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Wondering how to write the different combinations in a free group in a generic way, going off the definition:

The free group $FS$ over a given set $S$ consists of all expressions (a.k.a. words, or terms) that can be built from members of $S$.

Wikipedia doesn't offer a generic notation for the set of all expressions/words/terms that can be built from $S$. Instead they just give an example:

For example, if $S = \{a, b, c\}$, then $T = \{a, a^{−1}, b, b^{−1}, c, c^{−1}\}$, and $$ab^{3}c^{-1}ca^{-1}c$$ is a word in $S$.

But in this (arbitrary) paper on free groups found here, they seem to have some sort of generic notation (not sure):

...here we just give a couple of examples of test elements in the free group $F_n = gp\langle x_1,\dotsc,x_n\rangle : x_1^k,x_2^k,\dotsc x_n^k, k \geq 2 [x_1, x_2] \dot [x_3, x_4] \dot \dots \dot [x_{n-1},x_n]...$

What I'm wondering is how to define $ab^{3}c^{-1}ca^{-1}c$ generically, so something where it can handle:

  1. Arbitrary number of letters (a, b, c, ...)
  2. The power signs are unique to each letter in the expression.
  3. $\pm$

Something along the lines of:

$$ \langle x_i^p, \dotsc , x_n^q \mid p,q \in \mathbb{N} \rangle $$

Not sure if that is correct or there is a better more intuitive notation.

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The generic element is $$ x_1^{p_1} x_2^{p_2} \cdots x_n^{p_n}, $$ where $n \ge 0$, $x_i \in S$, $p_i \in \mathbb Z \setminus \{0\}$, and $x_i \ne x_{i+1}$ for all $i \le n-1$. This doesn't actually say much of anything though. It just says take an arbitrary words in the generators and their inverses, and group together adjacent factors involving the same generator.

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  • $\begingroup$ $n$ is used twice ($x_n$ and $p_n$), wondering if they should be distinguished like $x_n^{p_m}$. $\endgroup$ – Lance Pollard Mar 28 '18 at 17:11
  • $\begingroup$ @LancePollard, it's correct as is, I think. Example: $a^2 b^{-3} a^{-4} d^5$ where $n = 4$, $x_4 = d$ and $p_4 = 5$. $\endgroup$ – fredgoodman Mar 28 '18 at 17:18
  • $\begingroup$ ah that makes sense, thank you $\endgroup$ – Lance Pollard Mar 28 '18 at 17:21

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