Why does $\bigoplus_{i\in\kappa}\Bbb Z$ have cardinality $\kappa$ for an infinite $\kappa$?

Question

I was reading this post Does every set have a group structure? and I'd like some clarifications if possible.

If $X$ is uncountable, by the Axiom of Choice we'll have $|X| = \kappa$. I'm not sure why the direct sum $\bigoplus_{i \in \kappa} \mathbb{Z}$ has cardinality $\kappa$ and why that would make $X$ into a group.

Any set you put in bijection with a group can be made into a group, but I'm not seeing with the direct sum.

Answer 1

This is a cardinal arithmetic thing.

First, note that the union of $\kappa$ many sets which are countable and pairwise disjoint is itself a set of size $\kappa$.

Next, recall that if $X$ is infinite, then the set of finite sequences from $X$ has size $|X|$ again.

Finally, the direct sum can be represented as functions $f\colon\kappa\to\bigsqcup_{i<\kappa}\Bbb Z$ such that $f(\alpha)$ lies in the $\alpha$th copy of $\Bbb Z$, and $f(\alpha)=0$ for all but finitely many functions. This is equivalent to a subset of the finite sequences from this union, whose size is $\kappa$.

All this means that the size is at most $\kappa$, and I will leave you figuring out why it is at least $\kappa$.

Answer 2

Asaf has answered the cardinality part of your question. As to why the direct sum has a group structure, this is an instance of a more general fact: any direct sum of groups has a group structure. Namely, if $H=\bigoplus_{i\in I}G_i$, then there is a natural binary operation on $H$: given two elements $h_1,h_2$ of $H$, we "multiply them componentwise," using the $G_i$-multiplication on the $G_i$th coordinate, and this operation satisfies the group axioms.

I'm using "multiply" here since the word "add" is usually reserved for abelian groups; of course, in this instance the groups involved are abelian, but I think it's a bit better to be more general.

So what about your specific example? Well, $I$ here is just $\kappa$, and certainly $\mathbb{Z}$ has an obvious group structure ...

---

OK, now let me say a bit about the formal details, since if you haven't seen it before the above idea can seem a bit glib:

"Multiply componentwise" is easy to think about when our index set $I$ is $\mathbb{N}$, since then an element of $H$ is literally a sequence. In general, an element of $H$ is an "$I$-indexed sequence," and the right definition of these turns out to be that an $I$-indexed sequence is just a function with domain $I$ (the "$i$th term" is the value of the function on $i$). And the elements of $H$ are $I$-indexed sequences with appropriate values, that is, a function $f$ with domain $I$ is an element of $H$ if

• For each $i\in I$, $f(i)\in G_i$ (so it really is a "sequence of elements from the $G_i$s"), and

• For all but finitely many $i\in I$, $f(i)=e_{G_i}$ (this is the direct sum condition, that all but finitely many coordinates be trivial; if we drop this condition we get the direct product instead).

• And the "multiply componentwise" function on $H$ is now defined as follows: given $f_1, f_2\in H$, we let $f_1*f_2$ be defined by $$(f_1*f_2)(i)=f_1(i)*_{G_i}f_2(i).$$

It's now a good exercise to check that $H$ with this operation is indeed a group.