MHB Free Groups - Dummit & Fooote - Section 6.3

[Math Amateur]

I am reading Dummit and Foote's book: Abstract Algebra ... ... and am currently focused on Section 6.3 A Word on Free Groups ...

I have a basic question regarding the nature and character of free groups ...

Dummit and Foote's introduction to free groups reads as follows:https://www.physicsforums.com/attachments/5488In the above text, Dummit and Foote write the following:

" ... ... The basic idea of a free group $$F(S)$$ generated by a set $$S$$ is that there are no satisfied by any of the elements in $$S$$ ($$S$$ is "free"of relations.) ... ... "Dummit and Foote then show how to construct $$F(S)$$ as the set of all words (together with inverses) ... but they do not seem to prove that given that $$F(S)$$ contains all words in $$S$$ there are no relations satisfied by any of the elements in $$S$$ ...

Is the lack of a rigorous proof because the lack of any such relations is obvious ... ?

Can someone please help clarify this situation ...?

Peter

 

[Deveno]

Technically, what you wrote is not quite true-the set of all words in $S$ is $M(S)$, the free MONOID on the set $S$.

To create $F(S)$, we take $M(S \amalg S)$ first (here, $S \amalg S$ is the *disjoint union* of $S$ with itself, we regard this as "two copies of $S$ that don't overlap". One way to realize this (like many "categorical" constructions, unique only up to a unique isomorphism, in this case, an "isomorphism" is just a set-bijection) is as the set $S \times \{1,2\}$, so we have:

$(s,1) \neq (s,2)$).

The second copy of $S$ is usually thought of as "things we're going to tag as inverses" (usually they stick a prime mark, or an overline on them).

We're then going to take an equivalence relation on $M(S \amalg S)$, by declaring:

$(s,1)(s,2) \sim 0$ (where "0" is just a symbol for the "empty word", often represented by a blank space), for all $s \in S$.

This equivalence relation is usually called "word reduction", and allows us to write:

$a'bb'a = \ $", for example.

The reason for this particular peculiarity is that it is what we need to impose a group structure on a set containing $M(S)$, which is the set of all words in $S$ (you can think of it as a somewhat arbitrary "adjunction of inverses").

One of the reasons free groups are "difficult to understand", is that it can be unobvious that two distinct words are equivalent upon reduction (the words might be *really long*, and there may be many different paths to reduction depending on which adjacent pairs of words and their inverses you "cancel" first).

One upshot of this difficulty, is that two presentations (two distinct homomorphic images of $F(S)$) may, in fact, be isomorphic groups, and it may be impossible to know this, even if the set of relations and the homomorphic images are finite.

 

[Math Amateur]

Technically, what you wrote is not quite true-the set of all words in $S$ is $M(S)$, the free MONOID on the set $S$.

To create $F(S)$, we take $M(S \amalg S)$ first (here, $S \amalg S$ is the *disjoint union* of $S$ with itself, we regard this as "two copies of $S$ that don't overlap". One way to realize this (like many "categorical" constructions, unique only up to a unique isomorphism, in this case, an "isomorphism" is just a set-bijection) is as the set $S \times \{1,2\}$, so we have:

$(s,1) \neq (s,2)$).

The second copy of $S$ is usually thought of as "things we're going to tag as inverses" (usually they stick a prime mark, or an overline on them).

We're then going to take an equivalence relation on $M(S \amalg S)$, by declaring:

$(s,1)(s,2) \sim 0$ (where "0" is just a symbol for the "empty word", often represented by a blank space), for all $s \in S$.

This equivalence relation is usually called "word reduction", and allows us to write:

$a'bb'a = \ $", for example.

The reason for this particular peculiarity is that it is what we need to impose a group structure on a set containing $M(S)$, which is the set of all words in $S$ (you can think of it as a somewhat arbitrary "adjunction of inverses").

One of the reasons free groups are "difficult to understand", is that it can be unobvious that two distinct words are equivalent upon reduction (the words might be *really long*, and there may be many different paths to reduction depending on which adjacent pairs of words and their inverses you "cancel" first).

One upshot of this difficulty, is that two presentations (two distinct homomorphic images of $F(S)$) may, in fact, be isomorphic groups, and it may be impossible to know this, even if the set of relations and the homomorphic images are finite.

... thanks for the help, Deveno ... appreciate it ...

Just working through your post and reflecting on it ...

Peter