# Free groups: why are they significant in group theory?

1. Feb 18, 2014

### James MC

Mathematicians have produced a wide variety of long and complex proofs of the existence of free groups, and there appears to be a strong emphasis upon finding better proofs that involve a variety of techniques. (Examples are here and "www.jstor.org/stable/2317030" [Broken] and here.) So clearly, mathematicians see free groups as very important objects.

I don't understand why this is!

Definition: Let X be a set. A group F is free on X if there is an (inclusion) map ψ: X → F and for ANY group G and (inclusion) map δ: X → G, there is a unique homomorphism of groups β: F → G such that δ=β°ψ.

(I'm 80% sure inclusion maps are the relevant maps.) Why does it matter that for any given set, such an object exists? After all, given the wide variety of possible G's for any given X, F will have to be enormously complex, and so whatever additional structure a given G has that goes beyond the group-theory axioms, such structure will be wiped out in F if F is to bear homomorphisms to G's without those structures. Why would such a structureless object be of interest?

Last edited by a moderator: May 6, 2017
2. Feb 18, 2014

### jgens

The first function does not have be an inclusion in the most literal sense per se, but it can be interpreted that way. The second function does not need to be an inclusion in any sense however.

There are lots of reasons and here is one big one: Every group is a quotient of two free groups! In other words, given a group G, there are free groups F and K such that G = F/K. This implies all groups have a presentation <S|R> where S are the generators and R are the relations. Understanding these presentations is often helpful in understanding the group itself.

3. Feb 19, 2014

### mathwonk

duhhh, every group is a surjective image of a free group. that's enough for me. i.e. free groups are the canonical example of a group.

4. Mar 5, 2014

### James MC

I take it you mean that for any group G on set X, the free group F on X is such that every element in G has a corresponding element in F. If that's what you mean then this by itself hardly makes the free group canonical! Particularly in light of the category theoretic proof mentioned below, this just seems to fall out of the fact that the free group on X is just a big conglomeration of all the groups on X anyway. The question remains as to why we should care about such a beast.

5. Mar 5, 2014

### James MC

Are you able to point to proofs of the claims (i) that every group is a quotient of two free groups and (ii) that this entails that every group has a presentation? I've tried searching, closest I've found is the claim that every group generated by set X is isomorphic to a quotient of the group freely generated by X. And this latter claim has been related to the expression of a group's presentation in terms of generators and relations.

It's hard to see why (i) would make free groups significant, but I can see that (ii) might if it somehow illuminates how the presentation of all groups in terms of <S|R> arises.

It might help to say what's motivating my question: the standard proofs of free groups make it hard to see why they're of interest. The constructive proofs seem ad hoc because rather than showing that free groups involving familiar operations exist, they instead create the juxtaposition operation for the purpose of the proof to show that a free group with that operation exists. It's hard to see how objects constructed in such an ad hoc manner could be significant. And then regarding the category theoretic proof: we are simply shown that we can take the Cartesian product of all groups on X, and then of course there will be a (unique) homomorphism going from that group to any group on X, for there will be an isomorphism composed with a projection morphism! Again, hard to see what the point in showing that is.

I realise there is a point and that I'm just not getting it - so any further attempt to illuminate this point for me would be most welcome.

My understanding now is that neither have to be inclusions, both can be functions in the most general sense.

Last edited: Mar 5, 2014
6. Mar 5, 2014

### jgens

Most decent algebra texts will prove these results, but the arguments are actually quite simple so I can sketch them here. For (i) let G be any group and let FG be the free group generated by the elements of G. The universal property of this free group provides a homomorphism FG→G and let K denote its kernel. By the first isomorphism theorem it follows that FG/K = G and since subgroups of free groups are free this establishes the first claim. For (ii) the generators are given by the generators of FG and the relations are given by the generators of K. Pretty easy stuff.

Often times one can prove results about groups by first establishing them for free groups and then showing how it holds for the quotient of these groups. When these groups are abelianized (mod out by commutators) this fact has important consequences for computing things like Ext and Tor.

Exactly!

Well the first map is always an injection, and hence, can be interpreted as an inclusion. But the second map need not be.

7. Mar 12, 2014

### gufiguer

The structure of a free object F(X) depends uniquely on the cardinality of the set X. So, in a concrete category, there exists at most (up to iso) one free object for each cardinal number. In some categories some of these objects don't exists. In the usual algebraic categories, like groups, there exists exactly one free object (up to iso) for each cardinal. The proof of existence of free objects depends in which category we are working, so it uses individual constructions.
Free groups are quite interesting because it's elements are unable to mix together and forms another mix of elements. This is why we say they don't have relations. If you mix (by the group operation and inversion) some elements of F(X) to form a term T and you can't simplify the term T by using just the axioms of groups, then T is a new element, distinct from the formers and distinct from identity!!! I can imagine that some constructions will be enormously facilitated by that!

8. Mar 12, 2014

### gufiguer

We can build in this way, for example, tensor products of non-abelian groups, which are useful in homotopy theory.