Extending primitive systems in free groups

Question

It is a version of a question I asked on Math.StackExchange (no answers yet). Suppose $a_1,a_2,\ldots,a_n,b$ is a primitive system in a non-abelian free group $F$ (a part of a basis of $F$). Let $c \in F'$ be an element of the commutator subgroup of $F.$ Is there any simple condition on $c$ for the set $$ a_1,a_2,\ldots,a_n, bc $$ NOT to be a primitive system in $F$? An interesting case is the case where $bc$ is a primitive element of $F.$

Granted, for any given $c \in F',$ we can verify by means of the classical algorithm whether the set generates a free factor of $F,$ or we can reduce the problem to invertibility of a suitable matrix over the group ring $\mathbf Z[F].$

Answer 1

$\DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\IA}{IA} \newcommand\Z{\mathbb{Z}}$ Let me make a few comments about this. I'm going to call primitive systems "partial bases" since that's more standard terminology. I also am going to index things slightly differently than you to make it line up with my preferred notation.

Let $F_n$ be the free group on $\{x_1,\ldots,x_n\}$. Since $\Aut(F_n)$ acts transitively on bases for $F_n$, it also acts transitively on size-$k$ partial bases. You are interested in finding $c \in [F_n,F_n]$ such that $\{x_1,\ldots,x_k,x_{k+1} c\}$ is a partial basis. Ignoring the condition that $c \in [F_n,F_n]$ for the moment, for any $z \in F_n$ you can find some $c \in F_n$ with $x_{k+1} c = z$. Thus finding all $c \in F_n$ such that $\{x_1,\ldots,x_k,x_{k+1} c\}$ is a partial basis is equivalent to finding all $z \in F_n$ such that $\{x_1,\ldots,x_k,z\}$ is a partial basis.

Aside from some degenerate cases there isn't a nice answer to this question, but you can rephrase it as follows. Since $\Aut(F_n)$ acts transitively on size-$(k+1)$ partial bases, what you want to do is determine all elements of $F_n$ of the form $f(x_{k+1})$ for $f \in \Aut(F_n)$ satisfying $f(x_i) = x_i$ for $1 \leq i \leq k$. In other words, letting $\Aut(F_n,k)$ be the $\Aut(F_n)$ stabilizer of $\{x_1,\ldots,x_k\}$, you want to determine the $\Aut(F_n,k)$-orbits of $x_{k+1}$.

In his paper

J. McCool, Some finitely presented subgroups of the automorphism group of a free group, J. Algebra 35 (1975), 205–213.

McCool proves that $\Aut(F_n,k)$ is finitely presented and gives an algorithm for finding a presentation for it. I don't know of a paper that actually runs this algorithm and finds such a presentation, but if you can find generators that at least will give a way to systematically enumerate the $f(x_{k+1})$ and will give some hint as to their structure.

Now let's return to your original question: finding $c \in [F_n,F_n]$ such that $\{x_1,\ldots,x_k,x_{k+1} c\}$ is a partial basis. Note that $z \in F_n$ can be written in the form $z = x_{k+1} c$ with $c \in [F_n,F_n]$ if and only if $z$ maps to the same element of $\Z^n$ as $x_{k+1}$. Let $\IA_n$ be the subgroup of $\Aut(F_n)$ consisting of elements that act trivially on $\Z^n$. The letters "IA" stand for "identity on abelianization"; this is often called the Torelli subgroup of $\Aut(F_n)$. Also define $\IA_n(k)$ to be the $\IA_n$-stabilizer of $\{x_1,\ldots,x_k\}$. It is then not hard to see that what you want is equivalent to determining the orbit of $x_{k+1}$ under $\IA_n(k)$.

A classical theorem of Magnus says that $\IA_n$ is finitely generated and gives a generating set for it. See my paper

M. Day, A. Putman The complex of partial bases for Fn and finite generation of the Torelli subgroup of $\Aut(F_n)$, Geom. Dedicata 164 (2013), 139-153.

for references and additional information. I do not know if $\IA_n(k)$ is finitely generated, but I expect it is and that this can be proved using the same ideas that go into proving that $\IA_n$ is finitely generated (but starting with $\Aut(F_n,k)$ instead of $\Aut(F_n)$). Finding a finite generating set for $\IA_n(k)$ would give you a lot of insight into this problem.