# The Unapologetic Mathematician

## 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.

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