# The Direct Product of an Arbitrary Collection of Groups

Recall from The Direct Product of Two Groups page that if $(G, \cdot)$ and $(H, *)$ are groups then the direct product of these groups is another group, $G \times H$ with the operation defined for all $(g_1, h_1), (g_2, h_2) \in G \times H$ by:

(1)We will now extend this notion to an arbitrary collection of groups.

Let $\{ G_i: i \in I \}$ be an arbitrary collection of groups. The cartesian product $\displaystyle{\prod_{i \in I} G_i}$ is defined to be:

(2)For example, if $I = \{ 1, 2 \}$ then $\{ G_i : i \in I \} = \{ G_1, G_2 \}$, and an element $(g, h) \in G \times H$ can be thought of as the function $f : \{ 1, 2 \} \to G_1 \cup G_2$ given by $f(1) = g$, $f(2) = h$, so that $(g, h) = (f(1), f(2))$.

Note that if $I$ is a countably infinite set, i.e., if $I = \mathbb{N}$ then $\prod_{i=1}^{n} G_i$ can be thought of as all sequences $(g_n) \in \bigcup_{i=1}^{\infty}$ such that $g_n \in G_n$ for each $n \in \mathbb{N}$.

So we now construct the direct product for an arbitrary product of groups.

Proposition 1: Let $\{ G_i : i \in I \}$ be an arbitrary collection of groups. For all $f, g \in \prod_{i \in I} G_i$ define $fg : I \to \bigcup_{i \in I} G_i$ for each $i \in I$ by $(fg)(i) = f(i)g(i)$. Then $\prod_{i \in I} G_i$ with this operation is a group. |

**Proof:**For all $f, g \in \prod_{i \in I} G_i$ we have that for each $i \in I$ that $(fg)(i) = f(i)g(i) \in G_i$ since each $G_i$ is a group, and so $fg \in \prod_{i \in I} G_i$, i.e., $\prod_{i \in I} G_i$ is closed under this operation.

- Let $f, g, h \in \prod_{i \in I} G_i$. Then for each $i \in I$ we have by the associativity of the operation in $G_i$ that:

- Since this holds true for each $i \in I$ we see that $f(gh) = (fg)h$, so the operation on $\prod_{i \in I} G_i$ is associative.

- For each $i \in I$ let $e_{G_i}$ denote the identity in $G_i$. Let $e : I \to\bigcup_{i \in I} G_i$ be defined for each $i \in I$ by $e(i) = e_{G_i}$. Then $e \in \prod_{i \in I} G_i$ and for all $f \in \prod_{i \in I} G_i$ we have that for all $i \in I$:

- So $ef = f$ and $fe = f$, i.e., $e$ is the identity element for the operation on $\prod_{i \in I} G_i$.

- Lastly, for each $f \in \prod_{i \in I} G_i$ let $f^{-1} : I \to \prod_{i \in I} G_i$ be defined for all $i \in I$ by $f^{-1}(i) = [f(i)]^{-1}$. Then for each $i \in I$ we have that:

- So $ff^{-1} = f^{-1}f = e$. Thus every $f \in \prod_{i \in I} G_i$ has an inverse $f^{-1} \in \prod_{i \in I} G_i$. Therefore $\prod_{i \in I} G_i$ with this operation forms a group. $\blacksquare$

Definition: Let $\{ G_i : i \in I \}$ be an arbitrary collection of groups. The Direct Product of these groups is defined to be the set $\prod_{i \in I} G_i$ with the operation defined for all $f, g \in \prod_{i \in I} G_i$ by $fg : I \in \bigcup_{i \in I} G_i$ where $(fg)(i) = f(i)g(i)$ for all $i \in I$. |