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  "body": "\n## Proof\n\n1. Since $G$ is [[nilpotent|0DZPD2MG33]], it has a central series $\\{e\\} = N_0 \\subseteq N_1 \\subseteq \\cdots \\subseteq N_n = G$. Each $N_i$ is a [[normal subgroup|0W77CVBCHQ]] of $G$, and $N_{i+1}/N_i \\subseteq Z(G/N_i)$ for $i \u003c n$.\n2. Fix $i \u003c n$. Since $N_i$ is normal in $G$, it is normal in the subgroup $N_{i+1}$. So $\\{e\\} = N_0 \\trianglelefteq N_1 \\trianglelefteq \\cdots \\trianglelefteq N_n = G$ is a [[subnormal series|096HN851YZ]].\n3. The factor $N_{i+1}/N_i$ is the set of cosets $xN_i$ with $x \\in N_{i+1}$. Its product $(xN_i)(yN_i) = xyN_i$ is the product of the [[quotient group|0YYWCNTPWR]] $G/N_i$.\n4. By step 1, each such coset lies in the [[centre|0S7Y9Q990Q]] $Z(G/N_i)$. So for $x, y \\in N_{i+1}$, we get $(xN_i)(yN_i) = (yN_i)(xN_i)$.\n5. So each factor $N_{i+1}/N_i$ is abelian. By step 2, $G$ is [[solvable|0P3KZK0FTV]].\n\n## Idea\n\nA central series is already a subnormal series. Each of its factors sits inside the centre of a quotient group, and elements of a centre commute with each other. So every factor is abelian.\n",
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