Order in Abelian Groups

The first issue we shall address is the order of a product of two elements of finite order. Suppose G is a group and a,b ∈ G have orders m= |a| and n= |b|. What can be said about|ab|? Let’s consider the abelian case first.

Lemma 7.4.1. Let G be an abelian group and a,b ∈ G. Then for any n ∈ ℤ, (ab)ⁿ = aⁿbⁿ.

Proof. If ab != ba for some specific a and b, then (ab)² = abab, can't = a² b² = aabb since then you multiply on left by a⁻¹ and on right by b⁻¹ to get ba = ab. □

Lemma 7.4.2. Let G be an abelian group with the identity element 1. Let a,b be elements of G with order m and n, respectively. If m and n are relatively prime, then the order of the element ab is mn.

Proof. Let r be the order of the element ab. Since we have

(ab)^mn = a^mn b^mn (since G is an abelian group) =
(a^m)ⁿ(bⁿ)^m = 1. □

This implies that the order r of ab divides mn, that is, we have

r | mn

Now, sice r is the order of ab we have

1 = (ab)^r = a^rb^r

Then we have

1 = 1ⁿ = a^rnb^rn = a^rn

since b_n = 1. This yields that the order m of the element a divides rn.

Since m and n are relatively prime, this implies that we have

m | r

Similarly (switch the role of n and m), we obtain

n | r

Thus we have

mn | r

since m and n are relatively prime. From r | mn and mn | r we have r = mn, and hence the order of the element ab is mn. □

Lemma 7.4.3. Let G be an abelian group. Let a and b be elements in G of order m and n, respectively. Prove that there exists an element c in G such that the order of c is the least common multiple of m and n.

Proof. First, consider the case when m and n are relatively prime.

(When m and n are relatively prime) From lemma 7.4.2, the order of the product ab is mn. So if m,n are relatively prime, then we can take c = ab ∈ G and c has order mn, which is the least common multiple.

m = \prod_{i} p_{i}^{α_{i}} and n = \prod_{i} p_{i}^{β_{i}}

Now we consider the general case. Let p_i be the prime factors of either m or n (we take the union of prime factors). Then write prime factorizations of m and n as

Here α_i and β_i are nonzero integers (could be zero). Define

m^{'} = \prod_{i : α_{i} \geq β_{i}} p_{i}^{α_{i}} and n^{'} = \prod_{i : β_{i} > α_{i}} p_{i}^{β_{i}}

For example if m = 2³⋅3²⋅5 and n = 3²⋅7 then m′=2³⋅3²⋅5 and n′ = 7, p_i = {2, 3, 5, 7}, a_i = (3 2 1 0), (b_i) = (2 0 0 1). Note that m′∣m and n′∣n, and also m′ and n′ are relatively prime. The least common multiple l of m and n is given by

l = m′ n

Consider the element a′:= a^m/m′ We claim that the order of a′ is m′. Let k be the order of the element a'. Then we have

1 = a'^k = (a^m/m′)^k = a^mk/m′

where 1 is the identity element in the group G. This yields that m divides mk/m′ since m is the order of a. It follows that m′ divides k. On the other hand, we have

(a^m/m′)^m' = a^m = 1

and hence k divides m′ since k is the order of the element a^m/m′. As a result, we have k = m′. So the order of a′ is m′. Similarly, the order of b′:= b^n/n′ is n′. The orders of elements a′ and b are m′ and n′, and they are relatively prime. Hence we can apply the first case and we conclude that the element a′b′ has order

m′n′ = l

Thus, we can take c = a′b′. □

Remark 7.4.4 For nonabelian groups, basically nothing can be said about how to relate |ab| to |a| and |b|. For example, consider the symmetric group S₃ with three letters. Let

a = (123) and b = (12).

Then the order of a is 3 and the order of b is 2. The least common multiple of 2 and 3 is 6. However, the symmetric group S₃ has no elements of order 6. Hence the statement of the problem does not hold for non-abelian groups. ■

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