# Every free group admits a fixed-point-free involution automorphism

This comes from an exercise in Rotman's "An Introduction to the Theory of Groups":

11.6) Show that a free group $$F$$ of rank $$\geq 2$$ has an automorphism $$\phi$$ with $$\phi(\phi(w)) = w$$ for all $$w \in F$$ and with no fixed points ($$\phi(w) = w \implies w = 1$$).

I started out by working out some finitely generated cases, to make it a bit easier at first, and maybe get a general idea of $$\phi$$ for an arbitrary free group. The case where $$F$$ has rank $$2$$ is trivial: we can swap the generators. In fact, for any even rank, the same holds. This made me feel like there was some way to pick a permutation of the generators.

Enter rank $$3$$. The only permutations in $$S_3$$ that have order $$2$$ are transpositions - but these obviously have a fixed point, and thus don't work. So this is where I got stuck...

I also thought about how every free group (I think - I don't recall if it's true about those of uncountable rank) is a subgroup of the free group on $$2$$ generators, so maybe the switch of the two would induce an automorphism on its free subgroups, but I couldn't find anyway to make it work...

Could anyone please provide a hint as per how to procede, at least in the rank 3 case? And does this result still hold for any (even uncountable) rank? Rotman's "$$\geq 2$$" didn't quite make it clear if he only meant finite rank for me.

• Unfortunately, the function $f : F \to F$ which sends every element to its inverse is not an automorphism: $f(x) = x^{-1}$, $f(y)=y^{-1}$; and $f(xy)=(xy)^{-1}=y^{-1}x^{-1}$ which is not equal to $f(x)f(y)=x^{-1}y^{-1}$ if $x$ and $y$ do not commute (and a free group has plenty of noncommuting pairs of elements). @RocketMan Commented Jul 5, 2021 at 1:23
• @LeeMosher I can't see the comment you're responding to, but why not just send the generators to their inverses? Commented Jul 5, 2021 at 5:36
• @RaviFernando: This should have been the answer to the question in the OP. Commented Jul 5, 2021 at 12:29
• @RaviFernando For some reason, I originally thought this wouldn’t work, but once I read your comment, I tried it out and it does! I guess there are multiple answers, then Commented Jul 5, 2021 at 13:48
• And yes, there are multiple answers. An interesting question to ponder is: In the automorphism group $\text{Aut}(F_n)$, how many different conjugacy classes of order 2 elements are there? Commented Jul 5, 2021 at 15:10

Promoting my comment to an answer: let $$\phi$$ be the automorphism sending each generator of $$F$$ to its inverse. This clearly has order 2 (for free groups of arbitrary rank $$\geq 1$$), and since it sends each reduced word $$x_{i_1}^{e_1} \cdots x_{i_n}^{e_n}$$ to the reduced word $$x_{i_1}^{-e_1} \cdots x_{i_n}^{-e_n}$$, its only fixed point is the identity.
Here's an order 2 automorphism $$f : F \to F$$ of the rank 3 free group $$F=\langle x, y, z \rangle$$ with no fixed points (except the identity): \begin{align*} f(x) &= z y^{-1} \\ f(y) &= y^{-1} \\ f(z) &= x y^{-1} \end{align*} The way that I got this is by using topology. The rank $$3$$ free group is the fundamental group of any connected, finite graph of Euler characteristic $$-2$$. If you can cook up a connected, finite graph $$G$$ with $$\chi(G)=-2$$, and an order $$2$$ self-homeomorphism $$h : G \to G$$ which has a fixed vertex $$v$$ and no fixed loops, then when you write down an isomorphism $$\langle a,b,c \rangle \approx \pi_1(G,v)$$ you should be able to also write down the formula for the induced fundamental group isomorphism $$h_* : \pi_1(G,v) \to \pi_1(G,v)$$ which will be of order 2 (because $$h$$ has order 2) and has no fixed elements (because $$h$$ has no fixed loops).
I used the graph $$G$$ with two vertices $$p,q$$ and three edges $$\alpha,\beta,\gamma,\delta$$ each oriented with intial vertex $$p$$ and terminal vertex $$q$$. This graph has an order $$2$$ self-homeomorphism $$h$$ fixing each of $$p$$ and $$q$$, defined by \begin{align*} h(\alpha) &= \gamma\\ h(\gamma) &= \alpha\\ h(\beta) &= \delta\\ h(\delta) &= \beta \end{align*} The isomorphism $$\langle a, b, c \rangle \mapsto \pi_1(G,v)$$ is given by \begin{align*} a &\approx \alpha \, \bar\delta \\ b &\approx \beta \, \bar \delta \\ c &\approx \gamma \, \bar \delta \end{align*} and deriving the formula for $$h_*$$ is now just a computation. For example \begin{align*} h_*(a) &= [h(\alpha \, \bar \delta)] \\ &= [\gamma \, \bar \beta] \\ &= [\gamma \, \bar \delta \, \delta \, \bar \beta] \\ &= [\gamma \, \bar \delta] \, [\beta \, \bar \delta]^{-1} \\ &= z \, y^{-1} \end{align*}
And for groups of any infinite rank, just subdivide the free basis into subsets of size $$2$$ and do what you did for even ranks.