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To the best of my knowledge, whether 1-relator groups are coherent is still an open question. The group $F_2 \times F_2$ (where $F_2$ is the free group of rank $2$) is well-known to be non-coherent and can be presented using four relators. Are there known examples of non-coherent groups with two- or three-relator presentations?

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Interestingly the finitely presented HNN extension of the Grigorchuk group (which is not f.p) has 4 relators too. –  Mustafa Gokhan Benli Apr 15 '11 at 18:23

2 Answers 2

$F_2 \times F_2$ embeds in the two relator group $\langle t, x \ | \ [x, txt^{-1}] = 1 ,\ t^4xt^{-4} = x \rangle$.

Consider the presentation $\langle x_0, x_1, x_2, x_3 \ | \ [x_i, x_{i+1}] = 1, \ 0 \leq i \leq 3 \ \mathrm{mod} \ 4 \rangle$ of $F_2 \times F_2$. So the factors are generated by $\langle x_0, x_2 \rangle$ and $\langle x_1, x_3 \rangle$.

Form the HNN-extension $G$ given by the presentation

$\langle t, x_0, x_1, x_2, x_3 \ | \ [x_i, x_{i+1}] = 1 ,\ tx_it^{-1} = x_{i+1} , \ 0 \leq i \leq 3 \ \mathrm{mod} \ 4 \rangle$

But now, we may use the relations $tx_i t^{-1} = x_{i+1}$ to derive all the commutations from $[x_0,x_1] = 1$. This leaves us with

$ G = \langle t, x_0, x_1, x_2, x_3 \ | \ [x_0, x_1] = 1 ,\ tx_it^{-1} = x_{i+1} , \ 0 \leq i \leq 3 \ \mathrm{mod}\ 4 \rangle$.

But now we may eliminate the generators $x_1, x_2, x_3$ leaving us with

$G = \langle t, x_0 \ | \ [x_0, tx_0t^{-1}] = 1 ,\ t^4x_0t^{-4} = x_0 \rangle$.

Since $G$ contains $F_2 \times F_2$, it is not coherent.

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You can just use t^4=1 instead of the second relation (you're extending F_2xF_2 by an automorphism of order 4). –  Ian Agol Apr 17 '11 at 7:32
    
True. I wonder if that's the shortest presentation of an incoherent group? –  Richard Kent Apr 17 '11 at 15:14

Consider the group $G=\langle a, s, t\mid [a^t,a]=1, [s,t]=1, a^s=aa^t\rangle$. That group contains the wreath product $\langle a\rangle \wr \langle t\rangle$. Indeed, it is enough to show that $a^{t^n}$ commutes with $a$, $n\ge 1$. It is so for $n=1$. Suppose it is true for $\le n$, that is $[a^{t^n},a]=1$. Conjugate by $s$. We get, since $st=ts$, $[a^{st^n},a^s]=1$. Use the third relation twice: $[a^{t^n}a^{t^{n+1}},aa^t]=1$. Then using the induction hypothesis three times, we get $[a^{t^{n+1}},a]=1$, as desired. So this group has 3 relators and is not coherent. I do not know now an example with two relators but it should exist.

A little more details. The reason it is enough to prove commutativity is the following. Consider the ring of formal power series with integer coefficients $R=\mathbb{Z}[[t,t^{-1}]]$. It has a multiplicative subgroup $H$ generated by $t$ and $s=t+1$. Consider the action of that group on the additive group of $R$ by multiplication. Let $a=1\in R$. Then the group $G$ maps onto $\langle a,t,s\rangle$ under the natural map $f: a\to a, t\to t, s\to s$. The image certainly contains the wreath product $\langle a\rangle \wr \langle t\rangle$. The fact that the homomorphism is injective on $\langle a, t\rangle$ follows from the commutativity proved above.

This example can be found in Gilbert Baumslag. A finitely presented metabelian group with a free abelian derived group of infinite rank. Proc. Amer. Math. Soc., 35:61–62, 1972 . A similar construction was used by Remeslennikov a little earlier.

Update Although for 1-related group coherence in general is not known, almost all 1-related groups with at least 3 generators are coherent, see 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 .

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