Suppose $G$ is a solvable group and $|G|=n$, show that $G$ is isomorphic to a subgroup of group of upper triangular complex invertible matrices.

Question

I've proved that all subgroups of upper triangular complex invertible matrices is solvable, but I find it too hard to show the inverse proposition.

Answer 1

This is false. A group $G$ is isomorphic to a subgroup of the group of upper triangular $n \times n$ matrices iff it has a faithful $n$-dimensional representation $V$ which is an iterated extension of $1$-dimensional representations (since the upper triangular matrices stabilize a complete flag, namely $e_1 \subset \text{span}(e_1, e_2) \subset \dots $ and we can take successive quotients). If $G$ is a finite group and we are working over $\mathbb{C}$ then there are no nontrivial extensions so such a representation $V$ must in fact be a direct sum of $1$-dimensional representations, and for such a representation to be faithful $G$ must be abelian.

So any finite nonabelian solvable group is a counterexample, e.g. the smallest one $G = S_3$.

In a positive direction see the Lie-Kolchin theorem. It's also true that a finite $p$-group is isomorphic to a subgroup of upper triangular matrices over $\mathbb{F}_p$. The argument is easy: let $G$ be such a group and $V$ any faithful finite-dimensional representation of $G$ over $\mathbb{F}_p$ (e.g. the regular representation). Then $V \setminus \{ 0 \}$ has $p^{\dim V} - 1$ elements, which is not divisible by $p$, and so by the $p$-group fixed point theorem $G$ has a nonzero fixed point $v \in V$. Now we can consider the action of $G$ on $V/v$ and iterate this, obtaining a complete flag fixed by $G$.