The Direct Product of Two Groups

The Direct Product of Two Groups

Suppose that $(G, \cdot)$ and $(H, *)$ are two groups. We would like to obtain a new group on the cartesian product $G \times H = \{ (g, h) : g \in G, h \in H \}$. There is a rather natural way to do this as we prove in the following theorem.

Theorem 1: Let $(G, \cdot)$ and $(H, *)$ be two groups. Define an operation on $G \times H$ by $(g_1, h_1) (g_2, h_2) = (g_1 \cdot g_2, h_1 * h_2)$. Then $G \times H$ with this operation is a group.
  • Proof: Let $e_G$ denote the identity in $G$ and let $e_H$ denote the identity in $H$.
  • It is trivially clear that $G \times H$ is closed under $\cdot$.
  • Let $(g_1, h_1), (g_2, h_2), (g_3, h_3) \in G \times H$. Then by the associativity in the components of the following product we have that:
\begin{align} \quad [(g_1, h_1)(g_2, h_2)] (g_3, h_3) &= (g_1 \cdot g_2, h_1 * h_2)(g_3, h_3) \\ &= ([g_1 \cdot g_2] \cdot g_3, [h_1 * h_2] * h_3) \\ &= (g_1 \cdot [g_2 \cdot g_3], h_1 * [h_2 * h_3]) \\ &= (g_1, h_1)(g_2 \cdot g_3, h_2 * h_3) \\ &= (g_1, h_1)[(g_2, h_2) (g_3, h_3)] \end{align}
  • So this operation is associative on $G \times H$.
  • Let $e = (e_G, e_H) \in G \times H$. Then for all $(g, h) \in G \times H$ we have that:
\begin{align} \quad e (g, h) = (e_G, e_H)(g, h) = (e_G \cdot g, e_H \cdot h) = (g, h) \end{align}
\begin{align} \quad (g, h)e = (g, h) (e_G, e_H) = (g \cdot e_G, h * e_H) = (g, h) \end{align}
  • So an identity for $\cdot$ exists in $G \times H$.
  • Lastly, for all $(g, h) \in G \times H$ let $(g, h)^{-1} = (g^{-1}, h^{-1})$. Then:
\begin{align} \quad (g, h)(g, h)^{-1} = (g, h)(g^{-1}, h^{-1}) = (g \cdot g^{-1}, h \cdot h^{-1}) = (e_G, e_H) = e \end{align}
\begin{align} \quad (g, h)^{-1} (g, h) = (g^{-1}, h^{-1})(g, h) = (g^{-1} \cdot g, h^{-1} \cdot h) = (e_G, e_H) = e \end{align}
  • Therefore $(G \times H, \cdot)$ is indeed a group. $\blacksquare$

The group $G \times H$ with the operation above is well-defined for any two groups $(G, \cdot)$ and $(H, *)$ so we give it a special name.

Definition: Let $(G, \cdot)$ and $(H, \cdot)$ be two groups. Then the Direct Product of these two groups is the new group $G \times H$ with the binary operation on $G \times H$ defined for all $(g_1, h_1), (g_2, h_2) \in G \times H$ by $(g_1, h_1)(g_2, h_2) = (g_1 \cdot g_2, h_1 * h_2)$.

Sometimes the term "External Direct Product" is used in the above definition to avoid confusion with the "internal direct product" which we define later.

For example, if we consider the group $(\mathbb{Z}_2, +)$ then the external direct product of this group with itself is the group $\mathbb{Z}_2 \times \mathbb{Z}_2$ whose operation table is given by:

$\cdot$ $(0, 0)$ $(1, 0)$ $(0, 1)$ $(1, 1)$
$(0, 0)$ $(0, 0)$ $(1, 0)$ $(0, 1)$ $(1, 1)$
$(1, 0)$ $(1, 0)$ $(0, 0)$ $(1, 1)$ $(0, 1)$
$(0, 1)$ $(0, 1)$ $(1, 1)$ $(0, 0)$ $(1, 0)$
$(1, 1)$ $(1, 1)$ $(0, 1)$ $(1, 0)$ $(0, 0)$
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