Classification of finite abelian groups

With our present knwoledge we shall introduce briefly the fundamental theorem of finite abelian groups, which allows us to determine for each positive integer n, the number of abelian groups and whether two abelian groups with same order are isomorphic or not.

This powerful theorem provides an easy-to-understand recipe by which all such groups can be constructed, using the two familiar notions of cyclic group and direct product. The theorem is relatively difficult to prove, and so we will not prove it here.

We recall that a p-group is a group with cardinality a power of p, and each of its elements has order p (this latter statement can be take as definition of p-group). For example every cyclic group with order of prime order p is a p-group.

5.20.1 Theorem. Let G an abelian finite group. For every prime p such that p | |G| let

Σ_p := {x ∈ G | x^{p^s} = e for some s}

Then

every Σ_p is a subgroup of G (which is a p-subgroup of G);
G is a direct product of all Σ_p, with varying p among all prime divisors of p.

Proof. It results

Σ_p := {x ∈ G | x^{p^s} = e for some s}

a) Since the group is abelian, by multiplying the element x^p, s times we get x^p^s.

b) Let x an arbitrary element of G. Then x has a certain order n, with n | G. Let n = p₁^α₁ p₂^α₂ ⋅⋅⋅ p_r^α_r, p_i distinct primes and α_i ≥ 1, for every i = 1, ..., r. By letting q_i = n/(p₁^α_i), it results (q₁, q₂, ..., q_r) = 1; there exists thus the integers k_i such that

1 = k₁q₁ + k₂ + ⋅⋅⋅ + k_rq_r

Then

x = x¹ = x^{k₁q₁ + k₂q₂ + ... + k_rq_r} = x^k₁q₁ x^k₂q₂ ⋅⋅⋅ x^k_rq_r

Now

(x^q_i)^{p_i^α_i} = xⁿ = 1 ∀i = 1, ..., r

Thus x^q_i ∈ Σ_{p_i}. But then also x^k_iq_i ∈ Σ_{p_i}. We thus proved that every x in G can be written as product of elements of Σ_{p_i}, that is

G = Σ_p₁ Σ_p₂ ⋅⋅⋅ Σ_{p_k}

To prove that G = Σ_p₁ x Σ_p₂ x ⋅⋅⋅ x Σ_{p_k}, it sufficies to prove (see Theorem 7.17.16) that
$Σ_{p_{i}} \cap Σ_{p_{1}} Σ_{p_{2}} \dots \hat{Σ_{p_{i}}} \dots Σ_{p_{k}} = {e}$
which is obvious because there cannot be an element both in Σ_{p_i} and in Σ_p₁ Σ_p₂ ⋅⋅⋅ ̂̂Σ_{p_i} ⋅⋅⋅ Σ_{p_k}, having a period both a power of p_i and a number cooprime with p_i. □

We thus proved that every finite abelian group G is isomorphic to Σ_p₁ x Σ_p₂ x ... x Σ_{p_k}, where

Σ_{p_i} := {x ∈ G | x^{p_i^{^s_i}} = e for some s_i}

5.20.2 Theorem. Let Σ_p a finite and abelian p-group. Then Σ_p is a direct product of cyclic groups (with order a power of p).

Proof. Let s an element of Σ_p of maximum order p^α, an let T a maximal subgroup with respect to the condition of having trivial intersection with the subgroup ⟨s⟩ generated by s, i.e.

⟨s⟩ ∩ T = {e}

It can be proved (we omit the proof) that

⟨s⟩T = Σ_p

This assures us that

Σ_p = ⟨s⟩ x T

The proof of the theorem continues now by induction. We chose an element s' in T with maximum order. Let T' a subgroup of T maximial with respect to the condition

T ' ∩ ⟨s'⟩ = {e}

by repeating this reasoning we arrive to the conclusion that

Σ_p ≃ ⟨s₁⟩ x ⟨s₂⟩ x ... x ⟨s_l⟩

thus Σ_p is the direct product of cyclic groups. □

We just proved the following theorem.

