Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / B (Group Theory)

A composition series is a finite chain $1=G_0⊲G_1⊲\cdots ⊲G_r=G$ in which each inclusion is proper, each subgroup is normal in the next, and every quotient $G_i/G_{i-1}$ is a nontrivial simple group. These quotients are the composition factors.

For a finite group, use induction on its order. The trivial group has the empty series. Otherwise choose a maximal proper normal subgroup $N$; such a subgroup exists because there are finitely many subgroups and $1$ is proper. The correspondence theorem says normal subgroups of $G/N$ correspond to normal subgroups of $G$ containing $N$, so maximality makes $G/N$ simple. Inductively construct a composition series of the smaller group $N$ and append $G$. This proves existence.

A nonabelian example with abelian factors isBoth factors are simple because their orders are prime.

The cyclic group of order six has two different composition series:Their factor lists are $C_2,C_3$ and $C_3,C_2$ respectively; normality follows from commutativity.

The Jordan–Hölder theorem states that any two composition series of a group have the same length and the same multiset of composition factors up to isomorphism. Equivalently, the factors of one series can be reordered to match those of the other by group isomorphisms. The actual intermediate subgroups need not be the same.

Let $C_n=⟨g⟩$ and take $H=⟨g^p⟩$. Since $n=mp$, the order of $g^p$ is $m$, so $H\cong C_m$. All subgroups of a cyclic group are normal. The quotient is generated by $gH$ and has order $n/m=p$, givingA group of prime order has no nontrivial proper subgroup. By the subgroup correspondence theorem, no normal subgroup lies strictly between $H$ and $C_n$. Thus $H$ is a maximal proper normal subgroup.

Given a prime factorization $n=p_1\cdots p_r$, repeatedly apply part (c) to the cyclic group of order $n$: it has a chain of subgroups of successive ordersRead in increasing order, this is a composition series with factors isomorphic to the cyclic groups of orders $p_1,\dots ,p_r$. Any second prime factorization of the same integer produces another composition series of the same group $C_n$.

The Jordan–Hölder theorem equates the two multisets of simple factors. Since $C_p\cong C_q$ for primes $p,q$ implies $p=q$ by equality of group orders, the two prime multisets agree. HenceFor $n=1$ both factor lists are empty, corresponding to the trivial group.