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  "body": "\n## Proof\n\nWe argue by strong induction on $|G|$.\n\n1. Let $A$ be a group, $B$ a normal subgroup and $x \\in A$. The coset $xB$ lies in the [[centre|0S7Y9Q990Q]] of $A/B$ if and only if $x^{-1}g^{-1}xg \\in B$ for all $g \\in A$. Indeed, $xgB = gxB$ holds if and only if $(gx)^{-1}xg \\in B$.\n2. If $G = \\{e\\}$, then the series $N_0 = \\{e\\} = G$ of length $0$ is a central series. So $G$ is [[nilpotent|0DZPD2MG33]].\n3. Let $G \\neq \\{e\\}$ be a finite [[$p$-group|06DAH3VS6M]], and assume the claim for all $p$-groups of smaller order. Let $Z = Z(G)$.\n4. By [[this result|0RZCTT30H5]], $Z \\neq \\{e\\}$. The centre is a [[normal subgroup|0W77CVBCHQ]] of $G$, because $gzg^{-1} = z$ for $z \\in Z$.\n5. The [[quotient group|0YYWCNTPWR]] $\\bar G = G/Z$ has $|G|/|Z| \u003c |G|$ elements.\n6. For $g \\in G$ of [[order|0JPYSD5E0B]] $p^k$, we have $(gZ)^{p^k} = Z$. So the order of $gZ$ divides $p^k$ and is a power of $p$. Hence $\\bar G$ is a $p$-group.\n7. By the induction hypothesis, $\\bar G$ has a central series $\\{Z\\} = M_0 \\subseteq M_1 \\subseteq \\cdots \\subseteq M_k = \\bar G$.\n8. Let $\\pi \\colon G \\to \\bar G$ be the projection $\\pi(g) = gZ$. Put $N_0 = \\{e\\}$ and $N_{j+1} = \\pi^{-1}(M_j)$ for $0 \\leq j \\leq k$. Then $N_1 = Z$ and $N_{k+1} = G$.\n9. Each $N_{j+1}$ is a subgroup of $G$ containing $Z$. It is normal in $G$: for $x \\in N_{j+1}$ and $g \\in G$, $\\pi(gxg^{-1}) = \\pi(g)\\pi(x)\\pi(g)^{-1} \\in M_j$, since $M_j$ is normal in $\\bar G$.\n10. The sets $N_j$ increase with $j$, since the $M_j$ increase and $N_0 \\subseteq Z = N_1$.\n11. The first factor works: for $x \\in N_1 = Z$ and $g \\in G$, we get $x^{-1}g^{-1}xg = e \\in N_0$. By step 1, $N_1/N_0 \\subseteq Z(G/N_0)$.\n12. Let $1 \\leq j \\leq k$, let $x \\in N_{j+1}$ and $g \\in G$, and let $c = x^{-1}g^{-1}xg$. Then $\\pi(c) = \\pi(x)^{-1}\\pi(g)^{-1}\\pi(x)\\pi(g)$.\n13. Since $\\pi(x) \\in M_j$ and $M_j/M_{j-1} \\subseteq Z(\\bar G/M_{j-1})$, step 1 in $\\bar G$ gives $\\pi(c) \\in M_{j-1}$. So $c \\in \\pi^{-1}(M_{j-1}) = N_j$.\n14. By step 1, $N_{j+1}/N_j \\subseteq Z(G/N_j)$.\n15. By steps 8 to 14, $\\{e\\} = N_0 \\subseteq N_1 \\subseteq \\cdots \\subseteq N_{k+1} = G$ is a central series of $G$. So $G$ is nilpotent.\n\n## Idea\n\nA nontrivial finite $p$-group has a nontrivial centre $Z$. Dividing by $Z$ gives a smaller $p$-group, which is nilpotent by induction. Pulling its central series back to $G$ and putting the trivial group below $Z$ gives a central series of $G$.\n",
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