The definition of a group $G$ being polycyclic that I'm currently learning is: G has a normal series : $e = G_n \triangleleft G_{n-1} \triangleleft ... \triangleleft G_1 \triangleleft G_0 = G$ such that each factor $G_i / G_{i+1}$, $1 \le i \le n-1$ is cyclic.
I'm currently doing a problem that asks me to give an example of an abelian group $G$ which is not polycyclic.
I know that polycyclic groups are finitely generated so any abelian group that is not finitely generated, e.g. $\mathbb{R}$, works.
Although the intuition really goes fine, since the normal series is finite and every factor is cyclic thus can be generated by one element. But I don't really know how to prove the statement that polycyclic groups are finitely generated.
Any idea is appreciated.
The proof is by induction on the length of the normal series. The base case is that of a cyclic group (actually, the group of one element is even better), which is obviously finitely generated. The induction step is noting that $H$ is finitely generated and normal, and $G/H$ is cyclic, then the generating set of $H$ together with a generator of $G/H$ (meaning, an element $g\in G$ such that $gH$ generates $G/H$) is a generating set of $G.$ To answer the OP's comment: every element $x$ of $G$ is in some coset of $H,$ so $x = g^k h,$ for some $k \in \mathbb{Z}, h \in H.$