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}

(3)
\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}

(5)
\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:

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

$(0, 0)$ | $(0, 0)$ | $(1, 0)$ | $(0, 1)$ | $(1, 1)$ |

$(1, 0)$ | $(1, 0)$ | $(0, 0)$ | $(1, 1)$ | $(0, 1)$ |

$(0, 1)$ | $(0, 1)$ | $(1, 1)$ | $(0, 0)$ | $(1, 0)$ |

$(1, 1)$ | $(1, 1)$ | $(0, 1)$ | $(1, 0)$ | $(0, 0)$ |