5.20.3 Theorem. Every finite abelian group G is the direct product of cyclic groups whose orders are powers involved in the factorization of n = |G|. If n = p₁^α₁ p₂^α₂ ⋅⋅⋅ p_k^α_k, then

G = \underset{Σ_{p_{1}}}{\underset{⏟}{Z_{{p_{1}}^{a_{1, 1}}} \times Z_{{p_{2}}^{a_{1, 2}}} \times \dots Z_{{p_{2}}^{a_{1, s}}}}} \times \underset{Σ_{p_{2}}}{\underset{⏟}{Z_{p_{2}}^{a_{2, 1}} \times Z_{p_{2}}^{a_{2, 2}} \times \dots Z_{p_{2}}^{a_{2, t}}}} \times \dots \times \underset{Σ_{p_{k}}}{\underset{⏟}{\dots}}

where for every i = 1, ..., k the integers b_i,j are such that Σ_j b_i,j = α_i, and are ordered in such a way that b_i,j(i) ≥ ... ≥ b_i,2 ≥ b_i,1 < 0.

We proved an existence theorem for such a composition.

5.20.4 Definition. The integers p_i^b_i,h in the decomposition are called elementary divisors of G. For every fixed primary component Σ_p, i = 1, ..., k, the exponents b_i,j(i) are known as the invariants of Σ_p.

Such integers are determined uniquely by the group G and characterize it uniquely, in the sense that abelian groups having the same order are isomorphic if and only if have the same elementary divisors and the same invariants.

5.20.5 Theorem. (Fundamental Theorem of Finite Abelian Groups) Let G a finite abelian group of order n = p₁^α₁ p₂^α₂ ⋅⋅⋅ p_r^α_k. Then

G is isomorphic to a direct product of cyclic groups with order power of primes which appear in the factorization of G:

where for every i = 1, ..., k the integers b_i,j are such that Σ_j b_i,j = α_i and are ordered as b_i,j(i) ≥ ... ≥ b_i,2 ≥ b_i,1 < 0.
Such a factorization in unique, so that if two finite abelian groups are isomorphic then they have the same p_i^b_i,j factors (elementary divisors of G).

With the previous theorem we can know the structure of every finite abelian group. As a consequence of the Fundamental Theorem of Finite Abelian Groups, we have the following corollary, which shows that the converse of Lagrange’s Theorem is true for finite Abelian groups.

5.20.6 Corollary. If m divides the order of a finite Abelian group G, then G has a subgroup of order m.

Given an integer N to count the number of finite abelian groups with N has order, we must analyze in how many ways every Σ_p can be factorized as product of cyclic groups (with orders powers of p). That is one's has to count the different number of ways the following relation can be written

Σ_p = ℤ_p^n₁ x ℤ_p^n₂ x ... x ℤ_{p^n_s}

It must hold that

|Σ_p| = pⁿ = p^n₁ p^n₂ p^n_s = p^{n₁ + n₂ + ... + n_s}

We must count in how many different ways n can be writte as sum of n₁ + n₂ + ... + n_s with n₁ ≥ n₂ ≥ ... ≥ n_s. This number is exaclty the number of partitions p(n) of n. Thus every Σ_p with |Σ_p| = pⁿ can be written in p(n) ways as product of cyclic groups of order the different powers of p. Thus is order to determine the number of groups not isomorphic having the same order N we hat to:

Factorize N as product of powers of distinct primes.

N = p₁^α₁ p₂^α₂ ⋅⋅⋅ p_k^α_k

It results:

p₁^α₁ = |Σ_p₁|, p₂^α₂ = |Σ_p₂|, p_k^α_k = |Σ_{p_k}|
Count the different invariants for every Σ_{p_i}, which is equivalent to count the number of partitions of p(α_i) for every α_i.
The total number of non isomorphic groups with order N = p₁^α₁ p₂^α₂ ⋅⋅⋅ p_k^α_k is given by

p₁^α₁ p₂^α₂ ⋅⋅⋅ p₁^α_k

5.20.6 Example. How many Ablian groups are there of order 144? We have N = 144 = 2⁴ ⋅ 3² and there are five ways of partitioning 4 (namely 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1) and two ways of partitioning 2 (namely, 2 and 1 + 1),

Every G with 144 elements is decomposed in the direct product

G = Σ₂ x Σ₃

where

|Σ₂| = 2⁴ |Σ₃| = 3²

and

p(α₁) = p(4) = 5, p(α₂) = p(2) = 2

Thus the total number of abelian groups with order N = 144 is given by p(4)p(2) = 10.

