Hi all! Today I’ll talk about free products of groups.

Let and be groups. Then a *word* in and is a concatenation where each letter is an element of either or . Now suppose this concatenation does not have elements alternating from and . Then there must be at least one pair of elements from either or which are right next to each other. So let’s take their product in the group or (this includes the identity next to some other element). We can do this so that we get a word which has elements alternating from and . A word like this is called *reduced*. So if and are groups, we can define their *free product*, denoted to be the collection of all reduced words of and . This is actually a group under concatenation (and reduction, obviously).

For example, suppose . Then is cyclic, so is the free group on one generator. Thus is the free group on two generators. *However*, note that it’s not abelian. This is because it’s not necessarily true that where and . The free abelian group on two generators is the quotient of by the normal subgroup generated by . Another example would be the free product . This is isomorphic to the infinite dihedral group.

Alright, so suppose is a group, and is a generating set of . Then we can consider the free group over (the collection of all unreduced words over ). This has a natural map determined by for all (where is the set of all inverses of elements of ). Let be the kernel of this map. Let be a set which generates such that it is *normal*, i.e., such that is a generating set of .

OK, so now is isomorphic to , by the isomorphism theorems. The compact way of writing this is as , and is called a *presentation of .* Every group has a presentation. If and can be chosen to be *finite*, then the group is said to be *finitely presented*. Cool.

Now, suppose are groups. Write down presentations of and as follows: , and . Also suppose and are homomorphisms. Then define to be . This is called the *amalgamated product of and *(associated to the homomorphisms, etc.). If is the trivial group, then is simply the free product of and . (Do you see why?) So the free product is .

This is really all you need to know to understand a future post on algebraic topology. We’ll conclude this post with one simple remark. The idea behind free groups is that they are the coproduct in the category of groups. However, they do *not* form the coproduct in the category of *abelian* groups, because the free product of abelian groups need not be abelian (see the above example of ). (The coproduct in the category of abelian groups is the direct sum.)

Have fun,

SKD