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Let $G = \langle X \mid R \rangle$ be a group defined by a finite presentation, and let $F$ be the free group on $X$. If $w \in F$ represents the identity in $G$, then $w$ is equal in $F$ to (the free reduction of) an expression of the form $u_1r_1u_1^{-1} \cdots u_kr_ku_k^{-1}$, where each $r_i \in R \cup R^{-1}$. Call this a factorization of $w$ of length $k$. Then, if $w$ has a factorization of length $k$, we can find one in which $|u_1| < |w| + m/2$, where $|w|$ denotes the word-length of $w$, and $m$ is the length of the longest relator in $R$.

It is not hard to prove that statement using a reduced van Kampen diagram for $w$. Such a diagram must clearly have a boundary edge that also borders an interior face labelled by some $r \in R \cup R^{-1}$ (since otherwise the diagram would be a tree and would not have a freely reduced boundary label). So, we can choose a word of length less than $|w|$ going around the boundary from the base-point to that edge, and then at most $m/2$ to get to the starting point of $r$, to give a word $u$ with $|u| < |w|+m/2$ and $w$ equal in $F$ to $uru^{-1}w'$, where $w'$ has a factorization of length at most $k-1$.

I have no problems with that proof, but I was trying to come up with a direct proof that doesn't use van Kampen diagrams, but just uses the combinatorics of the free group. It is really just a statement about $F$ so I would expect there to be a direct proof in $F$, but I could not construct one. Can anyone help with this or provide a reference?

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    $\begingroup$ I wonder why you are looking for a proof which does not use van Kampen diagrams? -- And how would you distinguish a proof like you want it from a proof which just avoids using van Kampen diagrams literally, but is really only a reformulation of the proof you know? $\endgroup$
    – Stefan Kohl ♦
    Oct 7, 2013 at 18:47
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    $\begingroup$ @Stefan: perhaps direct proof would yield an algorithm working faster than that unwrapping from van Kampen diagrams. A good example of this is dealing with $C(4)$ monoids by Mark Kambites. And, of course, answering a question with different language is nothing unnatural to ask. $\endgroup$
    – Victor
    Oct 7, 2013 at 19:29
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    $\begingroup$ @Stefan: The diagram based methods are using some (admittedly rather elementary) planar geometry, and I am interested in knowing whether the geometry involved is somehow essential for the proof. I believe that a reformulation would make that clearer. $\endgroup$
    – Derek Holt
    Oct 7, 2013 at 20:07

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