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I know the definition of the direct product $G\times H$ of two random groups $G$ and $H$. It is also clear to me that this can be extended to a product of a finite number of groups $G_1\times \dots G_n$.

My question is whether it makes sense to do this for an infinite number of groups. For example, if $G_i$ are groups for $i\in \mathbb{Z}$ (or I guess $\mathbb{N}$) can one simply define $$ \times_{i\in \mathbb{Z}} G_i = \{ (g_i): g_i\in G_i\} $$

so that this is a group under componetwise operation?

I checked the definition of a group and I am pretty sure this would also be a groups, but I would still like to have confirmation.

My next questions is: Can this all be done if the index set is uncountable? In that case I am not sure how the operation would work since there isn't a first "coordinate".

John Doe
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  • Yes, you are right. You can also have an uncountable index set. See https://en.wikipedia.org/wiki/Direct_product_of_groups#Examples – saulspatz Feb 28 '19 at 21:18
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    One standard notation is $\prod_i G_i$ – YCor Feb 28 '19 at 21:51
  • Let $(X_i)_{i\in I}$ be a family of sets. Its cartesian product is the set of pairs $(i,x)$ for $i\in I$ and $x\in X_i$. There's no assumption that $I$ is countable. – YCor Feb 28 '19 at 22:10
  • @YCor Your description of the cartesian product isn't what you intended. It should be the set of all functions $f$ on $I$ such that $f(i)\in X_i$ for all $i\in I$. But you're right that countability is irrelevant. – Andreas Blass Feb 28 '19 at 22:15
  • @AndreasBlass and what is a function on $I$? it's a set of pairs $(x,y)$ with $x\in I$ s.t... so, I wrote what I intended, I think. – YCor Feb 28 '19 at 22:40
  • @YCor The product is a set of functions, and these functions are, as you said, sets of ordered pairs. So the product is a set of sets of ordered pairs, not a set of ordered pairs. (For example, if $X_1=\mathbb Z/2$ and $X_2=\mathbb Z/3$, then there are only 5 ordered pairs of the sort described in your earlier comment.) – Andreas Blass Feb 28 '19 at 22:57
  • @AndreasBlass Oh, you're right... I defined the disjoint union... my mistake, sorry. – YCor Feb 28 '19 at 23:16

2 Answers2

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Yes, it will work. It's simple enough to check each axiom. For an example, consider the functions from $\mathbb{R}$ to $\mathbb{R}$. These form a group under addition and could be recast in your style as an uncountable direct product.

In your notation, set each $G_i$ to $\mathbb{R}$ and have the index range over $\mathbb{R}$. The result should be clearly isomorphic to my function example.

In your notation:

$$ \times_{i\in \mathbb{R}} G_i = \{ (g_i): g_i\in G_i\} $$

Set each $G_i$ to $\mathbb{R}$. The operation is just the usual addition of $\mathbb{R}$. If you use $\mathbb{R}^*$ instead, you could use multiplication.

Any element of this product group can be mapped to a function from $\mathbb{R}$ to $\mathbb{R}$. Define $f(i) = g_i$. This is clearly a bijection and a homomorphism and hence an isomorphism.

badjohn
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  • Thank you. Can you add a bit more detail to the uncountable case? I am not sure that I understand what is written on Wikipedia. – John Doe Feb 28 '19 at 21:23
  • Has my added detail helped? – badjohn Feb 28 '19 at 21:28
  • Could you be even more specific? For example, what is the underluing set for this group and how specifically is the operation defined? – John Doe Feb 28 '19 at 21:29
  • So this is just a question about defining products of sets. – YCor Feb 28 '19 at 21:52
  • @YCor Products of groups is the what the OP asked for. So, not quite any sets. – badjohn Feb 28 '19 at 21:53
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    His question in his last comment is how to define the underlying set. Once it's defined, it's obvious to make it a group. – YCor Feb 28 '19 at 21:54
  • @YCor Ah, I wasn't quite sure what he meant by that but, of course, he could just forget the operation for a while if he is interested in the set. – badjohn Feb 28 '19 at 21:58
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A simple example: Let our index group be the positive reals (the poster child for uncountable), call an arbitrary index by some Greek letter, and let our collection of groups $G_\alpha$ be the collections of additive groups on a circle of circumference $\alpha$.

(Thus for instance $G_{2\pi}$ is the group of points on a unit circle, where adding two points means going to an angle that is the sum of the angles of those two points.)

Then in our group $$H = \prod_{\alpha \in \Bbb{R}^+} G_\alpha$$

  • An arbitrary member of $H$ can be described by an arbitrary function $$f : x \in \Bbb{R^+} \rightarrow f(x) \in [0, x]$$

  • The group operator for $H$ does $$ h_1 + h_2 = f : x \in \Bbb{R^+} \rightarrow f(x) \in [0, x] : f(x) = \left( f_1(x) + f_2(x) \right)\pmod{x} $$

  • The identity element in $H$ is $$f : x \in \Bbb{R^+} \rightarrow f(x) = 0$$

  • The inverse of an arbitrary member of $h \in H$ (with function $f$ associated with $h$) is $$ h^{-1} = f : x \in \Bbb{R^+} \rightarrow g(x) \in [0, x] : g(x) = \left\{ \begin{array}{cl} x - f(x) & x >0 \\ 0 &x=0\end{array} \right. $$

Mark Fischler
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