The Internal Direct Product of Two Groups

# The Internal Direct Product of Two Groups

Recall from The External Direct Product of Two Groups page that if $(G, *)$ and $(H, +)$ are two groups then the external direct product of these groups is the new group $(G \times H, \cdot)$ where $\cdot$ is the binary operation defined for all $(g_1, h_1), (g_2, h_2) \in G \times H$ by:

(1)
\begin{align} \quad (g_1, h_1) \cdot (g_2, h_2) = (g_1 * g_2, h_1 + h_2) \end{align}

We will now define another type of "direct product" known as the internal direct product of two groups.

 Definition: Let $(G, *)$ be a group and let $(H, *)$ and $(K, *)$ be two subgroups of $(G, *)$. Then $(G, *)$ is said to be the Internal Direct Product of $(H, *)$ and $(K, *)$ if: 1) $G = \{ h * k : h \in H, k \in K \}$. 2) $H \cap K = \{ e \}$ where $e$ is the identity element in $G$. 3) $h * k = k * h$ for all $h \in H$ and for all $k \in K$.

For example, consider the group $(\mathbb{Z}_6, +)$ and the following subgroups:

(2)
\begin{align} \quad H = \{ 0, 2, 4 \} \end{align}
(3)
\begin{align} \quad K = \{ 0, 3 \} \end{align}

Note that $\{ h * k : h \in H, k \in K \} = \{ 0, 1, 2, 3, 4, 5 \} = G$, so the first condition is met. Also the identity for $\mathbb{Z}_6$ is $e = 0$ and $H \cap K = \{ 0 \}$ so the second condition is met. Lastly $\mathbb{Z}_6$ is an abelian group so the third condition is met.