Let $\mathbb{F}_2$ be the free group generated by $a$ and $b$. Suppose we are given a homomorphism $\phi: \mathbb{F}_2 \to \mathbb{F}_2$ with the property that $\phi(aba^{-1}b^{-1}) = aba^{-1}b^{-1}$. Can I conclude that $\phi$ is surjective? Can I conclude $\phi$ is an isomorphism?
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3$\begingroup$ Only an answer to the second question. $\mathbb{F}_2$ is an Hopfian group (it cannot be isomorphic to a proper quotient group). Equivalently, any epimorphism of $\mathbb{F}_2$ onto $\mathbb{F}_2$ is an isomorphism. $\endgroup$– Clément GuérinMar 27, 2018 at 8:21
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$\begingroup$ With a lengthy and tedious computer calculation, I have checked that $\phi$ is surjective when ever $\phi(a)$ and $\phi(b)$ both have length at most $10$, so I am going perhaps rashly conjecture that the answer to your question is yes. $\endgroup$– Derek HoltMar 27, 2018 at 13:22
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$\begingroup$ Is there a non-trivial example of such $\phi$? I.e. except for the identity $\phi(x)=x$? $\endgroup$– freakishMar 27, 2018 at 13:56
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1$\begingroup$ Yes there were about 630 examples in which $\phi(x)$ and $\phi(y)$ both had length at most $10$. An easy example is $[ba,b] = [a,b]$ (using the definition $[a,b] = a^{-1}b^{-1}ab$ - why can't everyone use the same definition?) $\endgroup$– Derek HoltMar 27, 2018 at 17:02
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2 Answers
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It's a theorem of Nielsen known as "Nielsen commutator test" (Nielsen, J. Die Isomorphismen der aligemeinen unendlichen Gruppe mit zwei Erzeugenden. Math. Ann. 78 (1918), 385–397.) stating that automorpisms of $F$ = $\langle x, y \rangle$ are precisely those endomorphisms which take $[x, y]$ to any conjugate or inverse to conjugate — it's an easy consequense of the fact that any IA-automorphism of $F$ is inner; if you want, I can write proof here.
It's quite interesting that this result extends to some other 2-generated groups — for example, free 2-generated metabelian group (by V. Durnev) and "most" groups of type $F/[[R, R], F]$ (by N. Gupta and V. Shpilrain) also satisfy commutator test.
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$\begingroup$ It's worth pointing out that this means that automorphisms in $F_2$ can be defined using first order logic. IIRC, the same is not true for free groups of higher rank. $\endgroup$– user1729Aug 10, 2018 at 16:58
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I think this can be solved using Stallings' folds as follows.
First, note that if the image of $\phi$ were abelian, then $\phi([a,b]) = [\phi(a),\phi(b)] = 1$. So, the image of $\phi$ is a rank two free group.
Now suppose $\phi(a) = U$, a reduced word over $\{a,b\}$ and similarly $\phi(b) = V$. Let $|U|$ and $|V|$ denote their lengths.
Stallings' method is to represent $\phi$ as a map of graphs. Start with the graph consisting of a pair of circuits---with edges labeled and oriented according to the letters in each word---of length $|U|$ and $|V|$, respectively, and glue them at the vertex corresponding to where reading the edge labels reads off $U$ and $V$. Call this graph $\Gamma_0$.
Now perform Stallings folds. This means that if two labeled edges have the same initial vertex and orientation (both away from the initial vertex or both towards), then they are identified. After finitely many steps, there can be no more folds. Call the resulting graph $\Gamma_1$ and let $f_0: \Gamma_0 \to \Gamma_1$ be the composite of folds. Note that $f_0$ induces an isomorphism on fundamental groups because the image of $\phi$ is a rank two free group.
Finally, let $f_1:\Gamma_1 \to \Delta$ be the map (an immersion, by design---this is the point of Stallings folds) to the rose with two petals, labeled $a$ and $b$.
The path $f_0(U)f_0(V)f_0(U^{-1})f_0(V^{-1})$ is an immersed path. Its image under the immersion $f_1$ is another immersed path. But its image, by hypothesis, is $aba^{-1}b^{-1}$ in $\Delta$. Therefore, the lengths of $f_0(U)$ and $f_0(V)$ are 1. This implies that $f_0(\Gamma_0) = \Gamma_1$ is a rose with two petals. Consequently, $f_1$ induces an isomorphism onto the fundamental group of $\Delta$. Since $f_1 \circ f_0$ represents $\phi$, we have that $\phi$ is surjective (and an isomorphism).
