $B = B_s$, then we use the very handy result that a connected solvable group is the product of a torus and its unipotent part, but in this case the unipotent part is trivial, so $B$ itself is a torus and so $B$ is again nilpotent.

Now, we proceed by induction on the dimension of $G$. If $\dim G = 0$ then the result is clear, since $G$ is then trivial since it is connected. Now suppose $G$ has positive dimension, and so $B$ has positive dimension as well (since it is connected too, and the conjugates of $B$ cover $G$). In this case, as we have observed, $B$ is nilpotent and nontrivial, so it has a nontrivial center (this is true for any nontrivial nilpotent group, not just algebraic groups).

Now $Z(G)^\circ$, being solvable and connected, must be contained in a Borel subgroup, and since it is normal, it is contained in every Borel subgroup so we have the inclusions $Z(G)^\circ\subseteq Z(B)\subseteq Z(G)$. Since we have observed that $Z(B)$ is nontrivial, and hence has positive dimension, it also must be that $Z(G)^\circ$ (a normal subgroup) has positive dimension so that $G/Z(G)^\circ$ has strictly smaller dimension than $G$.

Since the surjective image of a Borel subgroup is a Borel (likewise for maximal tori, parabolics, and maximal connected unipotents), the group $B/Z(G)^\circ$ is a nilpotent Borel of $G/Z(G)^\circ$. But now we use the induction hypothesis which implies that $G/Z(G)^\circ = B/Z(G)^\circ$ and hence $B = G$. QED

Observe that the key to this proof really was the use of induction via the nontrivial $Z(G)^\circ$ contained in both $G$ and $B$, and that the homomorphic image of a nilpotent group is still nilpotent.