The m-mark dial
Reachable marks
Chosen x → A(x)
A symmetry?
Bonus lesson 12 · Symmetry · 65–80 minutes
The Hidden Symmetry Group
Lesson 08 found a mirror in the list of readings. Here we look for symmetries of the dial itself: ways to relabel the marks, by stretching and turning, that send the reachable marks onto the reachable marks. Lesson 03's square test, 8x + 1, turns out to explain all of them. You will find every such symmetry, see how they sort the reachable marks into orbits, and count both from the factorization of the dial.
Where this lesson is headed
A(x) = ux + c (mod m) is a symmetry when it sends Reach(m) onto Reach(m)
In words: stretch the dial by u and turn it by c; if every reachable mark lands on a reachable mark, that relabeling is a symmetry. On an odd prime dial p ≥ 5 there are exactly (p − 1)/2 of them, one for each nonzero square u, with c chosen so that 8x + 1 is simply multiplied by u. Setup as in Lessons 03 to 07: the clock with starting circumference m and growth 1 on an m-mark dial, so the readings are T(k) mod m.
u: the stretch, sharing no factor with m (so it can be undone)
c: the turn
m: number of dial marks
Reach(m): the marks T(k) ever reaches (two blocks, k = 0 to 2m − 1); its size is F(m) from Lesson 07
D(x) = 8x + 1: Lesson 03's square test
Orbit: marks that some symmetry can move onto each other
01. Stretch and turn the dial
Start with the 5-mark dial, where the reachable marks are 0, 1 and 3. The map A(x) = 4x + 1 sends 0 to 1, 1 to 0, and 3 to 13, which reads 3. Every reachable mark lands on a reachable mark, so this relabeling is a symmetry. A symmetry does not have to leave marks in place; it only has to keep the reachable set as a whole. Try other stretches and turns.
Guided experiments
Do these in order. "Set this up" picks the dial and the map.
The orbits
Two reachable marks are in the same orbit if some symmetry moves one onto the other. Each box is one orbit of the current dial.
Look for: on a prime dial p ≥ 5, one mark always sits alone in its own orbit. Which one? Compute 8x + 1 for it.
02. The group hiding in the square test
The plan: use D(x) = 8x + 1 to turn reachable marks into squares, find the stretches that keep squares as squares, then count orbits piece by piece and put the pieces together with CRT (Lesson 07). Each step has the general idea on the left and numbers on the right.
1
Change coordinates with the square test
From Lesson 03, 8T(k) + 1 = (2k − 1)², and on an odd dial b both x ↦ 8x + 1 and k ↦ 2k − 1 can be undone. So
x is reachable ⇔ D(x) = 8x + 1 is a square on b marks
Relabeling x as D(x) turns Reach(b) into the set of squares (including 0).
2
Stretches that keep squares as squares
What does a stretch-and-turn do to D?
D(ux + c) = 8ux + 8c + 1 = u·D(x) + (8c + 1 − u)
Choose the turn so the last bracket reads 0: c ≡ (u − 1)·8−1. Then D(A(x)) = u·D(x). If u = w² is a nonzero square, a square times u is still a square, and dividing by u undoes it. So A reshuffles the reachable marks: it is a symmetry.
These maps form a group: doing Av then Au gives Auv, A1 does nothing, and Au is undone by Au⁻¹. On a prime dial p ≥ 5 there is one for each nonzero square, (p − 1)/2 in all. The search in Stage 01 finds no others.
u = w²
c = (u − 1)/8
(p − 1)/2
| x | D(x) | A(x) | D(A(x)) |
|---|
3
The exception: 3 marks
On 3 marks the reachable set is {0, 1}. The only nonzero square is 1, so step 2 gives only the identity. But A(x) = 2x + 1 also works:
0 → 1reachable → reachable
1 → 0reachable → reachable
Here u = 2 is not a square, so this symmetry is extra. It appears whenever 3 divides the odd part of the dial exactly once. On 9 or 27 marks it disappears again.
Why not just call u = 2 a square?
On 3 marks, 1² = 1 and 2² = 4 reads 1, so 2 is not a square. The map 2x + 1 works for a different reason: with only two reachable marks, swapping them is always allowed. So the rule "only squares" needs p ≥ 5.
