Questions: Cyclic Groups

Question 1

In ℤ/12ℤ, which of the following elements is a generator of the entire group?

• A. 4 — because it divides 12 evenly
• B. 6 — because it is exactly half of 12
• C. 5 — because gcd(5, 12) = 1
• D. 3 — because 3 is prime

Answer: (C)

An element k generates ℤ/nℤ if and only if gcd(k, n) = 1. Here n = 12: gcd(5, 12) = 1, so 5 is a generator. gcd(4, 12) = 4, gcd(6, 12) = 6, gcd(3, 12) = 3 — none equal 1, so 4, 6, and 3 each generate a proper subgroup. Being prime doesn't guarantee generation of the whole group; the condition is coprimality with the group order.

Question 2

The Klein four-group V₄ = {e, a, b, ab}, where every non-identity element has order 2, is abelian. Why is it not cyclic?

• A. V₄ is not cyclic because cyclic groups can only have prime order
• B. V₄ is not cyclic because no single element generates all four elements — each non-identity element only generates a subgroup of order 2
• C. V₄ is not cyclic because it has too many elements of the same order
• D. V₄ is not cyclic because its generators must have prime order

Answer: (B)

A group is cyclic if and only if some element generates the entire group. In V₄, every non-identity element has order 2, so it generates only {e, itself} — a subgroup of order 2. No element produces all four elements, so V₄ is not cyclic. This shows that being abelian (which all cyclic groups are) does not imply being cyclic — the converse fails.

Question 3

Every subgroup of a cyclic group is itself cyclic.

• T. True
• F. False

Answer: True

This is a key structural theorem. If G = ⟨a⟩ is cyclic of order n, every subgroup consists of powers of a with a specific step size. The subgroups of ℤ/nℤ are exactly the sets generated by divisors of n: for each divisor d of n, there is exactly one subgroup of order d, namely ⟨d⟩. Each is cyclic. This clean structure has no analogue for general non-cyclic groups.

Question 4

If a group G is abelian, then G should be cyclic.

• T. True
• F. False

Answer: False

Being abelian is necessary but not sufficient for a group to be cyclic. Every cyclic group is abelian (since aᵐ · aⁿ = aᵐ⁺ⁿ = aⁿ · aᵐ), but the converse fails. The Klein four-group V₄ is a counterexample: abelian, but not cyclic because every non-identity element has order 2 and cannot generate all four elements.

Question 5

In ℤ/6ℤ, explain why 2 is not a generator of the whole group, and identify what subgroup it does generate.

Model answer: gcd(2, 6) = 2 ≠ 1, so 2 does not generate all of ℤ/6ℤ. Adding 2 repeatedly: 2, 4, 6 ≡ 0 — we cycle back after 3 steps, never reaching 1, 3, or 5. The element 2 generates the subgroup {0, 2, 4}, which has order 3.

When gcd(k, n) = d > 1, the element k generates the subgroup of order n/d, consisting of all multiples of d. Here d = 2, so 2 generates the unique subgroup of order 3. This illustrates how the subgroup structure of ℤ/nℤ is determined entirely by divisors of n.