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For a free group $F$, an element $w$ is primitive if it is part of some free basis for $F$.

Let $\pi:F[x_0,x_1,...,x_n]\rightarrow F[x_1,x_2,...,x_n]$ be defined $\pi (x_0)=1$ and $\pi (x_i)=x_i$ for $i\geq 1$.

My question is:

Is there an example of a cyclically reduced primitive element $w$ in $F[x_0,x_1,...,x_n]$ satisfying $\pi(w)=1$ whose length is greater than 1?

[Added for clarification]

Note: if $w$ is an example, then there must be more than one occurrence of $x_0$ in $w$ and the exponent sum of $x_0$ in $w$ must be $\pm 1$.

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  • $\begingroup$ There is a basis (xyzx^{-1}y^{-1}, xyz, y)$. $\endgroup$
    – Ilya Bogdanov
    Jun 15, 2022 at 16:18
  • $\begingroup$ It is true that this is a basis., therefore each element is primitive. But I don't think it provides an example. I assume you are taking $x_0 = z$. In this case, after applying $\pi$ to your set, we get $\{ xyx^{-1}y^{-1},xy,y \}$. $\endgroup$
    – Andrew Clifford
    Jun 15, 2022 at 17:38
  • $\begingroup$ Sorry, thanks!... $\endgroup$
    – Ilya Bogdanov
    Jun 15, 2022 at 19:06
  • $\begingroup$ Do you know the answer for $n=1$? $\endgroup$
    – YCor
    Jun 15, 2022 at 20:42
  • $\begingroup$ @YCor For $n=1$ I think no such $w$ can exist, since in this case the image of $\pi$ is abelian, so $w$ would be of the form $x_0^\pm$ times a commutator, and then the fact that $Out(F_2)\cong GL_2(\mathbb{Z})$ implies that $w$ is conjugate to $x_0^\pm$. But this doesn't work at all for $n>1$, since the image of $\pi$ is not abelian, and also there's a non-trivial Torelli group to worry about.... $\endgroup$
    – Matt Zaremsky
    Jun 16, 2022 at 1:50

1 Answer 1

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No, there cannot be a primitive element $w \in \ker \pi$ that is not conjugate to $x_0$ or $x_0^{-1}$.

The map $\pi$ factors through $F[x_0, x_1, \dots, x_n] / \langle \langle w \rangle \rangle$, which is isomorphic to $F_n$ since $w$ is primitive. As the free group is Hopfian, this means the induced surjection $F[x_0, x_1, \dots, x_n] / \langle \langle w \rangle \rangle \to F_n$ is in fact an isomorphism, so $w$ and $x_0$ have the same normal closure in $F_{n+1}$. A theorem of Magnus from 1931, citing from Lyndon and Schupp's book "Combinatorial Group Theory", states:

Proposition II.5.8. If two elements $r_1$ and $r_2$ in a free group $F$ have the same normal closure in $F$, then $r_1$ is conjugate to $r_2$ or to $r_2^{-1}$.

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