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?

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    $\begingroup$ What about $(2,1)$ in $\mathbb Z\times\mathbb Z$? $\endgroup$ – Gregory Grant May 5 '15 at 18:42
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    $\begingroup$ @GregoryGrant your group is not free. But anyway, what about $2\in\mathbb Z$? $\endgroup$ – Hagen von Eitzen May 5 '15 at 18:43
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    $\begingroup$ @GregoryGrant It is free-abelian, but not free. There is no homomorphism $\mathbb Z\times \mathbb Z\to S_3$ that maps $(1,0)\mapsto (1\,2)$ and $(0,1)\mapsto (1\,2\,3)$. $\endgroup$ – Hagen von Eitzen May 5 '15 at 18:45
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    $\begingroup$ Inverses tend to exist in groups. The problem with the argument is that it concludes e.g. that $-m+1$ is a basis of $\mathbb Z$. $\endgroup$ – Thomas Poguntke May 5 '15 at 18:47
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    $\begingroup$ You may have a typo somewhere: $\{(u_2\cdots u_m)^{-1},x_2,\ldots,x_n\}$ isn't a basis for the free group; $u$ itself isn't in it! (Consider $u=ab$ in the free group on $a$ and $b$; then the basis you claim is $\{b^{-1},b\}$...) $\endgroup$ – Steven Stadnicki May 5 '15 at 18:50

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.

  • $\begingroup$ Is it not true that the set $\{x_1, \dotsc, x_n\} \cup \{w\}$ generates the entire free group, for any free group element $w$? $\endgroup$ – argentpepper May 11 '15 at 20:41
  • $\begingroup$ @argentpepper: yes, but it's not a basis. $\endgroup$ – Qiaochu Yuan May 11 '15 at 20:50

$\newcommand{\GL}{\mathrm{GL}}$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 $x^{2}, z$.

  • $\begingroup$ Oh, dear hollie mollie! Exactly the same thing I posted as a comment some moments ago...hehe. Nice! Your explanation is fine, yet I'm not sure about mine...? $\endgroup$ – Timbuc May 5 '15 at 18:59
  • $\begingroup$ I'm a bit unfamiliar with this area, could you clarify your answer a bit? What is $F'$? What are $a$ and $b$? What does this matrix have to do with the elements $x^2$ and $z$? $\endgroup$ – argentpepper May 5 '15 at 19:03
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    $\begingroup$ @argentpepper $F'$ is the derived group; in this case you just need to see 'the abelianization', which is essentially the group that you get by taking $F$ and 'modding out' by element order (i.e., $xy=yx$; technically, $F'$ is the group generated by $xyx^{-1}y^{-1}$, so $F/F'$ is the group you obtain by setting all members of $F'$ equal to the identity). This argument is saying that if there were some two-element basis $x^2,z$ of $F$ then that basis would also be a basis of the abelianization of $F$. $\endgroup$ – Steven Stadnicki May 5 '15 at 19:15
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    $\begingroup$ But bases of the free abelian group on two elements (i.e., $\mathbb{Z}^2$) are related by a base change - i.e., an integer matrix of determinant 1 (this is classic linear algebra). The matrix here is the shape that such a base change would have to have, and the comment about the determinant being $2b$ is why such a base change can't exist. $\endgroup$ – Steven Stadnicki May 5 '15 at 19:19
  • $\begingroup$ Sorry for not intervening sooner, I was otherwise busy. And many thanks to @StevenStadnicki for filling in the dots in my too concise answer. Another way of looking at it is to consider the quotient $F/F^{2}F'$, which is the Klein four-group, the non-cyclic group of order $4$. Clearly the image of $x^{2}$ is trivial here, so that the image of $y$ alone is unable to generate it. $\endgroup$ – Andreas Caranti May 6 '15 at 7:05

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 $n$.

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    $\begingroup$ The subgroup generated by $\{aa, b\}$ works as a counterexample as well; this is the counterexample given in the answer by @AndreasCaranti below and in the comments on the original question. $\endgroup$ – argentpepper May 5 '15 at 19:07

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