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  "body": "\n## Proof\n\nWe lift a series of $G/N$ to $G$ and put a series of $N$ below it.\n\n1. Since $N$ is [[solvable|0P3KZK0FTV]], it has a [[subnormal series|096HN851YZ]] $\\{e\\} = N_0 \\trianglelefteq \\cdots \\trianglelefteq N_k = N$ with abelian factors.\n2. Since $G/N$ is solvable, it has a subnormal series $\\{N\\} = Q_0 \\trianglelefteq \\cdots \\trianglelefteq Q_l = G/N$ with abelian factors. Here $N$ is the identity of the [[quotient group|0YYWCNTPWR]].\n3. Let $\\pi \\colon G \\to G/N$ be the projection $\\pi(g) = gN$. It is a surjective [[homomorphism|0HV9Y7WG5R]] with kernel $N$.\n4. For $0 \\leq j \\leq l$, let $M_j = \\pi^{-1}(Q_j) = \\{g \\in G : \\pi(g) \\in Q_j\\}$. It is a subgroup of $G$, since $\\pi(xy^{-1}) = \\pi(x)\\pi(y)^{-1}$.\n5. We have $M_0 = \\pi^{-1}(\\{N\\}) = N$ and $M_l = G$.\n6. Fix $j$ with $1 \\leq j \\leq l$. Since $\\pi$ is surjective, $\\pi(M_j) = Q_j$.\n7. Define $\\psi_j \\colon M_j \\to Q_j/Q_{j-1}$ by $\\psi_j(x) = \\pi(x) Q_{j-1}$. It is a homomorphism, as a composite of homomorphisms. By step 6 it is surjective.\n8. The [[kernel|0Z424HK8CQ]] of $\\psi_j$ is $\\{x \\in M_j : \\pi(x) \\in Q_{j-1}\\} = M_{j-1}$.\n9. Since [[the kernel is a normal subgroup|0C9ZKYSPYR]], $M_{j-1} \\trianglelefteq M_j$.\n10. By the [[first isomorphism theorem|0CXXJNG5B6]], $M_j/M_{j-1} \\cong Q_j/Q_{j-1}$, which is abelian. So $M_j/M_{j-1}$ is abelian.\n11. Put the two series together: $\\{e\\} = N_0 \\trianglelefteq \\cdots \\trianglelefteq N_k = N = M_0 \\trianglelefteq M_1 \\trianglelefteq \\cdots \\trianglelefteq M_l = G$.\n12. Each term is [[normal|0W77CVBCHQ]] in the next, by steps 1 and 9. Each factor is abelian, by steps 1 and 10.\n13. So $G$ has a subnormal series with abelian factors, and $G$ is solvable.\n\n## Idea\n\nThe series of $G/N$ pulls back to a chain of subgroups of $G$ from $N$ up to $G$, with the same factors. Below $N$ we use the series of $N$. Stacking the two chains gives a series of $G$ with abelian factors.\n",
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