The different partitions of |Σ₂| are

ℤ_2⁴ ℤ_2³ x ℤ₂ ℤ₂ x ℤ₂ ℤ_2² x ℤ₂ x ℤ₂ ℤ₂ x ℤ₂ x ℤ₂ x ℤ₂

Those of Σ₃ are

ℤ_3² ℤ₃ x ℤ₃

Every abelian group with order 144 must be thus isomorphic to one of the following

Σ₂	x	Σ₃
ℤ_2⁴	x	ℤ_3²
ℤ_2⁴	x	ℤ₃ x ℤ₃
ℤ_2³ x ℤ₂	x	ℤ_3²
ℤ_2³ x ℤ₂	x	ℤ₃ x ℤ₃
ℤ₂ x ℤ₂	x	ℤ_3²
ℤ₂ x ℤ₂	x	ℤ₃ x ℤ₃
ℤ_2² x ℤ₂ x ℤ₂	x	ℤ_3²
ℤ_2² x ℤ₂ x ℤ₂	x	ℤ₃ x ℤ₃
ℤ₂ x ℤ₂ x ℤ₂ x ℤ₂	x	ℤ_3²
ℤ₂ x ℤ₂ x ℤ₂ x ℤ₂	x	ℤ₃ x ℤ₃

5.20.7 Example. Classify all abelian groups with order N = 105.

It results 105 = 3 ⋅ 5 ⋅ 7. G can be thus factorized in terms of primary components

G = Σ₃ x Σ₅ x Σ₇

with

|Σ₃| = 3, |Σ₅| = 5, |Σ₇| = 7

G can only be the following group

G = ℤ₃ x ℤ₅ x ℤ₇

G is thus the cyclic group of order 105. ■

The result obtained in the last example is a general one; for N = p₁p₂ ⋅⋅⋅ p_k with p_i distinct primes, there exists only one group of order N which is a cylic one.

5.20.7 Remark. For what we have seen the number of mutually non isomorphic abelian p-groups of order p^m equals the number p(m) of partitions of m. In particular, this number does not depend on the prime p, but only on the exponent m. Thus, for example, there are five non isomorphic abelian groups of order 16 = 2⁴ because there are five partitions of the number 4: to the partition 4 = 4 there corresponds the cyclic group ℤ₁₆, to the 4 = 1 + 1 + 2 the group ℤ₂ × ℤ₂ × ℤ₄, etc. For the same reason, there are five abelian groups of order 3⁴, 5⁴, 19⁴, etc.

However even if the number is the same the nature is clearly different depending on the prime p, e.g take 3⁴ and 7⁴.

In the first case the groups are

ℤ_3⁴, ℤ_3³ x ℤ₃, ...

In the second case

ℤ_7⁴, ℤ_7³ x ℤ₇, ... ■

Exercises

Find all abelian groups with the following orders

56, 2100 800, 45, 27

and for each group calculate the maximum order associated to their elements.
Let U(ℤ₃₀) the group of the invertible elements of ℤ₃₀. Calculate the decomposition of U(ℤ₃₀) in terms of cyclic groups, and check wheter it is a cyclic group.
Given an abelian group G with order 72, which has an element of order 36 and no subgroups isomorphic to ℤ₈, is it possible to find the group isomorphic to G?

Solutions

- 56 = 2³ ⋅ 7, G can be thus factorized in terms of primary components
  
  G = Σ₂ x Σ₇
  
  with
  
  |Σ₂| = 3, |Σ₇| = 7
  
  p(α₁) = p(3) = 3. 3, 1+1+1, 2+1,
  
  Thus the total number of abelian groups with order 56 is 3. These groups are
  
  ℤ₂ x ℤ₂ x ℤ₂x ℤ₇ (cyclic see Corollary 7.17.15)
  ℤ₈ x ℤ₇
  ℤ₄ x ℤ₂ x ℤ₇
  ℤ₂ x ℤ₂ x ℤ₂ x ℤ₇
  
  The maximum order of an element of ℤ₂ x ℤ₂ x ℤ₂ x ℤ₇ is lcm(2,2,2,7) = |2*2*2*7|/gcd(2,2,2,7) = 56, and similarly for the other groups.
- 2100 = 2² ⋅ 3 ⋅ 5² ⋅ 7, G can be thus factorized in terms of primary components
  
  G = Σ₂ x Σ₃ x Σ₅ x Σ₇
  
  p(α₁) = p(α₃) = p(2) = 2. 2, 1+1,
  
  Thus the total number of abelian groups with order 2100 is p(α₁) ⋅ p(α₃) = 4. These groups are
  
  ℤ₄ x ℤ₃ x ℤ₂₅ x ℤ₇
  ℤ₂ x ℤ₂ x ℤ₃ x ℤ₂₅ x ℤ₇
  ℤ₄ x ℤ₃ x ℤ₅ x ℤ₅ x ℤ₇
  ℤ₂ x ℤ₂ x ℤ₃ x ℤ₅ x ℤ₅ x ℤ₇
  
