Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 4 / 1

A nilpotent group has a finite central seriesEquivalently the upper central series, defined by $Z_{j+1}/Z_j=Z(G/Z_j)$, reaches $G$.

We first prove the normalizer condition for nilpotent groups. If $H<G$, choose the least $j$ with $Z_j⊈H$. Then $Z_{j-1}\le H$, and an element $z\in Z_j\setminus H$ satisfies $[z,H]\le Z_{j-1}\le H$. Thus $z$ normalizes $H$, giving $H<N_G(H)$.

Let $P$ be a Sylow subgroup of a finite nilpotent $G$, and write $N=N_G(P)$. If $N<G$, the normalizer condition gives some $g\in N_G(N)\setminus N$. The subgroups $P$ and $P^g$ are both Sylow subgroups of $N$, so Sylow theorem supplies $n\in N$ with $P^{gn}=P$. But then $gn\in N_G(P)=N$, forcing $g\in N$, a contradiction. Therefore every Sylow subgroup is a normal subgroup of $G$.

For normal Sylow subgroups $P,Q$ at different primes, $[P,Q]\le P\cap Q=1$. They commute, have coprime orders, and their product has order $|G|$. Multiplication consequently gives an isomorphism from their external direct product of groups onto $G$.

Conversely, a nontrivial finite p-group has nontrivial center of a group: its class equation has every noncentral conjugacy-class size divisible by $p$, so $|Z(G)|$ is a positive multiple of $p$. Induction on order, applied to $G/Z(G)$, gives a central series for $G$. Finite direct products of these central series give a central series for the product, padding shorter series by the full factor. Thus the finite nilpotent group decomposition is

For a finite group, define the Frattini subgroup bythe intersection is $G$ when there are no maximal subgroups, which for a finite group happens only for $G=1$. A second definition is the set of non-generators: an element $x$ such that $⟨S,x⟩=G$ always implies $⟨S⟩=G$.

If $x\in \Phi (G)$ and $H=⟨S⟩<G$, put $H$ in a maximal subgroup $M$. Then $x\in M$, so $⟨S,x⟩\le M<G$. Conversely, if $x\notin \Phi (G)$, some maximal subgroup $M$ omits it; maximality gives $⟨M,x⟩=G$, although $M\ne G$. This proves equivalence.

Every group automorphism permutes maximal subgroups, so $\Phi (G)$ is a characteristic subgroup, in particular a normal subgroup. Alsoa proper $H$ would lie in a maximal subgroup containing both factors.

Set $F=\Phi (G)$ and take a Sylow subgroup $P$ of $F$. For $g\in G$, normality of $F$ makes $P^g$ a Sylow subgroup of $F$. Conjugating inside $F$ therefore gives the Frattini argument factorization $G=F{N}_G(P)$. The preceding generating property forces $N_G(P)=G$. Thus every Sylow subgroup of $F$ is normal, and part (i) shows that $F$ is a nilpotent group.

Now suppose $G$ is a finite p-group. The normalizer condition makes every maximal subgroup $M$ normal: $M<N_G(M)$ and maximality force $N_G(M)=G$. The quotient $G/M$ has no proper nontrivial subgroup, hence is cyclic of prime order; its p-power order makes that prime $p$. Consequently every such $M$ containswhere the two factors are the commutator subgroup and the subgroup generated by pth powers. Thus $D\le \Phi (G)$.

The quotient $G/D$ is an elementary abelian group, hence a vector space over ${\mathbb{F}}_p$. If $x\notin D$, extend the nonzero vector $xD$ to a basis and choose a linear functional nonzero on it. Its kernel pulls back to a subgroup of index $p$, a maximal subgroup omitting $x$. Therefore $x\notin \Phi (G)$, provingThe two factors in the last formula are multiplied; the superscripted expression in the TeX aid is not the formula printed in the PDF.

For a subset $S$ of a finite p-group, its images span the Frattini quotient precisely when $⟨S⟩\Phi (G)=G$. By the generating property proved in part (ii), this is equivalent to $⟨S⟩=G$.

If $S$ is a minimal generating set of a group, its images must be an irredundant spanning set in the vector space $G/\Phi (G)$. A linear dependence would express one image using the others; those other images would still span, so the other elements would still generate $G$, a contradiction. Thus the images form a basis. Conversely lifts of a basis generate $G$, and deleting a lift destroys spanning and hence generation.

This proves the Burnside basis theorem, including equality of sizes for every inclusion-minimal generating set of a group:Inclusion-minimal generating sets of a group need not have equal sizes in arbitrary finite groups; the elementary-abelian Frattini quotient is the essential p-group feature.

Use rightmost-first composition of permutations, and letFor $n\ge 2$, direct evaluation gives $ab=(1n)$. Conjugating this transposition by powers of $a$ yields the transpositions between consecutive letters around the cycle, including $(12),(23),\dots ,(n-1n)$.

Put $s_i=(ii+1)$. For $i<j$,Thus the adjacent transpositions generate every transposition, and permutation cycle decomposition shows that transpositions generate the symmetric group. Therefore $⟨a,b⟩=S_n$.

For the final lower bound, associate a graph to a family of transpositions: its vertices are the letters and its edges are the swapped pairs. Every generator preserves every connected component, so generation of $S_n$ requires that graph to be connected. A connected graph on $n$ vertices needs at least $n-1$ edges: adding an edge can reduce the number of components by at most one, starting from $n$ isolated vertices. The adjacent transpositions supply exactly $n-1$ edges. Equivalently, transpositions on a connected graph generate the symmetric group. Hence