The External Direct Product of Two Groups

# The External Direct Product of Two Groups

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

Theorem 1: Let $(G, *)$ and $(H, +)$ be two groups. Define an operation $\cdot$ on $G \times H$ by $(g_1, h_1) \cdot (g_2, h_2) = (g_1 * g_2, h_1 + h_2)$. Then $(G \times H, \cdot)$ is a group. |

**Proof:**Let $e_1$ denote the identity in $G$ and let $e_2$ 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) \cdot (g_2, h_2)] \cdot (g_3, h_3) &= (g_1 * g_2, h_1 + h_2) \cdot (g_3, h_3) \\ &= ([g_1 * g_2] * g_3, [h_1 + h_2] + h_3) \\ &= (g_1 * [g_2 * g_3], h_1 + [h_2 + h_3]) \\ &= (g_1, h_1) \cdot (g_2 * g_3, h_2 + h_3) \\ &= (g_1, h_1) \cdot [(g_2, h_2) \cdot (g_3, h_3)] \end{align}

- So $\cdot$ is associative on $G \times H$.

- Let $e = (e_1, e_2) \in G \times H$. Then for all $(g, h) \in G \times H$ we have that:

\begin{align} \quad e \cdot (g, h) = (e_1, e_2) \cdot (g, h) = (e_1 * g, e_2 + h) = (g, h) \end{align}

(3)
\begin{align} \quad (g, h) \cdot e = (g, h) \cdot (e_1, e_2) = (g * e_1, h + e_2) = (g, h) \end{align}

- So an identity for $\cdot$ exists in $G \times H$.

- Lastly, for all $(g, h) \in G \times H$ let $(g, h)^{-1} = (g^{-1}, -h)$. Then:

\begin{align} \quad (g, h) \cdot (g, h)^{-1} = (g, h) \cdot (g^{-1}, -h) = (g * g^{-1}, h + (-h)) = (e_1, e_2) = e \end{align}

(5)
\begin{align} \quad (g, h)^{-1} \cdot (g, h) = (g^{-1}, -h) \cdot (g, h) = (g^{-1} * g, -h + h) = (e_1, e_2) = e \end{align}

- Therefore $(G \times H, \cdot)$ is indeed a group. $\blacksquare$

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

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

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

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

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

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

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

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