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.)

  • $\begingroup$ Briefly, if $A$ has ACC on normal subgroups, show that $A\ast \mathbb{Z}$ has also ACC on normal subgroups or give a counterexample. $\endgroup$
    – Sh.M1972
    Nov 28, 2012 at 18:35
  • 2
    $\begingroup$ 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. $\endgroup$ Nov 28, 2012 at 21:31

1 Answer 1


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.


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.

  • 3
    $\begingroup$ In fact, Greenberg proved that every normal subgroup of $F_2$ is trivial, of finite index or infinitely generated. $\endgroup$
    – HJRW
    Nov 28, 2012 at 20:53
  • $\begingroup$ 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. $\endgroup$
    – Sebastian
    Nov 28, 2012 at 21:04
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    $\begingroup$ 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). $\endgroup$
    – S. Carnahan
    Dec 4, 2012 at 14:21

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