The Unapologetic Mathematician

Mathematics for the interested outsider

Direct Products of Groups

There are two sorts of products on groups that I’d like to discuss. Today I’ll talk about direct products.

The direct product says that we can take two groups, form the Cartesian product of their sets, and put the structure of a group on that. Given groups G and H we form the group G\times H as the set of pairs (g,h) with g in G and h in H. We compose them term-by-term: (g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2). It can be verified that this gives us a group.

There’s a very interesting property about this group. It comes equipped with two homomorphisms, \pi_G and \pi_H, the “projections” of G\times H onto G and H, respectively. As one might expect, \pi_G(g,h)=g, and similarly for \pi_H. Even better, let’s consider any other group X with homomorphisms f_G:X\rightarrow G and f_H:X\rightarrow H. There is a unique homomorphism f_G\times f_H:X\rightarrow G\times H — defined by f_G\times f_H(x)=(f_G(x),f_H(x)) — so that \pi_G(f_G\times f_H(x))=f_G(x) and \pi_H(f_G\times f_H(x))=f_H(x). Here’s the picture.

Universal Property of Products

The vertical arrow from X to G\times H is f_G\times f_H, and I assert that that’s the only homomorphism from X to G\times H so that both paths from X to G are the same, as are both paths from X to H. When we draw a diagram like this with groups on the points and homomorphisms for arrows, we say that the diagram “commutes” if any two paths joining the same point give the same homomorphism between those two groups.

To restate it again, G\times H has homomorphisms to G and H, and any other group X with a pair of homomorphisms to G and H has a unique homomorphism from X to G\times H so that the above diagram commutes. This uniqueness means that has this property is unique up to isomorphism.

Let’s say two groups P_1 and P_2 have this product property. That is, each has given homomorphisms to G and H, and given any other group with a pair of homomorphisms there is a unique homomorphism to P_1 and one to P_2 that make the diagrams commute (with P_1 or P_2 in the place of G\times H). Then from the P_1 diagram with P_2 in place of X we get a unique homomorphism f_1:P_2\rightarrow P_1. On the other hand, from the P_2 diagram with P_1 in place of X, we get a unique homomorphism f_2:P_1\rightarrow P_2. Putting these two together we get homomorphisms f_1f_2:P_2\rightarrow P_2 and f_2f_1:P_1\rightarrow P_1.

Now if we think of the diagram for P_1 with P_1 itself in place of X, we see that there’s a unique homomorphism from P_1 to itself making the diagram commute. We just made one called f_2f_1, but the identity homomorphism on P_1 also works, so they must be the same! Similarly, f_1f_2 must be the identity on P_2, so f_1 and f_2 are inverses of each other, and P_1 and P_2 are isomorphic!

So let’s look back at this whole thing again. I take two groups G and H, and I want a new group G\times H that has homomorphisms to G and H and so any other such group with two homomorphisms has a unique homomorphism to G\times H. Any two groups satisfying this property are isomorphic, so if we can find any group satisfying this property we know that any other one will be essentially the same. The group structure we define on the Cartesian product of the sets G and H satisfies just such a property, so we call it the direct product of the two groups.

This method of defining things is called a “universal property”. The argument I gave to show that the product is essentially unique works for any such definition, so things defined to satisfy universal properties are unique (up to isomorphism) if they actually exist at all. This is a viewpoint on group theory that often gets left out of basic treatments of the subject, but one that I feel gets right to the heart of why the theory behaves the way it does. We’ll definitely be seeing more of it.

February 27, 2007 Posted by | Algebra, Group theory, Structure of Groups, Universal Properties | 13 Comments