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?
