Let $G=\langle X \mid R \rangle$ be a presentation where $X$ and $R$ are finite and $R$ contains words of length at most 3. Can we deduce that $G$ is either virtually abelian or virtually free?
Relators of length at most 3
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0Do you have any reason to believe that this is true? I can't see one, and a quick search turned up no references to this or similar results. In fact, I'm given to think that $X = \{a,b\}, R = \{ab\}$ might be a counterexample, as this definitely isn't virtually free and I can't think of a subgroup of finite index of the free group on two elements which contains $ab$. – 2011-12-25
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0Isn't your example an abelian group? – 2011-12-25
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0It would be even better to use $X = \{a,b,c\}$ I think. But I'm pretty sure group in my earlier comment is not abelian. It would be true if $R = \{ab,ba\}$, but in my example $ab = 1$ while $ba\neq 1$ as $ba$ is not in the subgroup generated by $ab$. The following computations show this, and are performed in $F_2$ rather than $
$ : $w(ab)^nw^{-1} = ba\implies [w,(ab)^n] = w(ab)^nw^{-1}(ab)^{-n} = ba(ab)^{-n}$ and this can only be true for $n=1$ as otherwise the image of $[w,(ab)^n]$ in the free abelian group would be nontrivial, but then $[w,ab] = 1$ so $wabw^{-1} = ab\neq ba$. – 2011-12-25 -
1Dear Alex, if in a group the product $ab$ of two elements $a$ and $b$ is equal to $1$, then the product $ba$ is also equal to $1$. (The important thing is, *is $ba$ in the **normal** subgroup generated by $ab$*?) – 2011-12-25
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0@MarianoSuárez-Alvarez Huh, I meant to say "normal" above but must have messed up somewhere in calculating the normal subgroup. – 2011-12-25
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1This has nothing to do, really, with that :D If $ab=1$, then $a=b^{-1}$, and obviously $a$ and $b$ commute! – 2011-12-25
2 Answers
If $r=x_1\cdots x_n$ is a relation of length $n>3$ in a presentation of a group, introducing a new generator $v$, and removing $r$ from the relations and adding in its place the two relations $x_1x_2u$ and $u^{-1}x_3\cdots x_n$ one gets a new presentation for the same group, which is closer to being «with relators of length at most $3$».
If we start with a finitely presented group, iterating this will give us a presentation in which all relations are of length at most $3$.
It follows that every finitely presented group has a presentation with relations of length at most $3$. (Of course, the same idea will also work for some infinite presentations...)
Any group has a presentation in which all relations have length 3. Take the group elements as generators, and all entries $ab=c$ of its multiplication table as relations. Proving this really is a presentation is often given as an early exercise in the theory of group presentations.
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0Is there an oh-duh way of seeing from this construction that if you start with a finite presentation you can produce a cubic finite presentation? – 2011-12-28
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0I think the method you described in your earlier post is the easiest way of doing this. – 2011-12-28