Let $G=\langle S\rangle$ be a finitely generated free group with at least two generators. Let $g_0\in G$ has the normal form $g_0= s_rs_{r-1}\ldots s_1$.
In a paper, author claimed that for every $t\in G$, there is $v\in G$, where $v$ is the element of minimal length, such that $t=vw$ for the some $w=tv^{-1}$ of the form $w=s_js_{j-1}\ldots s_1(s_rs_{r-1}\ldots s_1)^k$ or $w=s_{r-j+1}^{-1}\ldots s_r^{-1}(s_rs_{r-1}\ldots s_1)^{-k}$ for some $k\geq 0$ and $1\leq j\leq r$.
Please help me to know it.
Is it true that for every $t\in G$, there is a unique $v\in G$ with minimal length, such that $t=vw$?
Let for $t\in G$, element $v\in G$ has a minimal length such that $t=vw$. What can say about $st\in G$ for $s\in S$. We know that there exist $v'\in G$ and $w'\in G$ with $st=v'w'$. Is it true that the $w'=w$ and $v'=sv$?
I'll make a guess at how to correct an error in the original paper:
Given the normal form $t = t_k t_{k-1} ... t_1$, to find $v$ you simply search one-at-a-time through all factorizations of the normal form of $t$ into two words $$t = \underbrace{(t_k t_{k-1} ... t_j)}_{v_j} \underbrace{(t_{j-1} ... t_1)}_{w_j} $$ Starting with $j=k$ and working downward, examine each $w_j$ to see if it has one of the two required forms. Stop when you've found the first such $j$. The word you seek is $v=v_j$. It's possible you never find any such $j$, in which case $t=vw$ where $w$ is the empty word.