# The Unapologetic Mathematician

## Direct sums of Abelian groups

Let’s go back to direct products and free products of groups and consider them just in the context of abelian groups.

The direct product $G\times H$ of abelian groups $G$ and $H$ works as we expect it to, because the elements coming from $G$ and $H$ already commute inside $G\times H$. The free product of $G$ and $H$ as groups gives $G*H$ as before, but now this is not an abelian group. Let’s consider the property that defined free products a little more closely. Here’s the diagram.

We want to read it slightly differently now. The new condition is that for any abelian group $X$ and homomorphisms $f_G:G\to X$ and $f_H:H\to X$ there is a unique homomorphism of abelian groups from the free product to $X$ making the diagram commute. We know that there’s a unique homomorphism from $G*H$ already, but we need to “abelianize” this group. How do we do that?

We just move to the quotient of $G*H$ by its commutator subgroup of course! Recall that any homomorphism $f:G\to A$ to an abelian group $A$ factors uniquely through this quotient: $G\to G/[G,G]\to A$. So now $(G*H)/[(G*H),(G*H)]$ is an abelian group with a unique homomorphism to $X$ making the diagram commute, it works for a free product in the context of abelian groups. This sort of thing feels odd at first, but you get used to it: when you change the context of a property (here from all groups to abelian groups) the implications change too.

Okay, so $(G*H)/[(G*H),(G*H)]$ is like the free product $G*H$, but we’ve thrown in relations making everything commute. We started with abelian groups $G$ and $H$, so all we’ve really added is that elements coming from the two different groups commute with each other. And that gives us back (wait for it..) the direct product! When we restrict our attention to abelian groups, direct products and free products are the same thing. Since this is such a nice thing to happen and because we change all our notation when we look at abelian groups anyhow, we call this group the “direct sum” of the abelian groups $G$ and $H$, and write it $G\oplus H$.

Now I didn’t really talk about this much before in the context of groups, but I’m going to need it shortly. We can take the direct sum of more than two groups at a time. I’ll leave it to you to verify that the groups $(G_1\oplus G_2)\oplus G_3$ and $G_1\oplus(G_2\oplus G_3)$ are isomorphic (use the universal property), so we can more or less unambiguously talk about the direct sum of any finite collection of groups. Infinite collections (which we’ll need soon) are a bit weirder.

Let’s say we have an infinite set $S$ and for each of its elements $s$ an abelian group $G_s$. We can define the infinite direct sum $\bigoplus\limits_{s\in S}G_s$ as the collection of all “$S$-tuples” $(g_s)$ where $g_s\in G_s$ for all $s\in S$, and where all but a finite number of the $g_s$ are the zero element in their respective groups. This satisfies something like the free product’s universal property — each $G_s$ has a homomorphism $\iota_s:G_s\rightarrow\bigoplus\limits_{s\in S}G_s$, and so on — but with an infinite number of groups on the top of the diagram: one for every element of $S$.

The direct product $\prod\limits_{s\in S}G_s$, on the other hand, satisfies something like the product condition but with an infinite number of groups down on the bottom of the diagram. Each of the $G_s$ comes with a homomorphism $\pi_s:\prod\limits_{s\in S}G_s\to G_s$, and so on. We can realize this property with the collection of all $S$-tuples, whether there are a finite number of nonzero entries or not.

What’s really interesting here is that for finite collections of groups the free product comes with an epimorphism onto the direct product. Now for infinite collections of abelian groups, the free product (direct sum) comes with a monomorphism into the direct product. The free product was much bigger before, but now it’s much smaller. When all these weird little effects begin to confuse me, I find it’s best just to plug my ears and go back to the universal properties. They will never steer you wrong.

April 12, 2007 -