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 $(H \cap G_i)\cap G_{i-1}=H\cap G_{i-1}$.

Therefore, according to the Second Isomorphism Theorem, we get:


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

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

Because $(H \cap G_i)\cap G_{i-1}\subseteq G_i$, we can use the Corrsepondence Theorem to get:


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

However, because $G_i / G_{i-1}$ is Abelain, then according to Subgroup of Abelian Group is Abelian we get that $\frac {(H \cap G_i)\cap G_{i-1}} {G_{i-1}}$ is Abelian as well, and therefore $\frac {H\cap G_i} {H\cap G_{i-1}}$ is Abelian. Combining these facts together, we get that 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 $H$, and therefore $H$ is Solvable.