4
Powers of two: every stretch and turn works
On a 2a-mark dial, the two blocks reach every mark (Lesson 05). If every mark is reachable, any stretch that can be undone and any turn keeps the reachable set, so there are
2a × φ(2a) = 2a × 2a−1 = 22a−1
symmetries (2a turns, 2a−1 odd stretches), and all marks form one orbit.
For example
On 8 marks: 8 turns × 4 odd stretches = 32 symmetries. Check with the search in Stage 01 (m = 8).
5
Orbits: sort marks by how much p divides 8x + 1
On a pe-mark dial (p odd), a nonzero square D has p0, p2, p4, … as the exact power of p dividing it (the layers of Lesson 07). Multiplying by a unit square never changes the layer, and it can move any square in a layer to any other in the same layer. So each layer is one orbit, plus the orbit {D = 0}:
orbits on pe marks = ⌈e/2⌉ + 1
except on 3 marks, where the extra symmetry fuses the two into one. The mark with D(x*) = 0, that is x* ≡ −8−1, never moves (D(A(x*)) = u·0 = 0). On a prime dial it is the lone mark of Lesson 04, the only mark reached by just one rotation.
| p divides 8x + 1 exactly | Marks in this layer | Orbit |
|---|
6
Any dial: pieces multiply
By CRT (Lesson 07), a symmetry of the whole dial is one symmetry of each prime-power piece, chosen independently, and an orbit is one orbit from each piece. So both numbers multiply:
symmetries = ∏ (piece counts), orbits = ∏ (piece orbits)
with (p − 1)pe−1/2 symmetries on an odd piece pe (twice that on 3¹), 22a−1 on the power-of-two piece, and orbits as in steps 4 and 5. Adding up orbit sizes gives back F(m), the count from Lesson 07.
| Piece | Kind | Symmetries | Orbits |
|---|
Predicted symmetries
Search
Predicted orbits
7
Which other clocks share this structure?
Take a general clock S(k) = kN + gT(k) on an M-mark dial (Lesson 02). The identity T(k + t) − T(t) = T(k) + tk gives
g·[T(k + t) − T(t)] = gT(k) + (gt)k
If we can pick t with g·t ≡ N (mod M), the right side is S(k). Then the clock's readings are g × (T(k + t) − T(t)): Reach turned by −T(t), then multiplied by g. When g shares no factor with M, the multiplication can be undone, so the clock's reachable set is a stretched and turned copy of Reach(M), with the same numbers of symmetries and orbits. When d = gcd(g, M) > 1, the same happens on the smaller M/d-mark dial, as in Lesson 06. Such a t exists exactly when gcd(g, M) divides N.
The condition is needed. On M = 8 with g = 2 and N = 1, S(k) = k + 2T(k) = k², which reaches {0, 1, 4}: 3 marks. But g × Reach(4) = {0, 2, 4, 6} has 4. No turned copy can match.
03. Group, orbits, and limits
Answer each question, then read the explanation. The computer search helped find the statements; the arguments of Stage 02 are what prove them. One honest limit: Stage 02 shows the square-stretch maps are symmetries; that there are no others is confirmed here by complete search on every dial tested, but its general proof goes beyond this course.
Eight checkpoints
Computer check
For 21 dials, tries every stretch and turn directly and compares the number of symmetries and orbits with the formulas; also checks the square-stretch construction on nine prime dials.
Not run yet. Many matches are strong evidence, but only the arguments cover every dial.
Final exit problem · one dial, completely
Work on the 35-mark dial, 35 = 5 × 7.
Show the worked solution
On 5 marks there are (5 − 1)/2 = 2 square stretches; on 7 marks, (7 − 1)/2 = 3. CRT gives 2 × 3 = 6 symmetries. Each prime piece has two orbits (8x + 1 a nonzero square, or 0), so 2 × 2 = 4 orbits. Since 8 × 22 = 176 = 5 × 35 + 1, 8−1 is 22, so x* ≡ −22 ≡ 13. Check: 8 × 13 + 1 = 105 = 3 × 35.
Course conclusion
One identity, 8T(k) + 1 = (2k − 1)², has now answered three questions about the growing clock: which marks it reaches (Lesson 03), how many (Lessons 04 and 07), and how the reachable marks can be relabeled and sorted into orbits (this lesson). The special cases stay special: the 3-mark dial, powers of two, and growth that shares a factor with the dial each need their own stated conditions.