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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?

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Do 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 $ab$. –  Alex Becker Dec 25 '11 at 6:10
    
Isn't your example an abelian group? –  Mustafa Gokhan Benli Dec 25 '11 at 6:14
    
It 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 $<X,R>$: $w(ab)^nw^{-1} = ba\implies [w,(ab)^n] = w(ab)^nw^{-1}(ab)^{-n} = ba(ab)^{-n}$ 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 $wabw^{-1} = ab\neq ba$. –  Alex Becker Dec 25 '11 at 6:30
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Dear 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 $ab$?) –  Mariano Suárez-Alvarez Dec 25 '11 at 6:32
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This has nothing to do, really, with that :D If $ab=1$, then $a=b^{-1}$, and obviously $a$ and $b$ commute! –  Mariano Suárez-Alvarez Dec 25 '11 at 6:39
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2 Answers

up vote 4 down vote accepted

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...)

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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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Is 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? –  Mariano Suárez-Alvarez Dec 28 '11 at 4:44
    
I think the method you described in your earlier post is the easiest way of doing this. –  Derek Holt Dec 28 '11 at 12:10
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