Freiheitssatz implies a finitely generated one relator group embeds in a two-generator one relator group?

Question

I've read that every finitely generated one relator group embeds in a two generator one relator group, and that this fact follows from the Freiheitssatz.

Unfortunately, the only proof I can find of this fact applies B.H. Neumann's proof for denumerable n-relator groups, and it doesn't seem to use the Freiheitssatz. Further, I haven't found any mention of this in Lyndon and Schupp, but it's possible I overlooked a more general theorem from which this follows.

My question is: does this fact truly follow from the Freiheitssatz? Is the proof trivial? I apologize if it is; unfortunately I am new to one relator groups.

Answer 1

Yes, this follows from the Freiheitssatz. Assume that the 1-relator group is defined by $G=\langle g_1,\ldots, g_n | R(g_1,\ldots,g_n)\rangle$, such that the relator $R(g_1,\ldots, g_n)$ is cyclically reduced, and involves the generators $g_1$ and $g_n$ non-trivially. By the Freiheitssatz, the subgroups $\langle g_1,\ldots, g_{n-1}\rangle$ and $\langle g_2,\ldots g_n\rangle$ are free groups of rank $n-1$ freely generated by these elements. Then embed $G$ in an HNN extension $\langle g_1,\ldots, g_n, t | R(g_1,\ldots,g_n), tg_it^{-1} = g_{i+1}, i=1,\ldots ,n-1 \rangle = \langle g_1,t | R(g_1, tg_1t^{-1},\ldots, t^{n-1}g_1t^{1-n}\rangle$ by eliminating generators and relators.

By permuting the labels, one may guarantee that the relator involves $g_1, g_n$ unless the relator involves only one generator. In that case, if the relator is of the form $g_1^k$, then do a Nielsen transformation $h_1=g_1g_n^{-1},h_2=g_2,\ldots, h_n=g_n$. The group with this set of generators has presentation $\langle h_1,\ldots, h_n | (h_1h_n)^k=1\rangle$. One may apply the previous construction to this presentation.

Answer 2

I do not see how it follows from the Freiheitssatz, but it is proved here: Sapir, Mark; Špakulová, Iva Almost all one-relator groups with at least three generators are residually finite. J. Eur. Math. Soc. (JEMS) 13 (2011), no. 2, 331–343. It follows, essentially, from a result of Olshanskii about existence of subgroups of the free group satisfying the so-called congruence extension property. The fact is not very difficult, the proof using van Kampen diagrams is about 1/2 page long.

The idea is this: Take the free group $F(a,b)$ and a 1-relator group $G=\langle x_1,...,x_n\mid R=1\rangle$. Take $n$ words in $F(a,b)$ satisfying the small cancelation condition $C'(1/12)$ $u_1,...,u_n$ (say, take $n$ random sufficiently long words in $a,b$). Consider the group $H=F(a,b)/\ll R(u_1,...,u_n)\gg$. Note that $H$ is a 2-generated group with 1 defining relation. There is a natural homomorphism $\phi: G\to H$ sending $x_i$ to $u_i$. That homomorphism is injective. In order to prove it, consider a van Kampen diagram $\Delta$ with boundary label $\phi(U)$ over the presentation of $H$. We need to show that $\Delta=\phi(\Delta')$ where $\Delta'$ is a diagram over the presentation of $G$. Consider maximal subdiagrams with holes which are obtained by applying $\phi$ to diagrams over the presentation of $G$. Note that in fact each of these diagrams can be assumed simply connected (we can assume that $\Delta$ is a minimal counterexample, so the holes are in fact filled with diagrams of the form $\phi(\Delta'')$). These subdiagrams form a graph: two subdiagrams are connected if they share an edge on the boundary. That graph is planar. By the classical result about planar graphs, there exists a vertex of degree at most $5$. Therefore either there are two subdiagrams with many common boundary edges, or there exists a subdiagram with many common boundary edges with the boundary of the whole diagram. The first case is impossible because of $C'(1/12)$ and the assumption of maximality of the subdiagrams. In the second case, we can cut off the subdiagram to obtain a smaller diagram with the same properties as $\Delta$. This is basically the whole proof.

Update. This argument is of course more complicated than Ian Agol's (although I do not use Freiheitssatz which is a non-trivial statement). But "my" argument works for any (even infinite) number of relations, and proves that every finitely generated (in fact every countable) group embeds into a 2-generated group having the same number of defining relations.