Solvable subgroups in $GL(n,F)$ Is it true, that any solvable subgroup $G$ in $GL(n,F)$ is subgroup of upper triangular matrix in some basis?
 A: If $F$ is algebraically closed and the group is connected ( and algebraic)  then the answer is yes. This is Borel's theorem -- the search  term is Borel subgroup.
However, we need connected,  even for $F$ algebraically closed (@Derek Holt: thanks for pointing the example of finite solvable groups ) Indeed, any (solvable) non-abelian finite group has an irreducible representation of $\dim>1$, say $\dim n$ so take the image of that group in $GL(n,F)$. 
For  $F = \mathbb{R}$, you have an abelian subroup $SO(2,\mathbb{R})\subset GL(2, \mathbb{R})$ that is not upper triangular in any basis. 
Also note: any finite subgroup of the subgroup of upper triangular matrices  is abelian if $\text{char} F=0$. 
If $F$ is moreover algebraically closed then abelian groups are included in a conjugate of the upper triangular matrices. 
It's related to whether all the irreducible finite representations over $F$ of a given group are of dimension $1$. 
A: The group of upper triangular matrices over any field has a nilpotent normal subgroup (the unipotent matrices) with abelian quotient group. So any solvable group that does not have that property, such as $S_4$, can never be conjugate to an upper triangular group.
