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?
Best Answer
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 $a, b$,
And so forth.