  The maximum order of an element of ℤ₄ x ℤ₃ x ℤ₂₅ x ℤ₇ is lcm(4,3,25,7) = |4*3*25*7|/gcd(4,3,25,7) = 2100; Proceed similarly for the other groups.
- 180 = 2² ⋅ 3² ⋅ 5, G can be thus factorized in terms of primary components
  
  G = Σ₂ x Σ₃ x Σ₅
  
  p(α₁) = p(α₂) = p(2) = 2, 2, 1+1
  
  There are thus p(α₁) = p(α₂) = 4 groups of order 180, which are
  
  ℤ₄ x ℤ₉ x ℤ₅ ≃ ℤ₁₈₀
  ℤ₄ x ℤ₃ x ℤ₃ x ℤ₅ ≃ Z₆₀ x ℤ₃
  ℤ₂ x ℤ₂ x ℤ₉ x ℤ₅ ≃ Z₉₀ x ℤ₂
  ℤ₂ x ℤ₂ x ℤ₃ x ℤ₃ x ℤ₅ ≃ Z₃₀ x ℤ₆
  
  The maximum order of an element of ℤ₄ x ℤ₉ x ℤ₅ is lcm(4,9,5) = |4*9*5|/gcd(5,9,5) = 180; Proceed similarly for the other groups.
- 27 = 3³.
  
  p(α₁) = p(3) = 3, 3, 2 + 1, 1 + 1 + 1
  
  There are thus p(α₁) = 3 groups of order 27, which are
  
  ℤ₂₇ ℤ₉ x ℤ₃ ℤ₃ x ℤ₃ x ℤ₃
  
  The maximum order of an element of ℤ₂₇ -> 27, ℤ₉ x ℤ₃ is 9 and ℤ₃ x ℤ₃ x ℤ₃ is 3.
U(ℤ₃₀) = {[1]₃₀, [7]₃₀, [11]₃₀, [13]₃₀, [17]₃₀, [19]₃₀, [23]₃₀, [29]₃₀}. In the group (U(ℤ₃₀), ⋅), [7]₃₀, [13]₃₀, [17]₃₀ and [23]₃₀ have order 4.

[7]₃₀ ⋅ [7]₃₀ ⋅ [7]₃₀ ⋅ [7]₃₀ = [2401] ≡ [1] (mod 8)

The other elements have order 2 (except [1]₃₀). Finite G is cyclic iff it has at most one subgroup of each possible order (possible orders are the divisors of |G|). The G here has order 8, thus U(ℤ₃₀) is not cyclic.

8 = 2³. We resort to the fundamental theorem on finite abelian groups

G = Σ₂

since

p(α₁) = p(3) = 3, 3, 2 + 1, 1 + 1 + 1

The possible abelian groups of order 8 are ℤ₈, ℤ₄ x ℤ₂ and ℤ₂ x ℤ₂ x ℤ₂. Since ℤ₈ is cyclic we can exclude it. ℤ₂ x ℤ₂ x ℤ₂ = ( (0,0,0), (0,0,1), (0,1,1), (1,1,1), ..} it has no elements of order 4 but only of order 1 and 2. Then U(ℤ₃₀) = ℤ₄ x ℤ₂. ■
72 = 2³ ⋅ 3². Using the fundamental theorem on finite abelian groups

G = Σ₂ x Σ₃

with

|Σ₂| = 2³, |Σ₃| = 3².

since

p(α₁) = p(3) = 3, 2 + 1, 1 + 1 + 1 p(α₂) = p(2) = 2 2, 1 + 1

There are thus p(α₁)p(α₂) = 6 abelian groups of order 72, which are

ℤ₈ x ℤ₉
ℤ₄ x ℤ₂ x ℤ₉
ℤ₂ x ℤ₂ x ℤ₂ x ℤ₉
ℤ₈ x ℤ₃ x ℤ₃ x ℤ₃
ℤ₄ x ℤ₂ x ℤ₃ x ℤ₃
ℤ₂ x ℤ₂ x ℤ₂ x ℤ₃ x ℤ₃

excluding the groups containing ℤ₈ (even if there's ([2]₈, [1]₉) an element with order 36; the order of[1]₉ in ℤ₉ is 9, and hat of [2]₈ is 4), the only possibity for a group with an element of order 36 is G ≃ ℤ₄ x ℤ₂ x ℤ₉. ■

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