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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.

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    $\begingroup$ Maybe I do not understand the question, but consider words $ab^n$ where $n$ goes to infinity. Doesn't $b^n$ qualify as "big chunk"? $\endgroup$
    – Misha
    Commented Nov 21, 2012 at 13:34
  • $\begingroup$ By 'primitive' I suppose you mean 'not a proper power'? It's also used to mean 'an element of a basis' and 'maps to an element of a basis in the abelianization'. $\endgroup$
    – HJRW
    Commented Nov 21, 2012 at 14:27
  • $\begingroup$ (Misha's counter-example is primitive in any of these senses, of course.) $\endgroup$
    – HJRW
    Commented Nov 21, 2012 at 15:01
  • $\begingroup$ By the way, it seems to me plausible that something like your conjecture is true for elements which are shortest in their Aut(F)-orbit. Of course, this is trivial if by 'primitive' you mean 'a basis element'. $\endgroup$
    – HJRW
    Commented Nov 22, 2012 at 8:15

2 Answers 2

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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.

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To answer your last question, check out Cohen-Metzler-Zimmerman (1981) and references there in.

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