Past exam of the algebra course of the University of Oxford / 2009 / B / 6

The commutator subgroup is $G^{\prime }=⟨xy{x}^{-1}{y}^{-1}:x,y\in G⟩$. It is characteristic and hence normal, because every automorphism sends a commutator to a commutator. If $G^{\prime }\le H$, then $gh{g}^{-1}{h}^{-1}\in H$ for every $g\in G,h\in H$, so $H$ is normal. Every commutator becomes trivial in $G/H$, making that quotient Abelian. Conversely, the quotient homomorphism to an Abelian group kills each commutator, and hence kills $G^{\prime }$. ThereforeSet $G^{(0)}=G$, $G^{(i+1)}=(G^{(i)}{)}^{\prime }$. The group is soluble exactly when this derived series reaches the identity after finitely many steps.

If a nontrivial simple group is soluble, its normal subgroup $G^{\prime }$ is either $1$ or $G$. The latter would make the derived series constant and nontrivial, so $G^{\prime }=1$ and $G$ is Abelian. Every nonidentity element then generates $G$, since its cyclic subgroup is normal. An infinite cyclic group is not simple, and a finite cyclic group is simple exactly when its order is prime. This proves soluble simple group has prime order.

An Abelian group has the central series $1<G$, since $Z(G)=G$. In any central series, each factor is a subgroup of a center and hence Abelian. A finite normal series with Abelian factors makes the group soluble: inductively, if $N$ is soluble and $G/N$ is Abelian, then $G^{\prime }\subseteq N$, and successive derived groups eventually vanish. Hence every nilpotent group is soluble.

The group $S_3$ has the normal series $1<A_3<S_3$ with cyclic factors of orders three and two, so it is soluble. Its center is trivial, since a permutation commuting with both $(12)$ and $(23)$ must fix all three letters. Every nontrivial nilpotent group has nontrivial center: its first nonidentity central-series term is contained in $Z(G)$. Consequently $S_3$ is soluble but not nilpotent.

For a finite p-group, use induction on its order. The identity group is the base case. A nontrivial $G$ has nontrivial $Z=Z(G)$, so $G/Z$ has smaller order, still a power of $p$. By induction it has a central series. Lift that series to subgroups of $G$ and prepend $1<Z$. The first factor is central by definition of $Z$, and centrality of every lifted factor is exactly centrality of the corresponding quotient factor. This gives a central series for $G$. This proves finite p-groups are nilpotent: