Questions: Order of a Group Element

Question 1

A student claims there must be an element of order 6 in a group of order 15, reasoning that '6 is a small enough number.' Why is this impossible?

• A. Groups of order 15 are cyclic and cyclic groups only have elements whose orders are prime
• B. By Lagrange's theorem, the order of any element must divide the group order; since 6 does not divide 15, no element of order 6 can exist
• C. Elements of order 6 only exist in abelian groups, and groups of order 15 are nonabelian
• D. The converse of Lagrange's theorem shows that elements of composite order cannot exist in odd-order groups

Answer: (B)

By Lagrange's theorem, the order of any element a in a finite group G divides |G|. Since 6 does not divide 15 (15 = 3 × 5, while 6 = 2 × 3), an element of order 6 is impossible. The allowed element orders in a group of order 15 are exactly the divisors of 15: 1, 3, 5, and 15. Options A and C state false general claims; option D mischaracterizes Lagrange. This divisibility constraint is one of the most practically useful facts in finite group theory — it immediately rules out many orders without computation.

Question 2

An element a in a group has order 12. What is the order of a⁴?

• A. 4
• B. 48
• C. 3
• D. 12

Answer: (C)

If a has order n, then aᵏ has order n/gcd(n, k). Here n = 12 and k = 4, so gcd(12, 4) = 4, giving order 12/4 = 3. You can verify directly: (a⁴)³ = a¹² = e, and (a⁴)¹ = a⁴ ≠ e (since 4 < 12), (a⁴)² = a⁸ ≠ e (since 8 < 12 and 12 ∤ 8). Option A (order 4) is the tempting wrong answer — it confuses the exponent k with the resulting order. Option D (order 12) would only hold if gcd(12, 4) = 1, but they share the factor 4.

Question 3

In any finite group, the order of every element divides the order of the group.

• T. True
• F. False

Answer: True

This is a direct consequence of Lagrange's theorem. The subgroup ⟨a⟩ generated by any element a has order equal to the order of a. Since every subgroup's order divides the group's order (Lagrange), the order of a divides |G|. This applies to every element in every finite group — no exceptions. It is one of the most powerful elementary constraints in finite group theory, immediately restricting which element orders are possible in any group of known size.

Question 4

If k divides the order of a finite group G, then G is expected to contain an element of order k.

• T. True
• F. False

Answer: False

This is the false converse of Lagrange's theorem. Lagrange says element orders divide |G|; it does not say every divisor of |G| is achieved as an element order. The classic counterexample is A₄ (the alternating group on 4 elements, order 12): 6 divides 12, but A₄ has no element of order 6. (Cauchy's theorem does guarantee an element of order p for each prime p dividing |G|, but that's a separate and stronger result for prime divisors only.)

Question 5

An element a has order n in a group. Explain how to find the order of aᵏ for a positive integer k, and why the formula works.

Model answer: The order of aᵏ is n/gcd(n, k). The reasoning: (aᵏ)ᵐ = aᵏᵐ = e if and only if n divides km (since a has order n). The smallest positive m satisfying this is n/gcd(n, k), because dividing both n and k by their gcd d = gcd(n, k) gives n/d and k/d with gcd(n/d, k/d) = 1, so the smallest m with (n/d) | (k/d)m is m = n/d.

Intuitively: if k and n share a common factor d, then aᵏ cycles back to the identity in n/d steps rather than n steps — it 'shortcuts' the cycle. For example, a⁴ in a cyclic group of order 12 completes its cycle every 3 steps, not 12. This formula lets you navigate the subgroup structure of cyclic groups precisely: every element of ⟨a⟩ is a power of a, and its order is determined entirely by how k and n are related via their gcd.