What is a free group element that is not primitive?

Question

A primitive element of a free group is an element of some basis of the free group. I have seen some recent papers on algorithmic problems concerning primitive elements of free groups, for example, the papers on determining whether a subgroup of a free group contains a primitive element and determining whether a given element is primitive. However, I'm a little confused about the definition: it seems to me that every element of the free group on a finite set of generators is primitive.

Suppose $\{x_1, \dotsc, x_n\}$ is the set of generators for a free group on $n$ generators. Let $u$ be a word of length $m$ in the free group, and suppose $u = u_1 \dotsb u_m$, where each $u_i$ is one of the generators. I claim that $u$ is primitive because $u (u_2 \dotsb u_m)^{-1} = u_1$, hence $\{(u_2 \dotsb u_m)^{-1}, x_2, \dotsc, x_n\}$ is a basis of the free group, assuming without loss of generality that $u_1 = x_1$.

Where is the flaw in my argument?

Answer 1

Your argument would show that every element of $\mathbb{Z}$ is primitive. In fact the primitive elements are $1$ and $-1$. Do you see what goes wrong with your argument in this case?

The primitive elements of a free group $F_n$ have the special property that under a homomorphism $F_n \to G$ to some other group $G$, they can be sent to arbitrary elements of $G$. But most elements of a free group don't have this property. For example, in the free group $F_2$ on two generators $a, b$,

• $a^2$ doesn't have this property because it must be sent to a square, and for example $1 \in \mathbb{Z}_2$ is not a square.
• $[a, b]$ doesn't have this property because it must be sent to a commutator, and for example $1 \in \mathbb{Z}_2$ is also not a commutator.

And so forth.

Answer 2

$\newcommand{\GL}{\mathrm{GL}}$Let $F$ be free on $x, y$. Then $x^{2}$ is not primitive. If $x^{2}, z$ were a basis of $F$, then their images in the abelianization $F/F'$ should be a basis of $F/F'$. But with respect to the basis made of the images of $x, y$, we have that $x^{2}, z$ have matrix $$ \begin{bmatrix} 2 & a\\ 0 & b\\ \end{bmatrix} $$ which has determinant $2 b \ne \pm 1$, and thus is not in $\GL(2, \mathbb{Z})$.. So in $F/F'$ you cannot express the images of $x, y$ in terms of the images of $x^{2}, z$.

Answer 3

The problem with your argument as written is that you assume that $(u_2\cdots u_m)^{-1}\in\{x_2, x_3, \ldots, x_n\}$, but this isn't necessarily the case; it could be that $x_1$ is also among the $u_i$ and not just in the first place. For instance, consider the subgroup of the free group on two elements $a$ and $b$ generated by $\{aba, b\}$; it should be intuitively clear (and can be easily proven) that $a$ itself isn't a member of this subgroup.

Your argument does work in the case where the first 'basis letter' of your hypothetically-primitive element isn't repeated in the rest of the word; for instance, it's obvious that $\langle ab,b\rangle = \langle a,b\rangle$, and similarly $\langle ab^n,b\rangle = \langle a,b\rangle$ for all $n$.