# Ascending chain condition for 1-element normal closures in a free group

Let $$F$$ be a free group of finite rank. Does $$F$$ satisfy the ascending chain condition on normal subgroups each of which is a normal closure of one element?

In other words, can there exist elements $$r_1,r_2,\dots\in F$$ such that $$\langle r_i^F \rangle \subsetneqq \langle r_{i+1}^F \rangle$$, for all $$i \in \mathbb{N}$$?

I'm sure the answer should be known, but I have not yet been able to find a reference for it.

• B.H. Neumann [An essay on free products of groups with amalgamations. Philos. Trans. Roy. Soc. London Ser. A 246 (1954), 503–554.] proved that in the free group of rank $$3$$, $$F_3$$, there is an infinite strictly ascending sequence of normal subgroups $$N_1 \subsetneqq N_2 \subsetneqq \dots$$, such that each $$N_i$$ is the normal closure of two elements in $$F_3$$. Moreover, there is an automorphism $$\alpha \in Aut(F_3)$$ such that $$N_{i+1}=\alpha(N_i)$$, $$i \in \mathbb N$$.
• A.M. Brunner [On a class of one-relator groups. Canadian J. Math. 32 (1980), no. 2, 414–420.] constructed a sequence of (pairwise non-isomorphic and Hopfian) one-relator groups $$G_0,G_1,\dots$$, such that for all $$i, $$G_j$$ is a proper quotient of $$G_{i}$$ by the normal closure of a single element. (Unfortunately this does not answer my question since the group $$G_0$$ is not free.)
• @YCor, I'm not sure "locally $1$-relator" is the right terminology. For any non-Hopfian $1$-relator group $G$ we get a sequence of non-injective group epimorphisms $F \to G_1 \to G_2 \to \dots$, where each $G_i \cong G$, and the limit is an infinitely presented group. But I think that the kernel of the resulting map $F \to G_i$ cannot be the normal closure of a single element, for every $i \in \mathbb{N}$. – Ashot Minasyan Feb 4 at 15:10
• @Ashot: note that there are chains of arbitrary finite length: $[x,y, ,y] , ..., x$, I have never seen an infinite chain.. – user6976 Feb 4 at 18:02
• @Mark and Autumn: here is an even easier example: the relation $[x,y]=1$, of length $4$, follows from the relation $xy^n=1$, of length $n+1$. So arbitrarily long relators can have consequences of length $4$... – Ashot Minasyan Feb 5 at 13:53