Subgroup of Solvable Group is Solvable

Theorem
Let $G$ be a solvable group.

Let $H$ be a subgroup of $G$.

Then $H$ is solvable.

Proof
We will provide two proofs here, for the two equivalent definitions of solvable groups.

Firstly we, know that a group is solvable if and only if its derived series

$D(G)=[G,G] \, \ D^i(G)= [D^{i-1}(G),D^{i-1}(G)$

becomes trivial after finite iteration. Meaning

$D^j(G) = \{1\}$

for some finite j. Now it is trivial that

$D(H) \leq D(G)$

since H is smaller than G. Further since $D^i(H)$ is dominated by $D^i(G)$ it too has to become trivial after a finite amount of steps.

A different proof using the more common definition of solvability:

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.