# Infinite products of a (finite) group

So I'm having a little trouble understanding the concept of infinite (cartesian) products of a group -- specifically, my notes (and, of course, homework questions) have concepts of, say $S_3^\mathbb{Z}$ and $S_3^\mathbb{R}$ (where $S_3$ is the permutation group of 3 elements, of course), but I'm having a hard time conceptualising the difference between these.

If I assume that we're taking what seems to be the canonical approach: $G^I = \Pi_{i \in I} G$, how do I begin to understand the notion of, say, 3.141519 copies of $G$ (or in my case, $S_3$)? Is there a distinction between $\mathbb{Z}$ and $\mathbb{N}$ copies of a group? Or $\mathbb{Z}$ and $\mathbb{R}$?

It feels like there's something obvious I'm missing, but, of course, I don't know it!

• There aren't $\pi$ copies of $S_3$. There are uncountably many, one for every real number. – Cheerful Parsnip May 18 '13 at 7:17
• Of course; I was aiming to take one element of $\mathbb{R}$ for the sake of an illustrative example (given my understanding was flawed to begin with). I hope this didn't detract from the question! – Ben Stott May 18 '13 at 11:52
• It doesn’t detract from the question. On the contrary, it’s helpful, because it shows a fundamental misunderstanding of the concept. In the product $G^{\Bbb R}$ one of the factors is labelled $G_{3.141519}$, and another is labelled $G_\pi$, but these are just labels; they’re not counting anything, or giving you weird fractions of a group. – Brian M. Scott May 18 '13 at 18:17
• Yeah, right -- this makes far more sense now, coupled with the answer below! – Ben Stott May 19 '13 at 0:13

The direct product of copies of a single group $G$ over an indexing set $I$ can be thought of as the space of functions $f:I\to G$ with componentwise multiplication (i.e. $(f_1f_2)(i)=f_1(i)f_2(i)$ for all $i\in I$). This makes sense; any coordinate vector $(a_1,a_2,\cdots)$ can be thought of as a function from $\bf N$ into whatever, as $i\mapsto a_i$, and conversely any function $f:{\bf N}\to{\rm blah}$ can be thought of as a coordinate vector via $(f(1),f(2),f(3),\cdots)$. The utility of the functional approach is that it's kind of hard to write a "vector" with uncountably many coordinates (there are too many coordinates to enumerate), but functions are perfectly acceptable to talk about.

If $I$ and $J$ have the same size, then $\prod_IG\cong\prod_JG$. This can be accomplished by any bijection given by $\varphi:I\xrightarrow{\sim}J$; given $x\in\prod_IG$, define $\bar{x}\in\prod_JG$ to be the element whose $\varphi(i)$th coordinate is equal to the $i$th coordinate of $x$. So for example, let $\varphi:1,2,3,\cdots\mapsto0,1,-1,2,-2,\cdots$. Then

$$(a_1,a_2,a_3,\cdots)\mapsto (\cdots,a_5,a_3,a_1,a_2,a_4,\cdots)$$

is an isomorphism $\prod_{\bf N}G\cong\prod_{\bf Z}G$ for any group $G$. In the language of functions this means that the map $f\mapsto f\circ \varphi^{-1}$ is an isomorphism $\prod_IG\to\prod_JG$.

If $I,J$ are different sizes, then $G^I\not\cong G^J$ since in the first place they do not have the same sizes as sets. If we drop the condition that $G$ is finite though then I'm not so sure what can be said.

It does not make sense to take the direct product of a non-integer number of copies of $G$, any more than it makes sense to have a set of non-integer size. (Makes even less sense; you can define "fuzzy" sets, but I wouldn't know how to begin defining a direct product indexed by a fuzzy set.)

It's noteworthy that when working with an infinite number of nontrivial groups, direct sums and direct products are two different things. The direct product $\prod_I G_i$ can be formalized as the space of functions $f:I\to\bigcup_I G_i$ such that $f(i)\in G_i$ for all $i\in I$, with pointwise multiplication. The direct sum $\bigoplus_IG_i$ is a subgroup of the direct product consisting of those functions with "finite support," i.e. $f(i)$ is not an identity element of $G_i$ for only finitely many $i\in I$.

• Is it important to remark the OP about direct sum or weak direct product here? – mrs May 18 '13 at 8:04
• So the fact that we're taking $\mathbf{R}$ copies doesn't really matter, right (because of the isomorphism of sets) -- we only care for the cardinality of the set we're using as the indexing set? – Ben Stott May 19 '13 at 11:09
• @Ben: That’s correct. On rare occasions using a particular index set of a given cardinality makes some argument technically simpler, but that’s a matter of mere convenience. The product of $|\Bbb R|$ copes of $G$ is (up to an obvious isomorphism) the same group whether you use $\Bbb R$, $[0,1]$, $\wp(\Bbb N)$, or the set of continuous real-valued functions on $\Bbb R$ as the index set: all of these have the same cardinality. – Brian M. Scott May 19 '13 at 18:34
• For example, indexing by $\mathbb{N}$ is slightly neater than indexing by $\mathbb{Z}$, as you can just write $G\times G\times\ldots$ as opposed to, say, $\ldots\times G\times G\times\ldots$. Same thing, just one is kinda weird... – user1729 May 20 '13 at 11:20