Is there such a thing as direct product of an infinite number of groups?

Question

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".

Answer 1

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 $$ h^{-1} = f : x \in \Bbb{R^+} \rightarrow g(x) \in [0, x] : g(x) = \left\{ \begin{array}{cl} x - f(x) & x >0 \\ 0 &x=0\end{array} \right. $$

Answer 2

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.