Eight elements, ten subgroups, one octagon
Translations of the wrapped 2×2 grid together with the diagonal flip give a group of order 8. Small enough to draw completely — every element, every subgroup, and the pattern that picks each subgroup out.
Mod 2 the quarter turn r(a,b) = (−b, a) becomes the diagonal flip
s(a,b) = (b,a), because −1 ≡ 1. So there is no genuine
rotation here, and r² = 1. With the four translations this generates
Γ = (ℤ/2)² ⋊ ⟨s⟩ of order 8,
which is D₄ — acting on the four cells as a Sylow 2-subgroup of
S₄.
Writing X and Y for the two translations, the flip conjugates one
into the other: s X s⁻¹ = Y. So Y is redundant and
Γ = ⟨X, s⟩ — and since X² = s² = 1
with (Xs) of order 4, this is the dihedral presentation
⟨X, s | X² = s² = (Xs)⁴ = 1⟩.
Xorder 2sorder 2
X,
s. Girth 8, diameter 4.Xorder 2Yorder 2sorder 2
Y splits the group into two squares (the cosets of the
translation subgroup) joined by a perfect matching of s-edges. Girth 4,
diameter 3.
Dropping s altogether disconnects the graph into two components — the index of
the translation subgroup, visible as connected components.
Each card shows Cay(H, gens) with one colour per generator, next to the
orbit colouring: colour each orbit of H on the four cells a different
colour. That colouring is how you specify a subgroup by drawing — but it only works for
seven of the ten, and the three failures are labelled with their reason.
{X, XY, XYs, Xs, Y, Ys, e, s}
Xorder 2XYsorder 2
X, XYs[[0, 1, 2, 3]]{X, XY, Y, e}
Xorder 2XYorder 2
X, XY[[0, 1, 2, 3]]{XY, XYs, e, s}
XYorder 2XYsorder 2
XY, XYs[[0, 3], [1, 2]]{XY, Xs, Ys, e}
Xsorder 4
Xs[[0, 1, 2, 3]]{X, e}
Xorder 2
X[[0, 2], [1, 3]]{XY, e}
XYorder 2
XY[[0, 3], [1, 2]]{XYs, e}
XYsorder 2
XYs[[0, 3], [1], [2]]{Y, e}
Yorder 2
Y[[0, 1], [2, 3]]{e, s}
sorder 2
s[[0], [1, 2], [3]]{e}
[[0], [1], [2], [3]]
The subgroups above are single groups. A colour group records the whole triple
that a colouring produces:
G the symmetries permuting the colours,
H those fixing one chosen colour, and
K those fixing every colour. All three are stabilizers —
of the unordered partition, of one block, and of the labelled partition respectively —
and K ⊆ H ⊆ G. The colouring is regular when
|G/K| equals the number of colours, which happens exactly when
H = K; Conway then writes a single slash.
G2/KG = ⟨ X, s |
X2 = s2 = (Xs)4 = 1 ⟩
X | ↦ | (AB) | order 2 |
s | ↦ | 1 | order 1 |
signature X(AB) s1
short form X2 s1
ρ(X)2 = 1 ✓ ρ(s)2 = 1 ✓ ρ(Xs)4 = 1 ✓
| G | {X, XY, XYs, Xs, Y, Ys, e, s} | order 8 |
|---|---|---|
| H | {XY, XYs, e, s} | index 2 |
| K | {XY, XYs, e, s} | order 4 |
G2/KG = ⟨ X, Y |
X2 = Y2 = (XY)2 = 1 ⟩
X | ↦ | (AB) | order 2 |
Y | ↦ | 1 | order 1 |
signature X(AB) Y1
short form X2 Y1
ρ(X)2 = 1 ✓ ρ(Y)2 = 1 ✓ ρ(XY)2 = 1 ✓
| G | {X, XY, Y, e} | order 4 |
|---|---|---|
| H | {Y, e} | index 2 |
| K | {Y, e} | order 2 |
Both are regular, so H = K and a single slash suffices. The first is the
checkerboard, whose symmetry is all of Γ; the second is the striped colouring, where
the diagonal flip fails to act and G drops to the translation subgroup. Six
chromatic two-colourings in total — the colour classes must have equal size, so only
the three 2+2 partitions qualify, each with two labellings.
There are no chromatic three-colourings: the colour group permutes the
colour classes transitively, so they must all have the same size, and
3 ∤ 4. Only k dividing 4 is possible — one, two or
four colours.
G4//KG = ⟨ X, s |
X2 = s2 = (Xs)4 = 1 ⟩
X | ↦ | (AC)(BD) | order 2 |
s | ↦ | (BC) | order 2 |
signature X(AC)(BD) s(BC)
short form X2 s2
ρ(X)2 = 1 ✓ ρ(s)2 = 1 ✓ ρ(Xs)4 = 1 ✓
| G | {X, XY, XYs, Xs, Y, Ys, e, s} | order 8 |
|---|---|---|
| H | {e, s} | index 4 |
| K | {e} | order 1 |
Here |G/K| = 8 but there are only 4 colours, so H ≠ K and the
symbol carries a double slash. This is the rainbow colouring; the
non-regular page works through why H has
to be named separately.
| |H| | generators | index | normal | orbits | as a stabilizer |
|---|---|---|---|---|---|
| 8 | X, XYs | 1 | yes | 1 | drawable |
| 4 | X, XY | 2 | yes | 1 | not drawable |
| 4 | XY, XYs | 2 | yes | 2 | drawable |
| 4 | Xs | 2 | yes | 1 | not drawable |
| 2 | X | 4 | no | 2 | drawable |
| 2 | XY | 4 | yes | 2 | not drawable |
| 2 | XYs | 4 | no | 3 | drawable |
| 2 | Y | 4 | no | 2 | drawable |
| 2 | s | 4 | no | 3 | drawable |
| 1 | — | 8 | yes | 4 | drawable |
Seven of the ten subgroups are stabilizers of some pattern. The three exceptions fail for
two distinct reasons: the translation subgroup and the cyclic C₄ are
transitive, so they force the constant pattern; the centre
{e, XY} forces a pattern that carries strictly more symmetry than asked for.
Realizable subgroups are exactly the partition stabilizers, and they form a closure system.
Generated by scripts/generate_torus_cayley.py. Companion to the torus tutorial.