Confusion about the direct sum of abelian/nonabelian groups

Question

I know that there have been many questions on this site about the relationship between the direct product and direct sum of groups. But it seems they don't address the specific issue that I want to ask about.

My understanding of the direct sum is that it is defined only for abelian groups, and that the direct sum of abelian groups is the same as the direct product of the same groups assuming the number of groups is finite (and that, if the number is infinite, then the direct sum is restricted to those cases where a finite number of group elements from the sum are different from the identity, wheras the direct product is not restricted like this). This view is supported by the definition on Wikipedia here.

I have two questions:

1. Why in the world do we use two different words for essentially the exact same thing? Why call it something different when abelian groups are involved (namely, the direct sum) instead of always using the same word (namely, the direct product), whether they're abelian or not? Except for the fact that the direct sum (of finitely many groups) is said to only apply to abelian groups, the definitions of the direct sum and direct product seem to be exactly the same. There seems to be no reason why one couldn't use the direct sum even for nonabelian groups.

2. This got even worse when I noticed that Wikipedia had a separate definition of the direct sum of groups, namely here. Here, it is indeed NOT assumed that the two groups you sum together, namely $H_1$ and $H_2$ in the article, are abelian. Hence this definition does seem to be exactly equivalent (up to an isomorphism) to the direct product of groups, as described on Wikipedia here (specifically, go to the section called "Algebraic structure" to see how the direct product is isomorphic to the way the direct sum is defined in the second link). I am aware that there is a different question on this site about the two different defintions on Wikipedia (namely, here). But I understand why they're isomorphic. What I don't understand is why one definition assumes that the two groups to be summed are abelian, and the other doesn't. If people do use the direct sum even for nonabelian groups, and it is indeed equivalent to the direct product, then my first question is even more relevant, as that would make it even more pointless to have two separate words. I am very confused by all of this.

Answer 1

A little different point of view (from a very high vantage point). In category theory (which tries to generalize a lot of these constructions), we consider the category of abelian groups. In this category, the direct product is the "product", and the direct sum is the "coproduct". This isn't helpful unless I explain what this means.

Given two abelian groups $G_1, G_2$ we look for the following.

In a coproduct, we want a third abelian group $C$ with homomorphisms $i_1:G_1 \to C$ and $ i_2: G_2 \to C$ such that if I have a pair of homomorphisms $\phi: G_1 \to H$ and $ \phi_2: G_2 \to H$, then I get a unique homomorphism $\phi_3: C \to H$ such that the compositions line up (i.e. $\phi_1 = \phi_3 \circ i_1$ and $\phi_2 = \phi_2 \circ i_2$). Plainly speaking, any pairs of homomorphisms out of $G_1$ and $G_2$ factor through this third group. The direct sum satisfies this property (take the obvious maps $i_1: G_1 \to G_1 \oplus G_2$ and $i_2: G_2 \to G_1 \oplus G_2$ and this holds for any pairs of maps out of $G_1 $ and $G_2$. So the direct sum is a coproduct. See here for coproduct and examples.

In a product, we want a slightly different property. Given the same two abelian groups, this time with a pair of homomorphisms $h_1: H \to G_1$ and $h_2: H \to G_2$, we want a third abelian group $D$ with "projection" homomorphisms $p_1: D \to G_1$ and $p_2: D \to G_2$ such that the there is a unique homomorphism $h: H \to D$ so that all the compositions line up (i.e. $h_1 = p_1 \circ h$ and $h_2 = p_2 \circ h$). Plainly speaking, any pair of homomorphisms into $G_1$ and $G_2$ factor through this third group. The direct product satisfies this property (take the obvious projections $p_1: G_1 \times G_2 \to G_1$ and $p_2 : G_1 \times G_2 \to G_2$. So the direct product is a product. See here for product and examples.

Now, you are completely correct that in this category of Abelian groups, even though we want different properties for the coproduct and the product, they happen to be the same object (for finitely many summands/factors).

Key Idea: For general groups, it turns out the direct product doesn't actually satisfy the first definition property (in fact you need something else called the free product of groups there), so the distinction becomes important, but for abelian groups, they happen to be the same!

TL;dr: In category theory, we look for different properties in a direct sum (coproduct) and direct product (product). For abelian groups they happen to coincide and the distinction is not necessary, but for general groups there is a pretty big difference!

Also see some answers to this.

Edit for example Here is an example, it basically boils down to the fact that infinite addition of abelian groups isn't defined.

Let us see why $\Pi_{\mathbb{N}} \mathbb{Z}$ (the infinite direct product) is not also the coproduct of $\mathbb{N}$ copies of $\mathbb{Z}$. So, we have $G_i = \mathbb{Z}$ and let $H = \mathbb{Z}$ as well, maybe to keep things straight because of so many $\mathbb{Z}$s, let's call this codomain one actually $H = \mathbb{\tilde{Z}}$. Note that any map from $\Pi_i G_i \to \mathbb{\tilde{Z}}$ is just given by looking at the images in each factor. For example,

$$\phi(1,1,1,\dots,) = \phi(1,0,0,\dots) + \phi(0,1,0,\dots) + \phi(0,0,1,\dots) + \dots$$ If all of these are non-zero, this requires infinite sums which are not defined in abelian groups (recall we need some notion of convergence for infinite sums to hold). But if we restrict this map to elements such that all but finitely many terms are zero, then this is defined since it's just a finite sum then! This is all because you're mapping out of the infinite item into $\mathbb{Z}$ and so you're trying an infinite sum in $\mathbb{Z}$. But with the product, you're mapping from $\mathbb{Z}$ into the infinite item which is totally fine!