The set residue classes mod 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 group is considered cyclic if it is generated by a single element. commutative

### The Fundamental Theorem of Abelian Groups

Earlier we mentioned a broad goal of the theory of finite groups is to arrive at a classification of all finite groups. In this section we will achieve a much more modest goal: We will classify all finite abelian groups. Our main result will be the fundamental theorem of abelian groups (FT), which asserts that a finite abelian group is isomorphic to a direct product of cyclic groups. We will be able to pin down the structure of a finite abelian group further by specifying to some extent the cyclic groups which occur.

Theorem 1 (FT): Let G be a finite abelian group. Then G is isomorphic to a direct product of cyclic subgroups.

Proof: Suppose that G has order n. The theorem is clear if n = 1. Thus, we may assume that n > 1, and we may proceed by induction on n. Let x0 be an element of G of highest order, and let C0 = [x0]. Since G is abelian, C0 G. Moreover, G/C0 has order less than n, so the induction hypothesis implies that there exist cyclic subgroups H1,H2,...,Hs of G/C0 such that

(1)
G/C0 H1 H2 ... Hs.

Each Hi is of the form Ci/C0, where Ci is a subgroup of G which contains C0. Since Hi is cyclic, Hi = [xiC0] for some xi G. Let us prove that

(2)
G [x0] [x1,...xs].

for some choice of xi such that Hi = [xiC0]. Let us apply Theorem 3 of the section on direct products. Since G is abelian, every element of [x0] commutes with every element of [x1,...,xs]. If g G, then gC0 G/C0, so that by (1),

gC0 = (x1C0)a1...(xsC0)as = x1a1...xsasC0.

Thus,

g = c0x1a1...xsas

for some c0 C0. But c0 = s0a0 since C0 = [x0]. Therefore,

g = x0a0(x1a1...xsas),

and g is the product of an element of [x0] and an element of [x1...xs]. In order to apply Theorem 3 of the section on direct products, we must prove that

(3)
[x0] [x1,...,xs] = {1}.

This is the heart of the proof. Let ai denote the order of xi. Let us first show the order of xiC0 (in G/C0) is also ai. Let bi denote the order of xiC0. Since (xiC0)ai = xiaiC0 = C0, we see that

(4)
bi|ai

Since (xiC0)bi = C0, we see that xibi C0, so that xibi = x0c for some integer c > 0. Since the order of xiC0 is the same as the order of xix0dC0, and since xix0dC0 is a generator of Hi, we may replace xi by xix0d for any integer d. In doing so, we replace c + bid. By choosing d appropriately, we can guarantee that 0 < c + bid < bi. Thus, without loss of generality, we may assume that

(5)
xibi = x0c,    0 < c < bi.

The order of xibi is ai/(biai) = ai/bi [by(4)]. The order of x0c is r/(r,c), where r = the order of x0. Therefore, by (5),

ai = bir/(r,c).

On the other hand, since x0 has maximal order in G, ai < r. Therefore,

bi/(r,c) < r which is equivalent to bi < (r,c).

If c > 0, then (r,c) < c and thus we have

bi < c,

which contradicts (5). Therefore, c = 0 and xibi = x00 = 1. Thus ai|bi, which together with (4) implies that ai = bi. We have thus proved that the order of xiC0 is ai. Every element in G/C0is of the form

(6)
(x1C0)1...(xsC0)s = x11...xssC0,   0 < i < ai.

By what we have proved above, Hi = Ci/C0 has order ai, so that by (1), G/C0 has order a1a2...as. Therefore, all the elements of (6) are distinct. In particular, if x11...xss C0 (0 < i < ai), then 1 = ... = s = 0. Let us finally prove (3). If g [x0] [x1,...,xs], then g = x11...xss (0 < i < ai) and g C0. Therefore, as we have shown above, 1 = ... = s = 0, and g = 1. Thus, (3) is proved and with it (2). However, by induction,

(7)
[x1,...,xs] [y1] ... [yt],

so that by (2),

G [x1] [y1] [yt],

and G is a direct product of cyclic groups.

Corollary 2: Let G be a finite abelian group. Then there exist positive integers n1,...,nt such that

G Zn1 ... Znt.

Proof: Every cyclic group is isomorphic to Zn for some positive integer n.

Corollary 3: Let G be a finite abelian group. Then there exists a set of prime powers ,,..., (not necessarily distinct) such that

G Z Z ... Z.

Proof: Let n1,...nt be as in Corollary 2, and let

ni = ...  (1 < i < t).

Then by Corollary 5 of the section on direct products,

G Z Z ... Z ... Z.

Note that H K K H with respect to the mapping : H K K H defined by (h,k) = (k,h) (h H, k K). Therefore, we may rearrange factors in a direct product and still get a group isomorphic to the original one. Using this fact, we can normalize the decomposition of Corollary 3 as follows: Let q1,...,qv be the distinct primes among {p1,...,ps}, arranged so that

q1 < q2 < ... < qv

Let

(8)
qiai1,qiai2,...,qiaij(i)

be the powers of the prime qi (not necessarily distinct) appearing in the decomposition of Corollary 3, arranged so that

0 < ai1 < ai2 < ... < aij(i).

