## Section9.2Direct Products

Given two groups $$G$$ and $$H\text{,}$$ it is possible to construct a new group from the Cartesian product of $$G$$ and $$H\text{,}$$ $$G \times H\text{.}$$ 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 $$G\text{,}$$ it is often easier to study the component groups of $$G\text{.}$$

### SubsectionExternal Direct Products

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 $$h \in H\text{.}$$ 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 $$H\text{.}$$ We have specified the particular operations $$\cdot$$ and $$\circ$$ in each group here for the sake of clarity; we usually just write $$(g_1, h_1)(g_2, h_2) = (g_1 g_2, h_1 h_2)\text{.}$$

Clearly the binary operation defined above is closed. If $$e_G$$ and $$e_H$$ are the identities of the groups $$G$$ and $$H$$ respectively, then $$(e_G, e_H)$$ is the identity of $$G \times H\text{.}$$ The inverse of $$(g, h) \in G \times H$$ is $$(g^{-1}, h^{-1})\text{.}$$ The fact that the operation is associative follows directly from the associativity of $$G$$ and $$H\text{.}$$

#### Example9.14.

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

#### Example9.15.

Consider

\begin{equation*} {\mathbb Z}_2 \times {\mathbb Z}_2 = \{ (0, 0), (0, 1), (1, 0),(1, 1) \}\text{.} \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 $${\mathbb Z}_2 \times {\mathbb Z}_2$$ other than the identity has order $$2\text{,}$$ since $$(a,b) + (a,b) = (0,0)\text{;}$$ however, $${\mathbb Z}_4$$ is cyclic.

The group $$G \times H$$ is called the external direct product of $$G$$ and $$H\text{.}$$ 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 $$G = G_1 = G_2 = \cdots = G_n\text{,}$$ we often write $$G^n$$ instead of $$G_1 \times G_2 \times \cdots \times G_n\text{.}$$

#### Example9.16.

The group $${\mathbb Z}_2^n\text{,}$$ 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)\text{.} \end{equation*}

This group is important in coding theory, in cryptography, and in many areas of computer science.

Suppose that $$m$$ is the least common multiple of $$r$$ and $$s$$ and let $$n = |(g,h)|\text{.}$$ Then

\begin{gather*} (g,h)^m = (g^m, h^m) = (e_G,e_H)\\ (g^n, h^n) = (g, h)^n = (e_G,e_H)\text{.} \end{gather*}

Hence, $$n$$ must divide $$m\text{,}$$ and $$n \leq m\text{.}$$ However, by the second equation, both $$r$$ and $$s$$ must divide $$n\text{;}$$ therefore, $$n$$ is a common multiple of $$r$$ and $$s\text{.}$$ Since $$m$$ is the least common multiple of $$r$$ and $$s\text{,}$$ $$m \leq n\text{.}$$ Consequently, $$m$$ must be equal to $$n\text{.}$$

#### Example9.19.

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

#### 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)\text{.} \end{align*}

In this case, unlike that of $${\mathbb Z}_2 \times {\mathbb Z}_2$$ and $${\mathbb Z}_4\text{,}$$ it is true that $${\mathbb Z}_2 \times {\mathbb Z}_3 \cong {\mathbb Z}_6\text{.}$$ 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 $${\mathbb Z}_2 \times {\mathbb Z}_3\text{.}$$

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

We will first show that if $${\mathbb Z}_m \times {\mathbb Z}_n \cong {\mathbb Z}_{mn}\text{,}$$ then $$\gcd(m, n) = 1\text{.}$$ We will prove the contrapositive; that is, we will show that if $$\gcd(m, n) = d \gt 1\text{,}$$ then $${\mathbb Z}_m \times {\mathbb Z}_n$$ cannot be cyclic. Notice that $$mn/d$$ is divisible by both $$m$$ and $$n\text{;}$$ hence, for any element $$(a,b) \in {\mathbb Z}_m \times {\mathbb Z}_n\text{,}$$

\begin{equation*} \underbrace{(a,b) + (a,b)+ \cdots + (a,b)}_{mn/d \; \text{times}} = (0, 0)\text{.} \end{equation*}

Therefore, no $$(a, b)$$ can generate all of $${\mathbb Z}_m \times {\mathbb Z}_n\text{.}$$

The converse follows directly from Theorem 9.17 since $$\lcm(m,n) = mn$$ if and only if $$\gcd(m,n)=1\text{.}$$

Since the greatest common divisor of $$p_i^{e_i}$$ and $$p_j^{e_j}$$ is 1 for $$i \neq j\text{,}$$ the proof follows from Corollary 9.22.

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.

### SubsectionInternal Direct Products

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.

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

• $$H \cap K = \{ e \}\text{;}$$

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

Then $$G$$ is the internal direct product of $$H$$ and $$K\text{.}$$

#### Example9.24.

The group $$U(8)$$ is the internal direct product of

\begin{equation*} H = \{1, 3 \} \quad \text{and} \quad K = \{1, 5 \}\text{.} \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 \}\text{.} \end{equation*}

It can easily be shown that $$K \cong S_3\text{;}$$ consequently, $$D_6 \cong {\mathbb Z}_2 \times S_3\text{.}$$

#### 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 $$K\text{,}$$ then one of the subgroups, say $$H\text{,}$$ would have to have order $$3\text{.}$$ In this case $$H$$ is the subgroup $$\{ (1), (123), (132) \}\text{.}$$ The subgroup $$K$$ must have order $$2\text{,}$$ but no matter which subgroup we choose for $$K\text{,}$$ the condition that $$hk = kh$$ will never be satisfied for $$h \in H$$ and $$k \in K\text{.}$$

Since $$G$$ is an internal direct product, we can write any element $$g \in G$$ as $$g =hk$$ for some $$h \in H$$ and some $$k \in K\text{.}$$ Define a map $$\phi : G \rightarrow H \times K$$ by $$\phi(g) = (h,k)\text{.}$$

The first problem that we must face is to show that $$\phi$$ is a well-defined map; that is, we must show that $$h$$ and $$k$$ are uniquely determined by $$g\text{.}$$ Suppose that $$g = hk=h'k'\text{.}$$ Then $$h^{-1} h'= k (k')^{-1}$$ is in both $$H$$ and $$K\text{,}$$ so it must be the identity. Therefore, $$h = h'$$ and $$k = k'\text{,}$$ which proves that $$\phi$$ is, indeed, well-defined.

To show that $$\phi$$ preserves the group operation, let $$g_1 = h_1 k_1$$ and $$g_2 = h_2 k_2$$ and observe that

\begin{align*} \phi( g_1 g_2 ) & = \phi( h_1 k_1 h_2 k_2 )\\ & = \phi(h_1 h_2 k_1 k_2)\\ & = (h_1 h_2, k_1 k_2)\\ & = (h_1, k_1)( h_2, k_2)\\ & = \phi( g_1 ) \phi( g_2 )\text{.} \end{align*}

We will leave the proof that $$\phi$$ is one-to-one and onto as an exercise.

#### Example9.28.

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

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 $$G\text{,}$$ by requiring that

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

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

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

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