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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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    This has nothing to do, really, with that :D If $ab=1$, then $a=b^{-1}$, and obviously $a$ and $b$ commute!2011-12-25

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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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    I think the method you described in your earlier post is the easiest way of doing this.2011-12-28