Subgroup of Solvable Group is Solvable/Proof 3

Proof
Let $H \leq G$ and $G$ be solvable with normal series

$1 = G_0 \lhd G_1 \lhd \dots \lhd G_m = G$

such that $G_{i+1}/G_i$ is abelian for all i.

Define $N_i= G_i \cap H$. These $N_i$ will form a normal series with abelian factors.

Normality:

Let $x \in N_i$ and $y \in N{i+1}$ then $yxy^{-1} \in N$, since N is a group, and $yxy^{-1} \in G_i$, since $G_i$ is normal in $G_{i+1}$. Hence $N_i$ is stable under conjugation and therefore normal.

Abelian Factors:

We have:


 * $\dfrac{N_{i+1}}{N_i} = \dfrac{N_{i+1}}{N_{i+1}\cap G_i} \cong \dfrac{N_{i+1}G_i}{G_i} \leq \dfrac{G_{i+1}}{G_i}$

so the quotient $\dfrac{N_{i+1}}{N_i}$ is isomorphic to a subgroup of the abelian group $\dfrac{G_{i+1}}{G_i}$, due to the Second Isomorphism Theorem for Groups, hence its abelian, proving the theorem.