# Writing element of infinite direct product of abelian groups as an infinite sum

From Wikipedia, it states that

For an infinite sequence $$G_1, G_2, \ldots$$ of groups, this can be defined just like the finite direct product of above, with elements of the infinite direct product being infinite tuples.

Now let $$\mathbb{Z},\mathbb{Z},\dots$$ be an infinite sequence of groups and $$G$$ be the infinite direct product of $$\mathbb{Z}$$.
Consider the element $$(1,1,\dots)$$ in $$G$$.
Can it be written as an infinite sum of elements of $$G$$?
That is, $$g=\sum_{i=1}^{\infty}a_i$$ where for a fixed $$i$$, the $$i$$-th entry of $$a_i$$ is $$1$$ while all the other entries is $$0$$.

I find this confusing because when we talk about infinite sum/series, we need to consider about limit of partial sums, but it seems that the concept of limit cannot be applied here.

## 1 Answer

There is no notion of infinite sum in a group without extra structure. You can sort of write stuff like this down heuristically but you need to be careful when reasoning with it. For example, the image of $$g$$ under a homomorphism is not determined by the image of the $$a_i$$. As an extreme example, the quotient $$\prod G_i / \bigoplus G_i$$ exists and is nonzero iff infinitely many of the $$G_i$$ are nontrivial.

On the other hand, we can say the following. The discrete groups $$G_i$$ can be given the discrete topology, and then their infinite product $$\prod G_i$$ can be given the product topology, which in the infinite case will not be discrete (again iff infinitely many of the $$G_i$$ are nontrivial). If the $$G_i$$ are finite then this topology makes the infinite product a profinite group; in general it's only a "prodiscrete" group. A sequence of elements of $$\prod G_i$$ converges in the product topology iff it converges pointwise, so it is actually meaningful and true to say that an element $$g = (g_1, g_2, \dots )$$ of the infinite product is the limit, in the product topology, of the sequence of "partial sums"

$$(g_1, e, e, e, \dots)$$ $$(g_1, g_2, e, e, \dots)$$ $$(g_1, g_2, g_3, e, \dots)$$

since this sequence converges pointwise to $$g$$. Be careful that if you want to use this to conclude anything about homomorphisms out of the infinite product $$\prod G_i$$ then you need to ask for them to be continuous with respect to the product topology; there are discontinuous ones in general.