Questions: Direct Products of Groups

Question 1

Consider the group ℤ₄ × ℤ₆. What is the maximum order of any element, and is the group isomorphic to ℤ₂₄?

• A. Maximum order is 24 and ℤ₄ × ℤ₆ ≅ ℤ₂₄, since the group has order 24
• B. Maximum order is lcm(4,6) = 12, and ℤ₄ × ℤ₆ ≇ ℤ₂₄ because gcd(4,6) = 2 ≠ 1
• C. Maximum order is 24 because both factors are cyclic, so their product is cyclic
• D. Maximum order is 4 since the smaller factor has order 4 and limits the product

Answer: (B)

The order of (a, b) in G × H is lcm(ord(a), ord(b)). The maximum in ℤ₄ × ℤ₆ is lcm(4, 6) = 12 (achieved by e.g. (1,1)). For ℤ_m × ℤ_n ≅ ℤ_{mn}, the Chinese Remainder Theorem requires gcd(m, n) = 1. Since gcd(4, 6) = 2 ≠ 1, ℤ₄ × ℤ₆ is not cyclic and not isomorphic to ℤ₂₄. The group has order 24 but its maximum element order is only 12 — a cyclic group of order 24 would need an element of order 24. Option A is the classic error: confusing group order with maximum element order.

Question 2

A group G has two normal subgroups N and M such that N ∩ M = {e} and every element of G can be written as nm for some n ∈ N, m ∈ M. What can you conclude?

• A. G is abelian, since elements from N and M must commute with each other
• B. G ≅ N × M — G is internally the direct product of N and M
• C. N and M are both cyclic, since they generate G with trivial intersection
• D. G is simple, since N and M are the only proper normal subgroups

Answer: (B)

This is the internal characterization of a direct product. When G has two normal subgroups N and M with trivial intersection (N ∩ M = {e}) that together generate G (every element is a product nm), then G is isomorphic to the external direct product N × M. The normality ensures elements from N and M commute with each other (nm = mn), but G need not be abelian overall — N and M individually may be non-abelian. Option A is wrong: direct products of non-abelian groups are non-abelian. Option D is wrong: having two such normal subgroups is far from the definition of simplicity.

Question 3

ℤ₂ × ℤ₃ is isomorphic to ℤ₆.

• T. True
• F. False

Answer: True

Since gcd(2, 3) = 1, the Chinese Remainder Theorem for groups guarantees ℤ₂ × ℤ₃ ≅ ℤ₆. You can verify directly: the element (1, 1) in ℤ₂ × ℤ₃ has order lcm(2, 3) = 6, which equals the group order. Any group with an element whose order equals the group order is cyclic. Since ℤ₂ × ℤ₃ has order 6 and a generator, it is isomorphic to ℤ₆.

Question 4

ℤ₂ × ℤ₂ is isomorphic to ℤ₄ because both are groups of order 4.

• T. True
• F. False

Answer: False

Two groups of the same order need not be isomorphic — order alone does not determine structure. ℤ₂ × ℤ₂ (the Klein four-group) has no element of order 4: every non-identity element has order 2, since (1,0) + (1,0) = (0,0) in ℤ₂ × ℤ₂. ℤ₄ has an element of order 4 (namely 1). An isomorphism must preserve element orders, so they cannot be isomorphic. This also follows from the CRT criterion: gcd(2, 2) = 2 ≠ 1, so ℤ₂ × ℤ₂ ≇ ℤ₄.

Question 5

Explain why ℤ_m × ℤ_n is cyclic if and only if gcd(m, n) = 1, and what goes wrong when gcd(m, n) > 1.

Model answer: A group is cyclic iff it has an element whose order equals the group order (mn). In ℤ_m × ℤ_n, the order of (a, b) is lcm(ord(a), ord(b)), and the maximum possible order of any element is lcm(m, n). For the group to be cyclic, we need lcm(m, n) = mn. Since lcm(m, n) = mn / gcd(m, n), this holds iff gcd(m, n) = 1. When gcd(m, n) > 1, every element has order at most mn/gcd(m,n) < mn, so no element generates the whole group and the direct product is not cyclic.

The intuition is that when m and n share a common factor, the two cyclic components 'lap each other' before exhausting the group — they both return to the identity before mn steps have passed. Coprime orders ensure the two components cycle independently, reaching the identity simultaneously only after exactly mn steps.