Questions: Cycle Notation and Decomposition

Question 1

What is the order of the permutation (1 2 3)(4 5 6 7)?

• A. 3 — the length of the first cycle
• B. 4 — the length of the second cycle
• C. 7 — the total number of elements moved
• D. 12 — the LCM of the cycle lengths

Answer: (D)

The order of a permutation in disjoint cycle form equals the LCM of its cycle lengths, not the sum or product. Here LCM(3, 4) = 12: after 12 applications, the 3-cycle has completed 4 full rotations (back to start) and the 4-cycle has completed 3 full rotations (back to start), so every element is back to its original position. 12 is the smallest such number. The common wrong answers — 3 or 4 — confuse the order with one cycle's length, and 7 confuses it with a sum.

Question 2

A student writes: '(1 2)(2 3) = (2 3)(1 2) because disjoint cycles commute.' What is wrong with this reasoning?

• A. Nothing — both expressions produce the same permutation since order never matters for two-cycles
• B. The student forgot that cycles must first be converted to one-line notation before comparing
• C. These cycles are not disjoint — they share element 2 — so the commutative property does not apply
• D. Disjoint cycles do not actually commute; the student has misremembered the theorem

Answer: (C)

The commutative property applies only to DISJOINT cycles — cycles whose element sets have no overlap. (1 2) and (2 3) both contain element 2, so they are not disjoint. In fact, (1 2)(2 3) and (2 3)(1 2) produce different permutations: the first sends 1→3, while the second sends 1→2→... — they are not equal. The theorem that disjoint cycles commute is a precise exception to the general non-commutativity of permutation composition, not a blanket rule.

Question 3

The permutation (1 3 5)(2 4) and the permutation (2 4)(1 3 5) are equal, because disjoint cycles commute.

• T. True
• F. False

Answer: True

This is correct. (1 3 5) acts on {1, 3, 5} and (2 4) acts on {2, 4} — these sets are disjoint. Because the cycles act on completely separate elements, each cycle has no effect on the elements of the other. Applying them in either order produces identical results for every element in the set. This is the one important exception to the general rule that permutation composition is not commutative.

Question 4

The permutation (1 2 3 4 5)(6 7) has order 7, because 5 + 2 = 7.

• T. True
• F. False

Answer: False

The order equals the LCM of the cycle lengths, not their sum. LCM(5, 2) = 10, not 7. After 10 applications, the 5-cycle has completed 2 full rotations and the 2-cycle has completed 5 full rotations — every element is back to its start. After only 7 applications, the 5-cycle would have completed 1 full rotation plus 2 extra steps, so elements 1–5 would NOT be back to their original positions. The sum-of-lengths is a very common wrong answer; always use LCM.

Question 5

Why does the order of a permutation equal the LCM of its cycle lengths rather than the sum or product? Explain the reasoning in terms of what 'order' means.

Model answer: The order of a permutation is the smallest number of times you must apply it for every element to return to its starting position. Each disjoint cycle operates independently on its own set of elements. A k-cycle returns to the identity after exactly k applications. For all cycles to simultaneously return to their starting positions, you need a number of applications that is a multiple of each cycle length — the smallest such number is the LCM. The sum has no such interpretation, and the product is generally too large.

This is a direct consequence of how disjoint cycles interact (or rather, don't interact). Because they act on separate elements, each cycle's 'reset time' is independent. You need to wait until ALL cycles have simultaneously completed whole numbers of rotations — that's the definition of LCM. A concrete check: for (1 2 3)(4 5), after 3 steps the first cycle resets but the second hasn't (it needs 2 or 4 steps to reset); after 2 steps the second resets but the first hasn't; after LCM(3,2)=6 steps, both reset simultaneously.