# Polycyclic groups are finitely generated

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.

• I am not sure what the question is, since being finitely generated is (part of) one of the definitions of being polycyclic. Dec 6 '13 at 0:13
• @IgorRivin Thanks for point that out, I've edited the question. Dec 6 '13 at 0:21

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 $$k \in \mathbb{Z}, h \in H.$$

• Thank you. I am thinking about the last sentence of your proof. Is it because for any $k \in G$, there always exits a $n$ such that $kH = g^n H$. Therefore $k \in g^n H$? Dec 6 '13 at 0:44
• @user112564 basically, see the added sentence. Dec 6 '13 at 0:58
• Thank you so much for your help! :) Dec 6 '13 at 1:23