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

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: