Primitive subwords in a free group of rank 2 I am not sure yet about what I exactly need to prove, but I guess I can formulate a rough statement similar to the following:
Suppose $w\in F_2$ is a primitive word whose length is big enough. Then for every chunk of length greater than some big constant, there exists a "long" subword, contained in that chunk, which is primitive.
I suspect that the statement should be true, maybe in some similar form, and I thought about proving it by using some canonical form for primitive elements. Since the rank of the free group is just 2, I guess there should be an easy way for finding some good pattern in the primitive words, or something like that. 
Do you know any nice way of writing down primitive elements in $F_2$ that might be good in this context? 
Thank you very much.
 A: As Misha pointed out, your conjecture is false.
However, I thought I'd point out that there is a fairly nice way of parameterizing primitive elements of $F_2$.  Let $\pi : F_2 \rightarrow \mathbb{Z}^2$ be the abelianization map.  It is then standard that if $x \in F_2$ is primitive, then $\pi(x)$ is primitive (i.e. nonzero and not divisible by any integer other than $\pm 1$).  Moreover, if $v \in \mathbb{Z}^2$ is primitive, then there is an element $x \in F_2$ such that $\pi(x) = v$, and $x$ is unique up to conjugation.
It is easy to enumerate the primitive elements of $\mathbb{Z}^2$, so this leads to the question of determining for a primitive $v \in \mathbb{Z}^2$ the (unique up to conjugacy) primitive element $x \in F_2$ with $\pi(x) = v$.  There is an elegant geometric solution to this in the following paper.
MR0608526 (82i:20042)
Osborne, R. P.; Zieschang, H.
Primitives in the free group on two generators.
Invent. Math. 63 (1981), no. 1, 17–24.
A: To answer your last question, check out Cohen-Metzler-Zimmerman (1981) and references there in.
