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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.

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2 Answers 2

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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.

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@Ian: Yes, I missed it. Unfortunately this simple way cannot be used in our paper with Spakulova because your construction increases the Magnus level of the group (how many Magnus rewriting is needed to get a free group). –  Mark Sapir Apr 19 '11 at 12:15
    
I think I need a slight clarification, because I was promised this works for every one relator group, regardless of the relator. I know the Freiheitssatz extends to the case when the relator does not necessarily involve all generators, but the chosen subset still excludes a generator involved in $r$. However I'm not so sure this proof extends in a similar way. Take the case where $r = g_1^k$ involves a single generator, then it seems to me that any HNN extension (at least, the construction used here) would necessarily have at least three generators. Is there something I'm missing here? –  JeremyKun Apr 19 '11 at 18:49
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The point of the second paragraph is that one can apply the construction after doing a change of basis (I've added in a bit more explanation). There is no canonical choice of generating sets for a 1-relator group, but I think it's known that any two generating sets are Nielsen equivalent, so it's natural to allow Nielsen moves. –  Ian Agol Apr 19 '11 at 19:46

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.

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Mark, I am lost at the point "... there exists a vertex of degree at most 5. Therefore either there are two subdiagrams ...". How do you conclude from an upper bound on the degree of a vertex a lower bound on the number of edges between two vertices? –  Andreas Thom Apr 19 '11 at 7:56
    
Andreas: There is only one edge (at most) between two vertices by definition of the graph (otherwise the classical result about planar graphs cannot be aplied). But every edge on the boundary of the subdiagram is shared by two subdiagrams or by the subdiagram and the boundary of $\Delta$. Each of these edges "realize" an edge of the auxiliary graph. –  Mark Sapir Apr 19 '11 at 12:13
    
This is a very interesting proof! Unfortunately I am somewhat bound to use the Freiheitssatz in my exposition, but these stronger results are always interesting to note. Thank you! –  JeremyKun Apr 19 '11 at 22:23

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