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  "body": "\n## Proof\n\nWe intersect a series of $G$ with $H$.\n\n1. Since $G$ is [[solvable|0P3KZK0FTV]], it has a [[subnormal series|096HN851YZ]] $\\{e\\} = G_0 \\trianglelefteq G_1 \\trianglelefteq \\cdots \\trianglelefteq G_n = G$ in which each factor $G_i/G_{i-1}$ is abelian.\n2. For $0 \\leq i \\leq n$, let $H_i = H \\cap G_i$. It is a [[subgroup|0N0PXCTYA7]] of $H$, since an [[intersection of subgroups is a subgroup|0ZTGMDNVTB]]. Also $H_0 = \\{e\\}$ and $H_n = H$.\n3. 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|0YYWCNTPWR]] $G_i/G_{i-1}$.\n4. The map $\\varphi_i$ is a [[homomorphism|0HV9Y7WG5R]], because $(hk)G_{i-1} = (hG_{i-1})(kG_{i-1})$ for $h, k \\in H_i$.\n5. Its [[kernel|0Z424HK8CQ]] is $\\{h \\in H_i : h \\in G_{i-1}\\} = H \\cap G_{i-1} = H_{i-1}$.\n6. [[The kernel is a normal subgroup|0C9ZKYSPYR]] 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 $H$.\n7. By the [[first isomorphism theorem|0CXXJNG5B6]], $H_i/H_{i-1}$ is isomorphic to the image of $\\varphi_i$, which is a subgroup of $G_i/G_{i-1}$.\n8. 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.\n9. So $H$ has a subnormal series with abelian factors, and $H$ is solvable.\n\n## Idea\n\nCut the series of $G$ down to $H$ by intersecting each term with $H$. Each new factor embeds into the old factor at the same place. A subgroup of an abelian group is abelian, so the new factors are abelian too.\n",
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