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I'm a Bachelor student (third year) and I'm having trouble with the definition of word in group theory. From Wikipedia (our professor also gave us a similar definition):

Let G be a group, and let S be a subset of G. A word in S is any expression of the form ${\displaystyle s_{1}^{\varepsilon _{1}}s_{2}^{\varepsilon _{2}}\cdots s_{n}^{\varepsilon _{n}}}$ where $s_1,\ldots,s_n$ are elements of S, called generators, and each $ε_i$ is $±1$. The number $n$ is known as the length of the word. Each word in S represents an element of G, namely the product of the expression. By convention, the unique identity element can be represented by the empty word, which is the unique word of length zero.

My question is: how is defined $s^{-1}$ for a certain $s\in S$? Maybe it is just notation and I'm confused for nothing, but I don't get it: if $S$ is any subset (not necessarily a subgroup / group) of $G$, what exactly the inverse of an element is and why it exists. Is it just a formal notation and nothing else?

Sorry for the dumb question but I didn't find any information online, thanks in advance :)

Mati
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  • Be careful: When one defines presentations of groups, $S$ is (in general) not a subset of $G$. – Moishe Kohan Dec 07 '24 at 14:01
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    @MoisheKohan This is not about group presentations. You take it a step too far. Words are usually first introduced when one defines the subgroup generated by a subset. (and later used in order to define free groups) In this case, $S$ is a subset. – Mark Dec 07 '24 at 14:03
  • @Mark: I do not know how things "usually" are (when I was taking my first algebra class, the first example after the definition of a group was one of the semigroup of surfaces with connected sum as the binary operation) and I have no idea what student's teacher is doing (is the teacher talking about presentations or something else). It is quite possible, the teacher is making a common mistake or the student misunderstood what was said. All what I was saying that one needs to be careful when one defines group presentations. – Moishe Kohan Dec 07 '24 at 14:08
  • @Mark so basically a word has elements of S and G (the inverses of the elements in S)? Thanks for the answer, I'm not getting what exactly are we working with – Mati Dec 07 '24 at 14:09
  • @MoisheKohan we are giving some useful definitions before diving into presentations. The definition of presentation is given afterwards without the use of words. Yes in that case we work with the basis of a free group (and we know by a theorem that the group generated by the basis is in fact the group itself) – Mati Dec 07 '24 at 14:13
  • @Mati A word contains elements of $S$ and their inverses. You need to include inverses, as otherwise the words will not be a subgroup. – Mark Dec 07 '24 at 14:17
  • @Mark ok, thank you, now it is clear. – Mati Dec 07 '24 at 14:20
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    Please do not answer in the comments. – Shaun Dec 07 '24 at 18:55
  • To elaborate on @Derek's comment - it is essential to understand the distinction between syntax vs. semantics. Group words are purely syntactic entities - terms (expressions) in a formal language whose alphabet is a subset of a group, i.e. the letters of the alphabet are group elements (recall symbols [here letters] can be anything in math, e.g. see Hodges on naming). – Bill Dubuque Dec 08 '24 at 00:25
  • The meaning (semantics) of the word is given by evaluating the word into the group - mapping concatenation to product, the inverse symbol to the inverse operation, and the group element letters to themselves. To properly address such subtleties really requires seeing the exact details of your professor's presentation - which we don't have. – Bill Dubuque Dec 08 '24 at 00:25
  • @BillDubuque thank you very much! Yes I had difficulties in understanding if words were just synctatic entities or eventually what meaning they had (and the latter depends on what structures are we working with). Thanks for the useful link and clarifications – Mati Dec 09 '24 at 08:59

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Not sure I'm understanding the question, but...if $s$ is an element of $S$, then it's also an element of $G$, which means that the group structure of $G$ gives us an inverse $s^{-1}$.

In the same way, the group structure gives us a way to multiply elements, which is how we can get the "product of the expression".

Hew Wolff
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What might be confusing you is that we're working in $G$ not $S$. $S$ indeed doesn't have to contain the inverses of its elements - it's a subset, not a subgroup - but those inverses must exist somewhere in $G$, because all elements of $G$ have inverses in $G$ (as $G$ is a group).

If we stuck to non-negative powers of the elements of $S$, we would have a monoid, not a group, and that's not very useful (this is Group Theory, after all). By adding in all the inverses of the elements of $S$, the set of words forms a subgroup, which we can do much more with. So the convention is that the letters in the set (subgroup) of words over $S$ come from $S\cup S^{-1}$.

Possibly helpful fact: If all elements of $G$ have finite order, adding in the inverses like this makes no difference to the generated subset.

IanR
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  • Thanks! Sorry but I don't have enough reputation to upvote, but I've accepted your answer. Yes I didn't get that we were working in the group, so i made a dumb question. Thank you very much again :) – Mati Dec 09 '24 at 08:56
  • @Mati This answer doesn't address the source of your misunderstanding so you shouldn't have accepted it - cf. my comments above on syntax vs. semantcis. – Bill Dubuque Dec 09 '24 at 10:08
  • @BillDubuque your comment + his answer helped me understand all that I needed to know. If you had posted your comment as an answer I would have accepted that – Mati Dec 10 '24 at 07:24
  • @Mati If you hadn't accepted so quickly you might have received an answer from someone that said much more than what was in my comment. Once a question has an accepted answer many will skip past it. Accepting too quickly is one of the biggest mistakes made by beginners. – Bill Dubuque Dec 10 '24 at 07:35
  • @BillDubuque sorry, I didn't know that. Also, I realized my question was quite dumb, so I didn't want too many people to spend time reading and answering it. Thank you for your tip!! – Mati Dec 10 '24 at 13:30