Define

G(qi) = Z Z ....

Then Corollary 3 and the above remark imply that

(9)
G G(q1) G(q2) ...  G(qv).

It is clear that every element of G(qi) has order a power of qi. Moreover, from (9), we see that the set of all element of G(q1 ... G(qv) whose order is a power of qi is {1} {1} ... G(qi) ... {1} G(qi). Thus, we have

Corollary 4: Let H(qi) denote the subgroup of G consisting of all elements whose order is a power of qi. Then

H(qi) Z ... Z

and

G H(qi) ... H(qv).

If q is a prime, then H(q) is called the q-primary part of G. Let us consider an example. Let G = Z10 Z25. Then

Therefore,

G Z2 Z5 Z52.

The 2-primary part of G is Z2, while the 5-primary part of G is Z5 Z52.

Definition 5: The prime powers

qiaij   [1 < j < j(i), 1 < i < v]

are called the elementary divisors of G.

It is clear that if we are given the elementary divisors of G, then it is possible to determine G up to isomorphism using (9).

Theorem 6: The elementary divisors of G are uniquely determined by G. In other words, the decomposition (9) of G is unique.

Proof: The primes q1,...,qv are uniquely determined as the distinct primes dividing the order of G. Let q = qi for some i and let H = the q-primary part of G. Then H depends only on q and G. Suppose that we are given a decomposition of H of the form

It suffices to show that (1), (2),...,(a) depend only on G and q. If b is a positive integer, then set Hb = {hb | h H}. Note that qcZqb = {0} if c > b, Zqb-c if c < b. Thus,

so that order of Hqa-1 is q(a). Thus, (a) is determined only by G, q, and a. Next note that

Thus, the order of Hqa-2 is

q(a-1)+2(a).

Thus, (a - 1) depends only on G, q, and a - 1. Proceeding in this way, we see that (1),...,(a) are determined uniquely by G and q.

Example 1: Let us determine up to isomorphism all abelian groups of order 16. Since 16 = 24, the possible sets of elementary divisors are

(a) 24.

(b) 2, 23.

(c) 2, 2, 22.

(d) 2, 2, 2, 2.

(e) 22, 22.

The corresponding abelian groups are

(a) Z16.

(b) Z2 Z8.

(c) Z2 Z2 Z4.

(d) Z2 Z2 Z2 Z2.

(e) Z4 Z4.

None of the groups (a)-(e) are isomorphic to one another by Theorem 6. This example suggests the following generalization. Let p be a prime, m a positive integer. Let us determine the abelian groups of order pm. We know that every such groups is uniquely specified by its elementary divisors

pa1,pa2,...,pas,

where

a1 < a2 < ... < as,   a1 + a2 + ... + as = m.

The last property comes from the fact that

(10)

has order pa1·pa2...pas. Thus, to each abelian group of order pm is associated the set {a1,...,as} of positive integers such that a1 + ... + as = m, a1 < a2 < ... < as. Such a set is called a partition of m. Conversely, to each partition {a1,...,as} of m there corresponds an abelian group of order pm - the group (10). Moreover, by Theorem 6, distinct partitions of m correspond to nonisomorphic groups. Let p(m) denote the number of distinct partitions of m. Then, we see that the number of nonisomorphic abelian groups of order pm is p(m).

In the above example, p = 2, m = 4. The partitions of 4 are

{4}, {1,3}, {1,1,2}, {1,1,1,1}, {2,2}.

Thus, there are five nonisomorphic abelian groups of order 24, as we discovered.

We proved earlier that the number of nonisomorphic groups of order n is at most nn2. If n = pm, there are at most pmp2m. How does this number compare with the number of nonisomorphic abelian groups of order pm? In other words, how big is p(m)?

The problem of determining the law of growth of the partition function p(m) has been solved completely, but only in the twentieth century, and belongs to a chapter of contemporary mathematics called additive number theory. Although many results about partitions had been discovered as early as the eighteenth century by Euler and Lagrange, it was not until the invention of the circle method by Hardy and Ramanujan in 1918 that any real progress was made in determining the order of magnitude of p(m). Hardy and Ramanujan proved that

(11)

In other words, for large m, p(m) is approximately equal to

We leave it to the reader to demonstrate the this quantity is much smaller that pmp2m, in the sense that

The proof of (11) is very difficult and relies on analysis. Much more precise results than (11) are known. In fact in 1936 Rademacher found a beautiful exact formula for p(m) in terms of an infinite series.

Let us close this section with a very simple, but useful application of the fundamental theorem.

Proposition 7: Let G be an abelian group whose order is divisible by a prime p. Then G contains an element of order p.

Proof: Without loss of generality assume G to be of the form

Zpa1 Zpa2 ... Zpat Zpb1 ....

Then

(pa1-1,pa2-2,...,pat-1,0,0,...,0)

has order p.