## Definition of the n-ary direct product of groups

Let $I$ be any nonempty set and let $G_i$ be a group for each $i \in I$. The direct product of the groups $G_i$ is the set $G = \prod_{i \in I} G_i$ (the Cartesian product of the $G_i$) with a binary operation defined as follows: if $\prod a_i$ and $\prod b_i$ are elements of $G$, then their product is defined $(\prod a_i) (\prod b_i) = \prod (a_i b_i)$. I.e., the product is defined componentwise.

1. Show that this operation is well defined and associative.
2. Show that the element $\prod 1_i$ is an identity in $G$.
3. Show that the element $\prod a_i^{-1}$ is the inverse of $\prod a_i$.

Conclude that $G$ is a group.

1. Elements in a direct product are uniquely represented, so well-definedness is not an issue. Suppose $\prod a_i$, $\prod b_i$, and $\prod c_i$ are in $G$. Then $\left[(\prod a_i) (\prod b_i)\right] (\prod c_i) = (\prod a_ib_i)(\prod c_i)$ $= \prod (a_ib_i)c_i$ $= \prod a_i(b_ic_i)$ $= (\prod a_i)(\prod b_ic_i)$ $= (\prod a_i)\left[(\prod b_i)(\prod c_i)\right]$, so that the product in $G$ is associative.
2. We have $(\prod 1_i)(\prod a_i) = \prod 1_ia_i$ $= \prod a_i$; likewise $(\prod a_i)(\prod 1_i) = \prod a_i$. Thus $1 = \prod 1_i$ is an identity in $G$ under componentwise multiplication.
3. We have $(\prod a_i)(\prod a_i^{-1}) = \prod a_ia_i^{-1} = \prod 1_i$; likewise $(\prod a_i^{-1})(\prod a_i) = \prod 1_i$. Thus $(\prod a_i)^{-1} = (\prod a_i)$.

Thus $G$ is a group under componentwise multiplication.