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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.

YCor
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Nathaly O.
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2 Answers2

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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 $\kappa$.

Asaf Karagila
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  • Didn't you mean $f: \kappa \to \mathbb{Z}$ ? – Maxime Ramzi Dec 07 '17 at 19:21
  • No. I didn't mean that. – Asaf Karagila Dec 07 '17 at 19:21
  • @Max We can look at maps $\kappa\rightarrow\mathbb{Z}$ only, but that's a bad habit: it only works here since all the summands happen to be the same. (Actually, my quibble is that there is no need to take disjoint unions: an element of $\bigoplus_{i\in I} G_i$ is just a map $f$ from $I$ to $\bigcup G_i$ such that $f(i)=e_{G_i}$ for cofinitely many $i$, and $f(i)\in G_i$ always.) – Noah Schweber Dec 07 '17 at 19:25
  • @Noah: Using disjoint unions makes it simpler in this case, where all the summands are the same, in the sense that it is easier to see why $\kappa$ is an upper bound. – Asaf Karagila Dec 07 '17 at 19:26
  • I don't think it really does: in the non-disjoint version, one argues "each element of the direct sum is determined by finitely many pairs $p\in\kappa\times\mathbb{Z}$; there are $\kappa$-many of these." I don't think that's any harder than via disjoint unions (although of course it is ultimately a matter of taste, I just like having the summands be literal subgroups of the sum). – Noah Schweber Dec 07 '17 at 19:29
  • @Noah: And I like to think about direct sums as "finitary choice functions"... :) – Asaf Karagila Dec 07 '17 at 19:30
  • @AsafKaragila : it's the second time in a short span that I think you mean something when you really mean what you wrote, sorry about that ! – Maxime Ramzi Dec 07 '17 at 19:38
  • @Max: I assure you, I'm the problem... :) – Asaf Karagila Dec 07 '17 at 19:41
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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.

Noah Schweber
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