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Define the lower central series by ${\Gamma }_1(G)=G$ and ${\Gamma }_{j+1}(G)=[{\Gamma }_j(G),G]$. Its terms are characteristic normal subgroups and form a descending chain. Define the upper central series by $Z_0(G)=1$ andEquivalently, $Z_{i+1}(G)=\{x\in G:[x,G]\subseteq Z_i(G)\}$. These are characteristic normal subgroups in an ascending chain, and successive factors are central in the appropriate quotient.

For any length-$r$ central series $1=G_0\le \cdots \le G_r=G$, prove the two comparisons independently. Starting at ${\Gamma }_1(G)=G_r$, if ${\Gamma }_{r-i+1}(G)\le G_i$, thenDescending induction gives the lower containment. Starting at $G_0=Z_0(G)=1$, if $G_{i-1}\le Z_{i-1}(G)$, then $[G_i,G]\le G_{i-1}\le Z_{i-1}(G)$, which by definition gives $G_i\le Z_i(G)$. Thus the central-series comparison theorem yieldsIn particular, a nilpotent group satisfies ${\Gamma }_{r+1}(G)=1$ and $Z_r(G)=G$.

Conversely, if ${\Gamma }_n(G)=1$, reverse the finite lower chain to obtain $1={\Gamma }_n\le {\Gamma }_{n-1}\le \cdots \le {\Gamma }_1=G$. The defining commutator recurrence makes it a central series. If $Z_n(G)=G$, the finite upper chain is already a central series. This provesFinally, for an integer $c>0$, termination at ${\Gamma }_{c+1}=1$ gives a central series of length $c$ and hence $Z_c=G$ by the comparison. Conversely, $Z_c=G$ gives a central series of length $c$ and hence ${\Gamma }_{c+1}=1$. ThereforeRepeated terms cause no problem; the least such $c$ for a nontrivial group is its nilpotency class.