Let $F_n$ be a free group of rank $n$. Say that $w \in F_n$ is *non-reversible* if there does not exist any $f \in \text{Aut}(F_n)$ such that $f(w) = w^{-1}$.

**Original Question.** Intuitively, I expect that most elements of $F_n$ are non-reversible. However, I have trouble coming up with examples. Does anyone know any ways to produce them? To avoid just focusing on low-rank situations (where there might be tricks), I'd like to find examples that are not contained in any proper free factor of $F_n$.

A related question replaces $F_n$ with the fundamental group of a surface and $\text{Aut}(F_n)$ with the mapping class group. Here again it's not as easy as one would like to find examples, but in our paper here Margalit and I gave a geometric criterion that allowed us to find them. However, just like in the free group setting I expect that there are far more examples than our paper constructs, and that in some sense they should be "generic".

In this revision, I want to make a number of comments and pose some additional questions inspired in part by the comments on the original version.

Benjamin Steinberg pointed out that the answer to this math.stackexchange question claims to give examples for $n=2$. It is not surprising that this special case is more tractable than the case $n \geq 3$. Indeed, a classical theorem of Nielsen shows that $\text{Out}(F_2) \cong \text{GL}(2,\mathbb{Z})$, so it is often fairly easy to understand automorphisms of $F_2$ directly. Automorphisms of $F_n$ for $n \geq 3$ are much more complicated.

It is natural to try to promote examples for $n=2$ to higher $n$ via the standard inclusion $\iota\colon F_2 \hookrightarrow F_n$. As YCor pointed out, if an element $w \in F_2$ cannot be inverted by an endomorphism of $F_2$ (much stronger than being non-reversible; let me call this property

*non-endoreversible*), then $\iota(w)$ in $F_n$ is also non-endoreversible (and hence non-reversible). To see this, assume that for some $w \in F_2$ the endomorphism $\phi\colon F_n \rightarrow F_n$ takes $\iota(w)$ to $\iota(w)^{-1}$. Letting $r\colon F_n \rightarrow F_2$ be the retraction, we can then define $\phi'\colon F_2 \rightarrow F_2$ via the formula $\phi'(x)=r(\phi(\iota(x)))$. We then have $\phi'(w) = r(\phi(\iota(w))) = r(\iota(w)^{-1}) = \iota(w)^{-1}$.Carl-Fredrik Nyberg Brodda pointed out that there are algorithms to determine whether or not a word $w' \in F_n$ is the image of a word $w \in F_n$ under an endomorphism of $F_n$. This would give a finite procedure to verify that a specific element of $F_n$ is non-endoreversible. In particular, if you apply this for $F_2$ you can presumably get non-endoreversible elements of $F_n$ for all $n$.

However, note that by design this will not be able to answer the harder question from the original post of finding such elements that do not lie in any proper free factor of $F_n$.

Let me close by making a conjecture that makes precise the statement "Intuitively, I expect that most elements of $F_n$ are non-reversible" from the original version of the question. This actually does not hold for the non-endoreversible elements (see the comments for a proof), so let’s focus on just the non-reversible elements. Fix some $n \geq 2$, and let $G_k \subset F_n$ be all elements of word length at most $k$ and $N_k \subset F_n$ be the set of all non-reversible elements of length at most $k$. I then conjecture that $$\text{lim}_{k \mapsto \infty} \frac{|N_k|}{|G_k|} = 1.$$

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