Proposition 18

Let k be a commutative field. V a vector space of finite dimension > 0 over k and G a subgroup of $\mathbf{GL}(V)$ whose elements are unipotent. (i) There exists a non-zero element v of V such that gv = v for all g \in G. (ii) There exists a basis B of V such that, for all g \in G, the matrix of g with respect to B is lower triangular and has all its diagonal elements equal to 1. (iii) The group G is nilpotent. (a) Suppose first that k is algebraically closed and that the identity representation of G is simple. Let a, b be in G. Then

for $ab - 1$ and $a - 1$ are nilpotent. As the vector subspace of $\mathscr{L}(V)$ generated by G is $\mathscr{L}(V)$ (Algebra, Chapter VIII, § 4, Corollary 1 to Proposition 2), $\operatorname{Tr}(u(b-1)) = 0$ for all $u \in \mathscr{L}(V)$ and hence $b = 1$. Thus $G = \{1\}$. (b) We pass now to the general case. Let $\bar{k}$ be an algebraic closure of k, $\bar{V} = V \otimes_k \bar{k}$ and $\bar{G} \subset \mathbf{GL}(\bar{V})$ the set of $a \otimes 1$ for $a \in G$. Let W (resp. $W'$) be the set of elements of V (resp. $\bar{V}$) invariant under G (resp. $\bar{G}$). Then $W' = W \otimes_k \bar{k}$ for $W = \bigcap_{g \in G} \operatorname{Ker}(g-1)$ and $W' = \bigcap_{g \in G} \operatorname{Ker}(g-1) \otimes 1$. If $V_1$ denotes a minimal element in the set of non-zero vector subspaces of $\bar{V}$ which are stable under $\bar{G}$, then $V_1 \subset W'$ by part (a) of the proof; hence $W \neq \{0\}$, which proves (i). (c) By induction on $\dim V$, it follows from (i) that there exists an increasing sequence $(V_1, V_2, \ldots, V_n)$ of vector subspaces of V which are stable under G such that $V_n = V$ and the automorphism group of $V_i/V_{i-1}$ canonically derived from G reduces to $\{1\}$ for all $i$ (we make the convention that $V_r = \{0\}$ for $r \leq 0$). This implies first (ii) and consequently (iii) (Chapter II, § 4, no. 6, Remark).