Suppose $V$ is a vector space of dimension $n$ over the field $\mathbb{K}$ and
$$V_0 \subsetneq V_1 \subsetneq ...\subsetneq V_k =V $$
is a flag (not necessarily complete) preserved by a subgroup $G \leq GL(V)$. If moreover the projections $\rho_i : G \to GL(V_{i+1}/V_i)$, for $i \in \{0,...,k-1\}$ are solvable, show that G is solvable.
My attempt:
If the flag were exact, $G$ would be solvable. By induction, assume $k>1$ and the restriction of $G$ on $V_{k-1}$, say $G_{k-1} \leq GL(V_{k-1})$ is solvable. Now, consider the restriction homomorphism $\phi : G \to G_{k-1}$. Then, $G/ker(\phi) \cong G_{k-1}$, which is solvable. If we can show that $ker(\phi)$ is solvable, then we're done.
Take a basis $\{e_1,...,e_n\}$ of $V_{k-1}$ and extend it to a basis $\{e_1,...,e_n,f_1,...,f_m\}$ of $V_k$. If it were true that $ker(q) \subseteq ker(\rho_{k-1})$, then for any $h \in ker(q)$ we'd have $h(f_i) = f_i +v_i$, where $v_i$ is simply in the linear span of $\{e_1,...,e_n\}$, and so the matrix of $h$ in that basis would be (almost) upper-triangular. In this case we could make the induction work to find an upper-triangular matrix. But I don't think $ker(q) \subseteq ker(\rho_{k-1})$ is true and so I'm stuck.
I would appreciate some help with this. Thank you very much in advance!
You could do it by induction on $k$. It is true by assumption for $k=1$.
For $k>1$, the image of $\rho_k$ is solvable by assumption.
The kernel $K$ of $\rho_k$ acts on the flag $V_0 \subset V_1 \subset \cdots \subset V_{k-1}$ of length $k-1$ and the image of $\phi_i$ restricted to $K$ is solvable for $1 \le i \le k-1$, so $K$ is solvable by inductive hypothesis.
Now $G/K \cong {\rm Im}(\phi_k)$ with $K$ and ${\rm Im}(\phi_k)$ solvable, so $G$ is solvable.