## Free products of groups

On Tuesday I talked about how to put a group structure on the Cartesian product of two groups, and showed that it satisfied a universal property. For free products I’m just going to start from the property. First, the picture.

This is the same diagram as before, but now it’s upside-down. All the arrows run the other way. So we read this property as follows: Given two groups and , the free product is a group with homomorphisms and so that given any other group with homomorphisms and there is a unique homomorphism .

The exact same argument from Tuesday shows that if there is any such group , it is uniquely determined up to isomorphism. What we need is to have an example of a group satisfying this property.

First of all, has to be an monomorphism. Just put itself for , the identity on for , and the trivial homomorphism (sending everything to the identity) for . Now anything in the kernel of is automatically in the kernel of the composition of with the coproduct map. But this composite is the identity homomorphism on , which has trivial kernel, and so must . This means that a copy of must sit inside . Similarly a copy of must sit inside .

So how do those two copies interact? Let’s put in for and let send everything in to some power of the generator — effectively defining a homomorphism from to — while sends everything in to some power of the generator . If there’s *any* relation between the copies of and sitting inside , it won’t be respected by the function to required by the universal property. So there can’t be any such relation.

is actually very much like , only instead of alternating powers of and , we have alternating members of and . An arbitrary element looks something like . Of course it could start with an element of or end with an element of . The important thing is that the entries from the two groups alternate. We compose by just sticking sequences together like for a free group. If one sequence ends with an element of and the next sequence starts with an element of , we compose those elements in so the whole sequence is still alternating.

Does sit inside here? Of course! It’s just sequences with only an element of in them. The same goes for . And given any group and homomorphisms and , we can send to . That’s our . As I said above, any other group that has this property is isomorphic to , so we’re done.

If we compare the free product with the direct product , we see that the main difference is that elements of and don’t commute inside , but they do inside . We can check that . In fact, take a presentation of with generators and relations , and one of with generators and relations , and with and sharing no elements. Then has generators and relations , while has generators and relations .

Since we’ve only added some relations to the presentation to get from to , the latter group is a quotient of the former. There should be some epimorphism from to . I’ll leave it to you to show that some such epimorphism does exist in two ways: once by the universal property of and once by the universal property of .

[...] of diagrams. Check that in this property is satisfied by disjoint unions. In coproducts are free products. In a preorder, coproducts are least upper bounds. And, of course, the coproduct defines a functor [...]

Pingback by Products and Coproducts « The Unapologetic Mathematician | December 5, 2007 |

I thought the free product of groups did not necessarily yield a group. I read somewhere that the free product Z(2)*Z(2)(Z(2) = integers mod 2) is not a group. This was in the context of defining the free product from the “bottom up” – strings of representatives with concatenation and reduction. Does that give the same result as your top down method with morphisms etc?

Comment by GK | February 20, 2011 |

Sorry. I was confusing free products of groups with free groups. Dnnnng!

Comment by GK | February 20, 2011 |

Right: the product of free groups isn’t a free group. But the

freeproduct of free groups is a free group.Comment by John Armstrong | February 20, 2011 |

“Let’s put F_2 in for X and let f_G send everything in G to the generator a, while f_H sends everything in H to the generator b.”

Surely, then, f_G and f_H are not group homomorphisms? If we have an element g in G, by your definition we have f_G(g) = a, and f_G(g^2) = a, but we should have f_G(g^2) = a^2, which is not equal to a. Also, we can’t map the identity in G to a either. Similarly for H.

Comment by TC | May 29, 2012 |

Good catch, TC; I was overly glib there. Really I was thinking of sending everything to some expression in involving only one of the generators.

Comment by John Armstrong | May 29, 2012 |