# give an example of finitely generated not abelian group which has subgroup of infinite index and not finitely generated.

we know that if $G$ is finitely generated group and $H$ is subgroup of it which has finite index then $H$ is finitely generated, and we know that if $G$ is abelian and finitely generated then every subgroup of it ,is finitely generated.

give an example of finitely generated group $G$ which is not abelian and has subgroup which its index in $G$ is not finite,and also this subgroup won't be finitely generated.

I just need an example of a group with this property,thanks a lot.

• the commutator group of a free group on two generators is not finitely generated, so it would suffice to find a group which has such a free group as a subgroup of transfinite index – David Holden Oct 30 '14 at 19:49
• @DavidHolden, why $\;F_2\;,\;F_2'\;$ is not already a good example? $\;F_2\;$ is f.e., $\;F_2'\;$ is not and $\;F_2/F_2'\cong \Bbb Z^2\;$ . – Timbuc Oct 30 '14 at 19:52
• mathoverflow.net/q/74902/41862 – zibadawa timmy Oct 30 '14 at 19:53
• For proofs that the commutation subgroup of a free group is not finitely generated: math.stackexchange.com/q/983480/10513 Alternatively, consider the subgroup $\langle a^iba^i\mid i\in\mathbb{Z}\rangle$, which is rather easily seen to not be finitely generated. – user1729 Oct 30 '14 at 22:07
As mentionned, free groups have many non-finitely-generated subgroups. Commutator subgroups are such an example (see here), but more generally it is known that, in a free group of finite rank, a normal subgroup is finitely-generated if and only if it is a finite-index subgroup (see here). For instance, we know that the normal closure $\langle \langle a \rangle \rangle$ is a non-finitely-generated subgroup of the free group $\mathbb{F}_2= \langle a,b \mid \ \rangle$.
In fact, the commutator subgroup of any free product $A \ast B$ turns out to be free of infinite rank whenever $A$ or $B$ is infinite (see here).
More generally, any countable group is embeddable into a 2-generator group (see here for a proof). Therefore, any countable non-finitely-generated group (such that $\mathbb{Q}$, the free group $\mathbb{F}_{\infty}$ of infinite rank or the group $S_{\infty}$ of bijections $\mathbb{N} \to \mathbb{N}$ with finite support) appears as an infinite-index subgroup of a finitely-generated group.
Another source of examples comes from wreath products. For instance, let $L_2$ denote the lamplighter group. Then $$L_2 = \left( \bigoplus\limits_{n \in \mathbb{Z}} \mathbb{Z}_2 \right) \rtimes \mathbb{Z},$$ so the subgroup $\bigoplus\limits_{n \in \mathbb{Z}} \mathbb{Z}_2$ is clearly not finitely-generated whereas $L_2$ turns out to be finitely-generated.