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
Let $\left\{{e}\right\} = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$ be a normal series for $G$ with Abelian quotients.

For every $i = 1,\ldots,n$ we have $\left(H \cap G_i\right)\cap G_{i-1}=H\cap G_{i-1}$.

According to the Second Isomorphism Theorem:


 * $\displaystyle \frac {\left(H \cap G_i\right)\cap G_{i-1}} {G_{i-1}} \cong \frac {H\cap G_i} {H\cap G_{i-1}}$

Specifically, $H\cap G_{i-1}$ is a normal subgroup of $H\cap G_i$.

$\left(H \cap G_i\right)\cap G_{i-1}\subseteq G_i$ and so theCorrsepondence Theorem implies:


 * $\displaystyle \frac {\left(H \cap G_i\right)\cap G_{i-1}} {G_{i-1}} \leq G_i / G_{i-1}$

Because $G_i / G_{i-1}$ is Abelian, then according to Subgroup of Abelian Group is Abelian, $\frac {\left(H \cap G_i\right)\cap G_{i-1}} {G_{i-1}}$ is Abelian. Hence, $\frac {H\cap G_i} {H\cap G_{i-1}}$ is Abelian. Therefore, the series


 * $\left\{{e}\right\} = H\cap G_0 \lhd H\cap G_1 \lhd \cdots \lhd H\cap G_n = G$

is a normal series with Abelian quotients for $H$. Therefore, $H$ is Solvable.