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I am looking for an explicit injective homomorphism

$$ \phi:F_3=\langle x,y,z\rangle \longrightarrow F_2=\langle a,b\rangle $$

such that the induced map on abelianization $\phi^{ab}:\mathbb Z^3\to \mathbb Z^2$ is surjective.

What I’ve tried so far (and why it failed)

  1. Index-2 subgroups of $F_2$: By Nielsen–Schreier, any index-2 subgroup of $F_2$ is free of rank $3$, so inclusion $H\hookrightarrow F_2$ would give an injection $F_3\cong H\hookrightarrow F_2$. But: the abelianization of $H$ has image an index-2 sublattice of $\mathbb Z^2$, so $\phi^{ab}$ is not surjective.

  2. Short triples like $(a,\,bab^{-1},\,aba^{-1})$: These generate all of $F_2$ (since $b=a^{-1}(aba^{-1})a$), so the subgroup has rank $2$, not $3$.

  3. Conjugates of a single generator: Subgroups like $\langle a,\,bab^{-1},\,b^2ab^{-2}\rangle$ are free of infinite rank, but abelianization lands in the line spanned by $(1,0)$, so not surjective.

user1693987
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1 Answers1

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Take three consecutive Fibonacci numbers; $2,3,5$ will do. Note that the vectors $(2,3)$ and $(3,5)$ generate $\mathbb Z^2$, because $\det\begin{pmatrix} 2 & 3 \\ 3 & 5 \end{pmatrix} = 1$. (Fibonacci numbers are my favorite way of easily constructing large matrices of determinant $1$; in this case, it doesn't even have to be all that large).

The elements $x=a^2 b^3$, $y=a^3 b^5$, $z=ba \in F_2$ form a free basis of a rank 3 subgroup of $F_2$, as a simple calculation using Stallings fold method shows.

And since the subgroup generated by $x$ and $y$ alone maps surjectively to the abelianization of $F_2$, so does the one generated by $x,y,z$.

Lee Mosher
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