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commutative A group is considered cyclic if it is generated by a single element. A function f : AmapsB 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 intersection K = {1}.
(2) If h element of H, k element of K, then hk = kh.
(3) If g element of G, then there exist h element of H, k element of K such that g= hk.
Then G isomorphic with H cross K..
A function f : AmapsB is said to be surjective (or onto) if for every y element of B there exists x element of A such that f(x) = y. Let G1 and G2 be groups. An isomorphism from G1 to G2 is an injection phi:G1mapsG2 such that phi(a · b) = phi(a) · phi(b)  (a,b element ofG1). The set residue classes mod n.

Direct Product of Groups

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

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

It is clear that (1) defines a binary operation on G cross 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 cross H has an inverse with respect to the multiplication (1). Thus, with respect to the multiplication (1), G cross H is a group.

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

Example 1: Let G = H = Z2. Then

Z2 cross Z2 = {(a,b) | a,b element of 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 cross 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 cross 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 cross H = {(a,b) | a element of Z2, b element of Z3}
= {(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)}.

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

alpha2 = (0,2),
alpha3 = (1,0),
alpha4 = (0,1),
alpha5 = (1,2),
alpha6 = (0,0),

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

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

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

The verification that G1 cross ... cross 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 cross G2) cross G3, whose elements are of the form ((g1,g2),g3) (gi element of Gi, i = 1,2,3). It is trivial to see that the mapping

psi:(G1 cross G2) cross G3 mapsG1 cross G2 cross G3,
psi(((g1,g2),g3)) = (g1,g2,g3),

is a surjective isomorphism, so that

(G1 cross G2) cross G3 isomorphic with G1 cross G2 cross G3.

Similarly,

G1 cross (G2 cross G3) isomorphic with G1 cross G2 cross G3.

Subsequently, we will always identify the groups (G1 cross G2) cross G3, G1 cross (G2 cross G3), G1 cross G2 cross G3. Similar comments apply to direct products of more than three groups. For example, G1 cross (G2 cross G3) cross G4 will be identified with G1 cross G2 cross G3 cross 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 intersection K = {1}.

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

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

Then G isomorphic with H cross K.

.

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

h'-1h = k'k-1 element of H intersection K = {1},
implies h'-1h = k'k-1 = 1,
implies h = h', k = k'.

Let psi:G maps H cross K be the mapping described by psi(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 psi is surjective since psi(hk) = (h,k) for any h element of H, k element of K. It is also clear that psi is injective. Thus, it suffices to show that psi is a homomorphism for then G isomorphic H cross K. Let g = hk, g' = h'k'. Then, by (2), gg' = (hk)(h'k') = (hh')(kk'), so that

psi(gg') = (hh',kk')
= (h,k)(h',k')
= psi(g)psi(g')

Thus, psi 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 isomorphic withZm cross 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 isomorphic Zm,    K isomorphic with Zn.

Therefore, it suffices to prove that G isomorphic with H cross K. Let us apply Theorem 3. Let g element of H intersection 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 intersection K = {1G}. Every element of H commutes with every element of K since G is abelian. Finally, if g element of 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 isomorphic with H cross 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 isomorphic withZp1a1 cross ...cross 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 isomorphic with Zp1a1 cross Zn

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

(3)
Zn isomorphic with Zp2a2 cross ... cross Zptat.

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