Direct Product of Groups

Let G an H be groups. The Cartesian product G H consists of the ordered pairs (g,h) for g G, h H. Let us put a group structure on G H. We define multiplication in G H by (1)

It is clear that (1) defines a binary operation on G H. Associativity can be shown easily. Since

we see that (1_G,1_H) is an identity element. Moreover

so that every element of G H has an inverse with respect to the multiplication (1). Thus, with respect to the multiplication (1), G H is a group.

Definition 2: G H is called the direct product of G and H.

Example 1: Let G = H = Z₂. Then

Z₂ Z₂ = {(a,b) | a,b Z₂}

The identity element is (0,0), and if we set I = (0,0), A = (0,1), B = (1,0), C = (1,1), then we easily see that

Thus Z₂ Z₂ is isomorphic to the Klein 4-group, which was discussed in the section of group isomorphism. (Another way of establishing this fact is to observe that Z₂ Z₂ is a group of order 4 which is not cyclic. There is only one such group - the Klein 4-group.) Note that this example show that a direct product of a cyclic group need not be cyclic.

Example 2: Let G = Z₂, H = Z₃. Then

G H = {(a,b) | a Z₂, b Z₃}

Note that (0,0) is the identity and that if = (1,1), then

² = (0,2),

³ = (1,0),

⁴ = (0,1),

⁵ = (1,2),

⁶ = (0,0),

so that has order 6. Thus, Z₂ Z₃ is a cyclic group of order 6, with as a generator.

Let G₁,...,G_n be a finite collection of groups. The direct product G₁ ... G_n is the group whose elements are the ordered n-tuples (g₁,....g_n), where g_i G_i (1 < i < n), and whose multiplication is defined by

The verification that G₁ ... G_n is a group is similar to the proof given above for the special case n = 2.

If G₁,G₂, and G₃ are groups, we can form the direct product (G₁ G₂) G₃, whose elements are of the form ((g₁,g₂),g₃) (g_i G_i, i = 1,2,3). It is trivial to see that the mapping

:(G₁ G₂) G₃ G₁ G₂ G₃,

(((g₁,g₂),g₃)) = (g₁,g₂,g₃),

is a surjective isomorphism, so that

(G₁ G₂) G₃ G₁ G₂ G₃.

Similarly,

G₁ (G₂ G₃) G₁ G₂ G₃.

Subsequently, we will always identify the groups (G₁ G₂) G₃, G₁ (G₂ G₃), G₁ G₂ G₃. Similar comments apply to direct products of more than three groups. For example, G₁ (G₂ G₃) G₄ will be identified with G₁ G₂ G₃ G₄, and so on.

The next result is a convenient one for decomposing a given group into a direct product of two subgroups.

Theorem 3: Let G be a group. Let H < G, K < G. Assume that H and K satisfy the following:

(1) H K = {1}.

(2) If h H, k K, then hk = kh.

(3) If g G, then there exist h H, k K such that g= hk.

Then G H K.

Proof: First let us prove that if g G, then g can be uniquely written in the form g = hk (h H, k K). In deed, if g = hk = h'k', then

h'^-1h = k'k^-1 H K = {1},

h'^-1h = k'k^-1 = 1,

h = h', k = k'.

Let :G H K be the mapping described by (g) = (h,k), where g = hk. This mapping is well defined by (3) and the fact that the representation of g is unique. It is clear that is surjective since (hk) = (h,k) for any h H, k K. It is also clear that is injective. Thus, it suffices to show that is a homomorphism for then G H K. Let g = hk, g' = h'k'. Then, by (2), gg' = (hk)(h'k') = (hh')(kk'), so that

(gg') = (hh',kk')

= (g)(g')

Thus, is a homomorphism.

A simple, but important application of Theorem 3 is the following.

Theorem 4: Let m and n be positive integers such that (m,n) = 1. Then if G is a cyclic group of order mn, we have

G Z_m Z_n.

Proof: Let a be a generator of G. By Theorem 16 of the section of subgroups, the order of a^m is n and the order of aⁿ is m. Let H = [aⁿ], K = [a^m]. Then H and K are cyclic groups of orders m and n, respectively. Therefore, by Theorem 3 of the section on group isomorphism, we see that

H Z_m,    K Z_n.

Therefore, it suffices to prove that G H K. Let us apply Theorem 3. Let g H K. Then the order of g divides the orders of both H and K, so that g is a common divisor of m and n. But since (m,n) = 1, we see that f is of order 1 - that is, g = 1_G and H K = {1_G}. Every element of H commutes with every element of K since G is abelian. Finally, if g K, then g = a^r for some integer r. Since (m,n) = 1, we see that there exist integers c and d such that mc + nd = 1. Therefore, r = mcr + ndr, and

Therefore (1)-(3) of Theorem 3 hold, and G H K.

Corollary 5: Let G be a cyclic group of order p₁^a₁p₂^a₂...p_t^a_t, where p₁,...,p_t are distinct primes and a₁,...,a_t are positive integers. Then

G Z_p1^a1 ... Z_pt^at.

Proof: Let us proceed by induction on t. The assertion is true for t = 1. Assume t > 1 and the assertion t - 1. Set m = p₁^a₁, n = p₂^a₂...p_t^a_t. Then by Theorem 4 above,

G Z_p1^a1 Z_n

But Z_n is a cyclic group of order p₂^a₂...p_t^a_t. Therefore, by the assertion for t - 1, we see that

(3)

Z_n Z_p2^a2 ... Z_pt^at.

From (2) and (3), we conclude the assertion for t and the induction step is proved.