Proposition 22

Suppose that $K = \mathbf{R}$ or $\mathbf{C}$. Let $V$ be a finite-dimensional vector space and $G$ a connected solvable subgroup of $\mathbf{GL}(V)$. Suppose that the identity representation of $G$ is simple. (i) If $K = \mathbf{R}$, then $\dim V \leq 2$ and $G$ is commutative. (ii) If $K = \mathbf{C}$, then $\dim V = 1$. (i) Suppose that $K = \mathbf{R}$. Then the closure $H$ of $G$ in $\mathbf{GL}(V)$ is a solvable connected Lie subgroup of $\mathbf{GL}(V)$ (no. 1, Corollary 2 to Proposition 1). Hence $L(H)$ is solvable (Proposition 19). The identity representation of $L(G)$ is simple (§ 6, no. 5, Corollary 2 to Proposition 13). Hence $\dim V \leq 2$ and $L(G)$ is commutative (Chapter I, § 5, Corollaries 1 and 4 to Theorem 1). Hence $G$ is commutative. (ii) Suppose that $K = \mathbf{C}$. Let $W$ be a minimal element among the non-zero real vector subspaces of $V$ which are stable under $G$. The complex vector subspace of $V$ generated by $W$ is equal to $V$ since the identity representation of $G$ is simple. By (i), $G|W$ is commutative. Hence $G$ is commutative. Therefore every element of $G$ is a homothety (Algebra, Chapter VIII, § 4, Corollary 1 to Proposition 2), so that $\dim V = 1$.