Questions: Classification of Finite Abelian Groups

Question 1

How many non-isomorphic abelian groups of order 8 exist?

• A. One — only Z₈, since 8 = 2³ is a prime power and prime-power order groups are cyclic
• B. Two — Z₈ and Z₄ × Z₂
• C. Three — Z₈, Z₄ × Z₂, and Z₂ × Z₂ × Z₂
• D. Four — one for each divisor of 8

Answer: (C)

Since 8 = 2³, there is only one prime to handle. The theorem says we count integer partitions of the exponent 3. The partitions are {3} → Z₈, {2,1} → Z₄ × Z₂, and {1,1,1} → Z₂ × Z₂ × Z₂ — exactly three. Option A is the most tempting error: not every prime-power order group is cyclic. Z₂ × Z₂ has order 4 = 2² but is not cyclic (every non-identity element has order 2). The theorem's value is precisely enumerating all possibilities, not assuming the cyclic group is the only one.

Question 2

A student claims that Z₄ × Z₃ and Z₁₂ are non-isomorphic abelian groups of order 12 because one is expressed as a direct product and the other as a cyclic group. What does the classification theorem say?

• A. She is correct — a direct product of cyclic groups is never isomorphic to a single cyclic group
• B. Z₄ × Z₃ ≅ Z₁₂ because their elementary divisors are both {4, 3} and gcd(4, 3) = 1, making the direct product cyclic
• C. The theorem requires converting both groups to invariant factor form before comparison; direct product form is insufficient
• D. They are non-isomorphic because Z₄ × Z₃ has an element of order 4 while Z₁₂ has elements of order 12

Answer: (B)

When the elementary divisors are coprime — gcd(4, 3) = 1 — the direct product Z₄ × Z₃ is isomorphic to the cyclic group Z₁₂. Two finite abelian groups are isomorphic iff they have identical lists of elementary divisors; both Z₄ × Z₃ and Z₁₂ have elementary divisors {4, 3}. Option D is wrong: Z₁₂ has elements of every order dividing 12, including order 4 (the element 3), so the argument fails. The 'looks different' intuition is exactly what the classification theorem corrects by reducing the isomorphism question to comparing two lists of integers.

Question 3

Nearly every abelian group of prime-power order p^n is cyclic, isomorphic to Z_{p^n}.

• T. True
• F. False

Answer: False

Cyclic is only one of the possibilities. For p² there are already two non-isomorphic abelian groups: Z_{p²} (cyclic) and Z_p × Z_p (non-cyclic — every non-identity element has order p). For p³ there are three. The Fundamental Theorem's power is enumerating all possibilities via partitions, not just the cyclic case. Assuming prime-power groups must be cyclic is the most common error when first applying the classification theorem.

Question 4

The Fundamental Theorem of Finite Abelian Groups states that two finite abelian groups are isomorphic if and only if they have the same list of elementary divisors (prime-power cyclic factors).

• T. True
• F. False

Answer: True

This is the uniqueness half of the theorem — arguably more powerful than the existence half. It gives a complete, checkable criterion for isomorphism: compute the elementary divisors of both groups (the prime-power cyclic factors in the primary decomposition) and compare the lists. If they match, the groups are isomorphic; if they don't, they aren't. Without this criterion, determining whether two large groups are isomorphic would require checking every possible bijection — computationally intractable. With it, the question reduces to integer arithmetic.

Question 5

Describe the procedure for determining how many non-isomorphic abelian groups of order 72 exist, without listing them — just explain the method.

Model answer: Factor 72 into prime powers: 72 = 2³ × 3². Handle each prime independently. For prime 2 with exponent 3, count integer partitions of 3: {3}, {2,1}, {1,1,1} — three options, corresponding to Z₈, Z₄ × Z₂, and Z₂ × Z₂ × Z₂. For prime 3 with exponent 2, count partitions of 2: {2}, {1,1} — two options, corresponding to Z₉ and Z₃ × Z₃. The total number of non-isomorphic abelian groups of order 72 is 3 × 2 = 6, one for each combination of a choice at prime 2 and a choice at prime 3. The theorem guarantees these six are distinct and exhaustive.

The method generalizes immediately: for any order n = p₁^{a₁} × p₂^{a₂} × ... × p_k^{a_k}, the number of non-isomorphic abelian groups of order n is the product of the number of integer partitions of each exponent a_i. This reduces a seemingly complex group-theoretic question to elementary combinatorics.