###### Proposition9.13

Let \(G\) and \(H\) be groups. The set \(G \times H\) is a group under the operation \((g_1, h_1)(g_2, h_2) = (g_1 g_2, h_1 h_2)\) where \(g_1, g_2 \in G\) and \(h_1, h_2 \in H\text{.}\)

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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 \(G\text{.}\)

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 \((g_1, h_1)(g_2, h_2) = (g_1 g_2, h_1 h_2)\text{.}\)

Let \(G\) and \(H\) be groups. The set \(G \times H\) is a group under the operation \((g_1, h_1)(g_2, h_2) = (g_1 g_2, h_1 h_2)\) where \(g_1, g_2 \in G\) and \(h_1, h_2 \in H\text{.}\)

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 \(H\text{.}\)

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 \((-a, -b)\text{.}\)

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 \((a,b) + (a,b) = (0,0)\text{;}\) 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

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 \(G_1 \times G_2 \times \cdots \times G_n\text{.}\)

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.

Let \((g, h) \in G \times H\text{.}\) If \(g\) and \(h\) have finite orders \(r\) and \(s\) respectively, then the order of \((g, h)\) in \(G \times H\) is the least common multiple of \(r\) and \(s\text{.}\)

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 \(n\text{.}\)

Let \((g_1, \ldots, g_n) \in \prod G_i\text{.}\) If \(g_i\) has finite order \(r_i\) in \(G_i\text{,}\) then the order of \((g_1, \ldots, g_n)\) in \(\prod G_i\) is the least common multiple of \(r_1, \ldots, r_n\text{.}\)

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; hence, \((8, 56)\) has order 15 in \({\mathbb Z}_{12} \times {\mathbb Z}_{60}\text{.}\)

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.

The group \({\mathbb Z}_m \times {\mathbb Z}_n\) is isomorphic to \({\mathbb Z}_{mn}\) if and only if \(\gcd(m,n)=1\text{.}\)

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 \(\gcd(m,n)=1\text{.}\)

Let \(n_1, \ldots, n_k\) be positive integers. Then

\begin{equation*} \prod_{i=1}^k {\mathbb Z}_{n_i} \cong {\mathbb Z}_{n_1 \cdots n_k} \end{equation*}if and only if \(\gcd( n_i, n_j) =1\) for \(i \neq j\text{.}\)

If

\begin{equation*} m = p_1^{e_1} \cdots p_k^{e_k}, \end{equation*}where the \(p_i\)s are distinct primes, then

\begin{equation*} {\mathbb Z}_m \cong {\mathbb Z}_{p_1^{e_1}} \times \cdots \times {\mathbb Z}_{p_k^{e_k}}. \end{equation*}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.

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 \(K\text{.}\)

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

\begin{equation*} H = \{1, 3 \} \quad \text{and} \quad K = \{1, 5 \}. \end{equation*}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, \(D_6 \cong {\mathbb Z}_2 \times S_3\text{.}\)

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

Let \(G\) be the internal direct product of subgroups \(H\) and \(K\text{.}\) Then \(G\) is isomorphic to \(H \times K\text{.}\)

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.

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.

Let \(G\) be the internal direct product of subgroups \(H_i\text{,}\) where \(i = 1, 2, \ldots, n\text{.}\) Then \(G\) is isomorphic to \(\prod_i H_i\text{.}\)