Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 7 / 1

A central series is a finite chain of normal subgroupswith $G_i/G_{i-1}\le Z(G/G_{i-1})$ for every $i\ge 1$. Equivalently, writing $[x,y]=x^{-1}{y}^{-1}xy$ and $[A,B]$ for the subgroup generated by these group commutators, the condition is $[G_i,G]\le G_{i-1}$. Centrality is relative to the whole quotient group, not just to the next subgroup in the chain.

A group is nilpotent if it admits a finite central series. The trivial group admits a series of length zero. For a nontrivial nilpotent group, the minimum possible length is its nilpotency class.

Use induction on the order of the finite p-group. The trivial group is nilpotent. For $G\ne 1$, the permitted center theorem gives $Z=Z(G)\ne 1$, so the quotient group $G/Z$ is a smaller p-group. By induction it has a central seriesLet $H_i$ be the inverse image of ${\overline{H}}_i$ under the quotient map, so $H_0=Z$ and $H_s=G$. Centrality in the quotient gives $[H_i,G]\le H_{i-1}$ for $i\ge 1$: the image of every such group commutator lies in ${\overline{H}}_{i-1}$. Also $[Z,G]=1$. Thereforeis a central series of $G$. This proves every finite p-group is nilpotent, including the case $Z=G$, when the quotient is trivial and the series is simply $1\le G$.

Assume $[H,G]\le K$. For $k\in K\le H$ and $g\in G$,Thus $g^{-1}Kg\le K$ for every $g$; applying this also to $g^{-1}$ gives equality. Hence $K$ is a normal subgroup of $G$, even though initially it was assumed normal only in $H$. Now the quotient $G/K$ is defined, and for $h\in H$, $g\in G$,Every element of $H/K$ consequently lies in the center of a group $Z(G/K)$.

Conversely, if $K⊴G$ and $H/K\le Z(G/K)$, each displayed quotient group commutator is trivial, so $[h,g]\in K$. Taking the subgroup generated by these elements gives $[H,G]\le K$. ThereforeThe forward argument also implies $H⊴G$, since $g^{-1}hg=h[h,g]\in H$.

Let $1=A_0\le \cdots \le A_r=A$ and $1=B_0\le \cdots \le B_s=B$ be central series. Pad the shorter chain by repeating its terminal group, so both have length $t=\max (r,s)$. In the direct product of groups, group commutators are coordinatewise, henceThe equality holds because commutators supported in either coordinate generate the corresponding factor independently. Thereforeis a central series, proving that a direct product of two nilpotent groups is nilpotent. More precisely, the coordinatewise recurrence for the lower central series gives ${\Gamma }_j(A\times B)={\Gamma }_j(A)\times {\Gamma }_j(B)$, so the nilpotency class of a direct product is the maximum of the two factor classes.

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.

Choose a representative Sylow subgroup with generatorsOn the first four points, $r^4=s^2=1$ and $srs=r^{-1}$, so $⟨r,s⟩$ is the dihedral group of order eight. The disjoint transposition $t$ commutes with it and generates an independent factor of order two. Thus $G=⟨r,s,t⟩\cong D_8\times C_2$ has order 16. Since the power of two dividing $7!$ is $2^{\lfloor 7/2\rfloor +\lfloor 7/4\rfloor }=2^4$, this is a Sylow 2-subgroup of $S_7$. Any other choice is conjugate to it and has conjugate series.

The dihedral relation gives $[r,s]=r^2$, while $t$ commutes with both generators. Modulo $⟨r^2⟩$ all three generators commute, so the commutator subgroup is exactly $⟨r^2⟩$, and $r^2$ is central. Hence the lower central series isWithin the dihedral factor only $1,r^2$ commute with both $r$ and $s$. The center of $G$ is therefore $⟨r^2,t⟩\cong C_2\times C_2$. Its quotient has order four and is abelian, so the upper central series isBoth series show that the nilpotency class is two.