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.

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