Past exam of the algebra course of the University of Oxford / 2010 / B / 5 / a

First establish the two general claims about polycyclic groups. Intersect a subnormal cyclic-factor series $(G_i)$ with $H\le G$. Then $H\cap G_i⊴H\cap G_{i+1}$, and the inclusion induces an injective homomorphismA subgroup of a cyclic group is cyclic, so this is a cyclic-factor subnormal series for $H$, after omitting repetitions. This proves subgroups of polycyclic groups are polycyclic.

To prove the generator bound, choose $g_i\in G_i$ whose coset generates $G_i/G_{i-1}$ whenever that factor is nontrivial. Inductively $G_i=⟨G_{i-1},g_i⟩$, so $G=⟨g_1,\dots ,g_n⟩$. This is the generator bound from a cyclic-factor series: at most $n$ generators suffice.

For the first example, take $G=C_2^m$ with its coordinate flagIt has length exactly $m$ and each factor is $C_2$, so all factors are cyclic.

TakeStart with $1<A_3<S_3$, whose factors are $C_3$ and $C_2$, and then append the $m-2$ coordinate factors of the elementary Abelian group $C_2^{m-2}$. More explicitly, append $S_3\times C_2^j$ for $1\le j\le m-2$. Each subgroup is normal in the next, and the factors consist of one $C_3$ and $m-1$ copies of $C_2$. Thus there are exactly $m$ finite cyclic factors. The group is non-Abelian because it contains $S_3$.

Take the free Abelian groupIts coordinate flag $0<\mathbb{Z}<{\mathbb{Z}}^2<\cdots <{\mathbb{Z}}^m$ is a subnormal series of length $m$, with every factor isomorphic to the infinite cyclic group $\mathbb{Z}$. Hence it has exactly $m$ infinite cyclic factors.