Given two groups $G$ and $H$, it is possible to construct a new group from the Cartesian product of $G$ and $H$, $G \times H$. Conversely, given a large group, it is sometimes possible to decompose the group; that is, a group is sometimes isomorphic to the direct product of two smaller groups. Rather than studying a large group $G$, it is often easier to study the component groups of $G$.

If $(G,\cdot)$ and $(H, \circ)$ are groups, then we can make the Cartesian product of $G$ and $H$ into a new group. As a set, our group is just the ordered pairs $(g, h) \in G \times H$ where $g \in G$ and $h \in H$. We can define a binary operation on $G \times H$ by \begin{equation*}(g_1, h_1)(g_2, h_2) = (g_1 \cdot g_2, h_1 \circ h_2);\end{equation*} that is, we just multiply elements in the first coordinate as we do in $G$ and elements in the second coordinate as we do in $H$. We have specified the particular operations $\cdot$ and $\circ$ in each group here for the sake of clarity; we usually just write $(g_1, h_1)(g_2, h_2) = (g_1 g_2, h_1 h_2)$.

##### Example9.14

Let ${\mathbb R}$ be the group of real numbers under addition. The Cartesian product of ${\mathbb R}$ with itself, ${\mathbb R} \times {\mathbb R} = {\mathbb R}^2$, is also a group, in which the group operation is just addition in each coordinate; that is, $(a, b) + (c, d) = (a + c, b + d)$. The identity is $(0,0)$ and the inverse of $(a, b)$ is $(-a, -b)$.

##### Example9.15

Consider \begin{equation*}{\mathbb Z}_2 \times {\mathbb Z}_2 = \{ (0, 0), (0, 1), (1, 0),(1, 1) \}.\end{equation*} Although ${\mathbb Z}_2 \times {\mathbb Z}_2$ and ${\mathbb Z}_4$ both contain four elements, they are not isomorphic. Every element $(a,b)$ in ${\mathbb Z}_2 \times {\mathbb Z}_2$ has order 2, since $(a,b) + (a,b) = (0,0)$; however, ${\mathbb Z}_4$ is cyclic.

The group $G \times H$ is called the external direct product of $G$ and $H$. Notice that there is nothing special about the fact that we have used only two groups to build a new group. The direct product \begin{equation*}\prod_{i = 1}^n G_i = G_1 \times G_2 \times \cdots \times G_n\end{equation*} of the groups $G_1, G_2, \ldots, G_n$ is defined in exactly the same manner. If $G = G_1 = G_2 = \cdots = G_n$, we often write $G^n$ instead of $G_1 \times G_2 \times \cdots \times G_n$.

##### Example9.16

The group ${\mathbb Z}_2^n$, considered as a set, is just the set of all binary $n$-tuples. The group operation is the “exclusive or” of two binary $n$-tuples. For example, \begin{equation*}(01011101) + (01001011) = (00010110).\end{equation*} This group is important in coding theory, in cryptography, and in many areas of computer science.

##### Example9.19

Let $(8, 56) \in {\mathbb Z}_{12} \times {\mathbb Z}_{60}$. Since $\gcd(8,12) = 4$, the order of 8 is $12/4 = 3$ in ${\mathbb Z}_{12}$. Similarly, the order of $56$ in ${\mathbb Z}_{60}$ is $15$. The least common multiple of 3 and 15 is 15; hence, $(8, 56)$ has order 15 in ${\mathbb Z}_{12} \times {\mathbb Z}_{60}$.

##### Example9.20

The group ${\mathbb Z}_2 \times {\mathbb Z}_3$ consists of the pairs \begin{align*} & (0,0), & & (0, 1), & & (0, 2), & & (1,0), & & (1, 1), & & (1, 2). \end{align*} In this case, unlike that of ${\mathbb Z}_2 \times {\mathbb Z}_2$ and ${\mathbb Z}_4$, it is true that ${\mathbb Z}_2 \times {\mathbb Z}_3 \cong {\mathbb Z}_6$. We need only show that ${\mathbb Z}_2 \times {\mathbb Z}_3$ is cyclic. It is easy to see that $(1,1)$ is a generator for ${\mathbb Z}_2 \times {\mathbb Z}_3$.

The next theorem tells us exactly when the direct product of two cyclic groups is cyclic.

##### Proof

In Chapter 13, we will prove that all finite abelian groups are isomorphic to direct products of the form \begin{equation*}{\mathbb Z}_{p_1^{e_1}} \times \cdots \times {\mathbb Z}_{p_k^{e_k}}\end{equation*} where $p_1, \ldots, p_k$ are (not necessarily distinct) primes.

The external direct product of two groups builds a large group out of two smaller groups. We would like to be able to reverse this process and conveniently break down a group into its direct product components; that is, we would like to be able to say when a group is isomorphic to the direct product of two of its subgroups.

Let $G$ be a group with subgroups $H$ and $K$ satisfying the following conditions.

• $G = HK = \{ hk : h \in H, k \in K \}$;

• $H \cap K = \{ e \}$;

• $hk = kh$ for all $k \in K$ and $h \in H$.

Then $G$ is the internal direct product of $H$ and $K$.

##### Example9.24

The group $U(8)$ is the internal direct product of \begin{equation*}H = \{1, 3 \} \quad \text{and} \quad K = \{1, 5 \}.\end{equation*}

##### Example9.25

The dihedral group $D_6$ is an internal direct product of its two subgroups \begin{equation*}H = \{\identity, r^3 \} \quad \text{and} \quad K = \{\identity, r^2, r^4, s, r^2s, r^4 s \}.\end{equation*} It can easily be shown that $K \cong S_3$; consequently, $D_6 \cong {\mathbb Z}_2 \times S_3$.

##### Example9.26

Not every group can be written as the internal direct product of two of its proper subgroups. If the group $S_3$ were an internal direct product of its proper subgroups $H$ and $K$, then one of the subgroups, say $H$, would have to have order 3. In this case $H$ is the subgroup $\{ (1), (123), (132) \}$. The subgroup $K$ must have order 2, but no matter which subgroup we choose for $K$, the condition that $hk = kh$ will never be satisfied for $h \in H$ and $k \in K$.

##### Example9.28

The group ${\mathbb Z}_6$ is an internal direct product isomorphic to $\{ 0, 2, 4\} \times \{ 0, 3 \}$.

We can extend the definition of an internal direct product of $G$ to a collection of subgroups $H_1, H_2, \ldots, H_n$ of $G$, by requiring that

• $G = H_1 H_2 \cdots H_n = \{ h_1 h_2 \cdots h_n : h_i \in H_i \}$;

• $H_i \cap \langle \cup_{j \neq i} H_j \rangle = \{ e \}$;

• $h_i h_j = h_j h_i$ for all $h_i \in H_i$ and $h_j \in H_j$.

We will leave the proof of the following theorem as an exercise.