
## 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) \}. \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 $2\text{,}$ 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). \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). \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). \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). \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 \}. \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 $K \cong S_3\text{;}$ 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 ). \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.