Proof that subgroups of solvable groups are solvable

Proof

We intersect a series of $G$ with $H$.

1. Since $G$ is solvable, it has a subnormal series $\{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_n = G$ in which each factor $G_i/G_{i-1}$ is abelian.
2. For $0 \leq i \leq n$, let $H_i = H \cap G_i$. It is a subgroup of $H$, since an intersection of subgroups is a subgroup. Also $H_0 = \{e\}$ and $H_n = H$.
3. Fix $i$ with $1 \leq i \leq n$. Define $\varphi_i \colon H_i \to G_i/G_{i-1}$ by $\varphi_i(h) = hG_{i-1}$. Since $h \in G_i$, this lands in the quotient group $G_i/G_{i-1}$.
4. The map $\varphi_i$ is a homomorphism, because $(hk)G_{i-1} = (hG_{i-1})(kG_{i-1})$ for $h, k \in H_i$.
5. Its kernel is $\{h \in H_i : h \in G_{i-1}\} = H \cap G_{i-1} = H_{i-1}$.
6. The kernel is a normal subgroup of the domain. So $H_{i-1} \trianglelefteq H_i$, and $\{e\} = H_0 \trianglelefteq \cdots \trianglelefteq H_n = H$ is a subnormal series of $H$.
7. By the first isomorphism theorem, $H_i/H_{i-1}$ is isomorphic to the image of $\varphi_i$, which is a subgroup of $G_i/G_{i-1}$.
8. A subgroup of an abelian group is abelian, and a group isomorphic to an abelian group is abelian. So $H_i/H_{i-1}$ is abelian.
9. So $H$ has a subnormal series with abelian factors, and $H$ is solvable.