commutative A group is considered cyclic if it is generated by a single element. A function f : AB is said to be injective (or one to one) if whenever f(x) = f(y), we have x = y. Let a have order n. then am has order n/(n,m). In particular if (m,n) = 1, then am has order n. 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..
A function f : AB is said to be surjective (or onto) if for every y B there exists x A such that f(x) = y. Let G1 and G2 be groups. An isomorphism from G1 to G2 is an injection :G1G2 such that (a · b) = (a) · (b)  (a,b G1). The set residue classes mod n.

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)

(g,h)(g',h') = (gg',hh').

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

(g,h)(1G,1H) = (g·1g, h·1H) = (g,h),
(1G,1H)(g,h) = (1G·g,1H·h) = (g,h),

we see that (1G,1H) is an identity element. Moreover

(g,h)(g-1,h-1) = (g-1,h-1)(g,h) = (1G,1H),

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}
= {(0,0),(0,1),(1,0),(1,1)}.

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

AB = BA,    A2 = B2 = I,    C = AB.

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}
= {(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)}.

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,...,Gn be a finite collection of groups. The direct product G1 ... Gn is the group whose elements are the ordered n-tuples (g1,....gn), where gi Gi (1 < i < n), and whose multiplication is defined by

(g1,...,gn)(g1',...,gn') = (g1g1',...,gngn').

The verification that G1 ... Gn 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) (gi Gi, 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')
= (h,k)(h',k')
= (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 Zm Zn.

Proof: Let a be a generator of G. By Theorem 16 of the section of subgroups, the order of am is n and the order of an is m. Let H = [an], K = [am]. 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 Zm,    K Zn.

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 = 1G and H K = {1G}. Every element of H commutes with every element of K since G is abelian. Finally, if g K, then g = ar 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

g = ar = amcr+ndr
= (am)cr·(an)dr.

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

Corollary 5: Let G be a cyclic group of order p1a1p2a2...ptat, where p1,...,pt are distinct primes and a1,...,at are positive integers. Then

G Zp1a1 ... Zptat.

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 = p1a1, n = p2a2...ptat. Then by Theorem 4 above,

(2)
G Zp1a1 Zn

But Zn is a cyclic group of order p2a2...ptat. Therefore, by the assertion for t - 1, we see that

(3)
Zn Zp2a2 ... Zptat.

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