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6: Direct Products of Groups

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    74643
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    Recall that the Cartesian product \(X_1 \times X_2 \times \dots \times X_n\) of \(n\) sets \(X_1, X_2, \dots, X_n\) is the set of all ordered \(n\)-tuples \((x_1,x_2, \dots, x_n)\) where \(x_i \in X_i\) for all \(i \in \{ 1, 2, \dots, n\}\). Equality for \(n\)-tuples is defined by \[(x_1,x_2, \dots, x_n)=(y_1,y_2, \dots, y_n) \Longleftrightarrow \mbox{ $x_i = y_i$ for all $i \in \{1,2, \dots, n \}$}.\]

    Definition 6.1:

    If \(G_1, G_2, \dots, G_n\) is a list of \(n\) groups we make the Cartesian product \(G_1\times G_2 \times \dots \times G_n\) into a group by defining the binary operation \[(a_1,a_2, \dots, a_n) \cdot (b_1, b_2, \dots, b_n) = (a_1 \cdot b_1, a_2 \cdot b_2, \dots, a_n \cdot b_n).\] Here for each \(i \in \{ 1, 2, \dots, n \}\) the product \(a_i \cdot b_i\) is the product of \(a_i\) and \(b_i\) in the group \(G_i\). We call this group the direct product of the groups \(G_1, G_2, \dots, G_n\).

    As an example, consider the direct product \(\mathbb{Z}_2 \times \mathbb{Z}_3\) of the two groups \(\mathbb{Z}_2\) and \(\mathbb{Z}_3\). \[\mathbb{Z}_2 \times \mathbb{Z}_3 = \{ (0,0), \ (0,1), \ (0,2), \ (1,0), \ (1,1), \ (1,2) \}.\] We add modulo 2 in the first coordinate and modulo 3 in the second coordinate. Since the binary operation in each factor is addition, we use \(+\) for the operation in the direct product. So, for example, in this group \[(1,1) + (1,1) = (1 + 1, 1 + 1) = (0,2).\] The identity is clearly \((0,0)\) and, for example, the inverse of \((1,1)\) is \((1,2)\). It is clear that this is a group. More generally we have the following result.

    Theorem \(\PageIndex{1}\)

    If \(G_1, G_2, \dots, G_n\) is a list of \(n\) groups the direct product \(G=G_1\times G_2 \times \dots \times G_n\) as defined above is a group. Moreover, if for each \(i\), \(e_i\) is the identity of \(G_i\) then \((e_1,e_2, \dots, e_n)\) is the identity of G, and if \[\mathbf{a} = (a_1,a_2, \dots, a_n) \in G\] then the inverse of \(\mathbf{a}\) is given by \[\mathbf{a}^{-1} = (a_1^{-1},a_2^{-1}, \dots, a_n^{-1})\] where \(a_i^{-1}\) is the inverse of \(a_i\) in the group \(G_i\). \(\blacksquare\)

    Problem 6.1 Prove the above theorem for the special case \(n=2\).

    Problem 6.2 Find the order of each of the following groups. Also give the identity of each group and the inverse of just one element of the group other than the identity.

    1. \(\mathbb{Z}_2 \times \mathbb{Z}_2\)
    2. \(\mathbb{Z}_3 \times S_3 \times U_5\)
    3. \(\mathbb{Z}\times \mathbb{Z}_3 \times \mathbb{Z}_2\)
    4. \(GL(2,\mathbb{Z}_2) \times \mathbb{Z}_4 \times U_7 \times \mathbb{Z}_2\)


      This page titled 6: Direct Products of Groups is shared under a not declared license and was authored, remixed, and/or curated by W. Edwin Clark via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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