Do Borel subgroups make sense in an arbitrary algebraic group. When is $G/B$ projective? Let us be over $\Bbb C$.
It seems that in an arbitrary algebraic group, it makes sense to talk about solvable subgroups and connected subgroups. Then, it seems that the concept of a Borel subgroup holds in an arbitrary algebraic group.
Is $G/B$ projective then, if we consider an arbitrary algebraic group $G$, or does this not make sense unless $G$ is atleast reductive? When does it make sense? Is there something weaker than reductivity that gives us this? The proofs I have seen for projectivity of $G/B$ rely it seems on affine algebraic groups embedding into $\text{GL}(n,\Bbb C)$, so it seems we do at least need $G$ to be an affine algebraic group.
 A: Let $G$ be reduced, connected but not necessarily affine algebraic group over a perfect field(and hence smooth), Barsotti-Chevalley theorem says that there exists a (unique) connected normal affine algebraic subgroup say $H$ of $G$ such that $G/H$ is an abelian variety. Now Let $B$ be a borel subgroup of $H$, then $H/B$ is a projective variety. We now claim that $B$ is the Borel subgroup in the sense that $G/B$ is complete.
Claim 1 : $ G(\mathbb{C}) \rightarrow (G/H)(\mathbb{C}) $ is a locally trivial fiber bundle with fibers $H(\mathbb{C})$.
Claim 2 : $ (G/B)(\mathbb{C}) \rightarrow (G/H)(\mathbb{C}) $ is a locally trivial fiber bundle with fibers $(H/B)(\mathbb{C})$.
Note that claim 1 and claim 2 follow once we know that the quotients in the algebraic sense also turn out to be quotients in the sense of lie groups on $\mathbb{C}-$points. Note that 
Claim 3 : $(G/B)(\mathbb{C})$ is compact.
This follows from here Fiber bundle is compact if base and fiber are
and the fact that $A, H/B$ is a projective variety and hence $A(\mathbb{C}), H/B(\mathbb{C})$ is compact.
Claim 4 : $G/B$ is a proper over $\mathbb{C}$
Claim 4 follows from chow's lemma. For a proof look at section 7 of GAGA by Serre.
So $G/B$ will be projective once we know the fact that the quotient of a group is quasi-projective.
Let $B'$ be a maximal connected solvable subgroup of $G$, then $B' \supset B$ and hence $G/B \rightarrow G/B'$ is surjective and thus $G/B'$ is also proper over $\mathbb{C}$. Once again So $G/B'$ will be projective once we know the fact that the quotient of a group by a closed subgroup is quasi-projective.
I realize this lacks a lot of details but hopefully this gives an idea as to how this problem can be approached.
