# Normal Subgroups of Free Products

Let $$G=A\ast \mathbb{Z}$$ be the free product of a group $$A$$ and the cyclic group $$\mathbb{Z}$$ and suppose $$K$$ is a subgroup of $$G$$. By Kurosh Subgroup Theorem we know that $$K=F\ast (\ast_{i\in I}(K\cap A^{u_i}))$$, where $$F$$ is free group and $$u_i$$ are some representatives of double cosets $$KxA$$ in $$G$$.

Now suppose further that $$A$$ has ACC on normal subgroups and $$K$$ is normal.

Is it true that $$K$$ is finitely generated?

(This will be true if we can show that $$|I|$$ and $$\operatorname{rank}\ F$$ are finite.)

• Briefly, if $A$ has ACC on normal subgroups, show that $A\ast \mathbb{Z}$ has also ACC on normal subgroups or give a counterexample. Nov 28 '12 at 18:35
• In fact a non-trivial free product $G=A*B$ (with $|A|\ge 3$ and $|B|\ge 2$) never has ACC on normal subgroups. This is because $G$ is a non-elementary relatively hyperbolic group, and there is a versions of small cancellation theory over such groups, which, in particular, implies that every non-elementary rel. hyperbolic group possesses a proper non-elementary rel. hyperbolic quotient. Nov 28 '12 at 21:31

Set $A$ equal to $\mathbb{Z}$, which satisfies the ascending chain condition ("ACC", every strictly ascending chain of (normal) subgroups eventually terminates). Then $G=\mathbb{Z}\ast\mathbb{Z}=F_2$ and $F_2$ contains normal subgroups that are not finitely generated.

Examples:

1) The commutator subgroup is normal and not finitely generated.

2) The subgroup generated by $\left\{b^k a b^{-k}\ |\ k\in\mathbb{Z}\right\}$ is normal and not finitely generated.

• In fact, Greenberg proved that every normal subgroup of $F_2$ is trivial, of finite index or infinitely generated.
– HJRW
Nov 28 '12 at 20:53
• This is true for all finitely generated free groups (Hatcher, p.87, problem 7): If $N\leq F_n$ is a nontrivial normal subgroup of infinite index then $N$ is not finitely generated. This is an easy exercise that involves covering theory. Nov 28 '12 at 21:04
• In a now-deleted answer, the OP says: Thank you for answers. I was trying to prove some kind of Hilbert basis theorem for "Algebraic Geometry over Groups". Now, it is clear that there is no such a generalization: It is not true to say that if a group $G$ has ACC on normal subgroups, then $G[X] = G \ast F(X)$ is so. Therefore there may exist $G$-groups which are not Equationally Noetherian. For Algebraic Geometry over Groups, see J. Alg. (Baumslag, Miasnikov, Remeslennikov). Dec 4 '12 at 14:21