# I Free Groups - Dmmit & Fooote - Section 6.3

1. Apr 7, 2016

### 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:

In 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 ... ?

Peter

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2. Apr 7, 2016

### Samy_A

That there is no relation satisfied by the elements of S is a given, it can't be proved.
It means that a word like abc-1b-1a can't be simplified (assuming a, b, c stand for different elements of S).

3. Apr 7, 2016

### Math Amateur

Thanks for the help Samy ...

I must say that I am a little surprised the D&F assert something (which does not have the status of an axiom) as true ... but it cannot be proven ...

Peter

4. Apr 7, 2016

### Samy_A

You seem to be overthinking this.

They simply mean that the elements of S are to be treated as abstract symbols, without any intrinsic meaning. That's why a word like $abc^{-1}b^{-1}$ is just that.
The only simplification (or reduction) you have to make to get a well defined free group is reducing a word like $acc^{-1}b^{-1}$ as follows: $acc^{-1}b^{-1}=ab^{-1}$.

Last edited: Apr 7, 2016
5. Apr 7, 2016

### lavinia

Given any group, any product $g^{k_{1}}_{1}...g^{k_{n}}_{n}$ of its elements determines some other element of the group. .

In an arbitrary group, not all such products represent different elements. For instance in an abelian group $a^{n}b^{m}$ is the same as $b^{m}a^{n}$. When two different products or "words" represent the same element of the group, this is called a relation. A relation can always be expressed as a product word that equals the identity. For instance for the abelian group $b^{m}a^{n}b^{-m}a^{-n}$ is a relation.

As Sammy_A pointed out one excludes from the definition of relation trivial expressions that are true for all groups.

So abtractly one can think of a group as the group of all finite length words $g^{k_{1}}_{1}...g^{k_{n}}_{n}$ modulo the normal subgroup of its relations.

If the subgroup of relations is trivial, then the group is just all finite length words with multiplication given by concatenation of words. This is called a free group.

Last edited: Apr 7, 2016
6. Apr 7, 2016

### mathwonk

the essential property of a free group F(S) on the set S, is its "mapping property", the fact that every function from S to any group G extends to a group homomorphism from F(S) to G. If you prove that, it is equivalent to saying there are no non trivial relations among the elements of S. I presume they do prove that.

actually the statement you reference above from DF about "no relations" is a bit imprecise. indeed as pointed out here there are relations, but they are the ones imposed by the group axioms. a relation is simply given by two different words that are equal. by definition such pairs are given by "reducing" one word to another. so a more correct statement would be that there are no "non trivial" relations, in the sense that no two reduced words are equal. but this is true by definition of F(S) since DF define F(S) to be the set of all reduced words.

so the hard part of their approach is proving associativity, since by definition, the product of two reduced words is the reduction of their concatenation. and if you have three words, there are two orders in which to concatenate them, i.e. you must show that Red(Red(AB).C) = Red(A.Red(BC)).

I think Mike Artin makes it much clearer in his book Algebra, by simply proving that each word has a unique reduced form, no matter what order you reduce it in. Then since concatenation of words is obviously associative, you get associativity of reduced words as well. I think others have said essentially the same thing above. The treatment in DF is so hard to read my eyes just blur looking at it.

Last edited: Apr 7, 2016
7. Apr 7, 2016

### Math Amateur

Samy, Lavinia, mathwonk ... ...

Thanks so much for your help ...

I now have a much better understanding of what is happening regarding free groups ...

Am still reflecting on what you have written ...

Peter

8. Apr 7, 2016

### WWGD

They're giving away free groups!!! :).

9. Apr 8, 2016

### lavinia

Yeah but you can't commute there to get them.