Subgroup of Solvable Group is Solvable/Proof 3

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

$\set e = G_0 \lhd G_1 \lhd \dots \lhd G_m = G$

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

Define $N_i = G_i \cap H$.

We show that these $N_i$ will form a normal series with abelian factors.


 * Normality:

Let $x \in N_i$ and $y \in N_{i + 1}$.

Then $y x y^{-1} \in N$ since $N$ as a group is closed.

We also have $y x y^{-1} \in G_i$ since $G_i$ is normal in $G_{i + 1}$.

Hence $N_i$ is invariant under conjugation and therefore normal.


 * Abelian Factors:

Note that:

Thus:

so the quotient $\dfrac {N_{i + 1}} {N_i}$ is isomorphic to a subgroup of the abelian group $\dfrac{G_{i + 1}} {G_i}$.

Hence it is abelian, proving the theorem.