Let $V$ be a $k$-vector space of dimension $n$.
Take a flag $$0 = V_0 \subsetneq V_1 ...\subsetneq V_{m-1} \subsetneq V_{m} = V , $$ and a subgroup $G \leq GL(V)$ which stabilizes this flag, so that $\forall g \in G$, $gV_i = V_i$. For each $ i \in \{0,1,...,m-1\}$ we have projection maps $p_i: G \to GL(V_{i+1}/V_i)$. If each $p_i(G)$ is solvable, why is $G$ solvable?
What I've been trying so far:
We work by induction on the length of the flag.
For a flag of length $ m =1$ we get that $G / ker(p_0) \cong p_0(G)$ which is solvable. Since $ker(p_0) = \{g \in G : g(v) = v \} = \{Id\}$ is solvable we get that $G$ is too.
Next for a flag of length $m + 1$, assume the statement for flags of length $ \leq m$; We can define $G_m = \{g:V_m \to V_m\ $ $ | g \in G\}$ the restriction. Then $G_m$ stabilizes the shorter flag up to $V_m$ and its projection maps are the same on those indices. So the induction assumption gives that $G_m$ is solvable. How can relate $p_m(G)$ to $G_m$?
Also tried to look at $ker(p_m)$ but I can't see why this would be solvable either (if it was then $G/ ker(p_m) \cong p_m(G)$ so $G$ is solvable)
What do you think?
The extension of solvable groups is solvable, let $G_i$ be the restriction of $G$ to $V_i$, consider $q_{i+1}:G_{i+1}\rightarrow G_i$ such that $q_{i+1}(g)$ is the restriction of $g$ to $V_i$. Let $H_i$ be the kernel of $q_i$, $H_i$ is solvable, to see this, consider a basis of $V_{i+1}$ obtained by completing a basis $(e_1,..,e_{n_i})$ by $(f_1,...,f_{m_i})$, let $h$ be an element of $h$ in the kernel of $p_i$, you have $h(f_j)=f_j+v_{f_i}$ where $v_{f_i}\in V_i$, this implies the matrix of $h$ is a triangular superior, so $H_i$ is solvable.
We have an exact sequence:
$1\rightarrow H_i\rightarrow G_{i+1}\rightarrow G_i\rightarrow 1$, a recursive argument implies that $G$ is solvable.