Where the double slash comes from

Non-regular colourings, and why H must be named

The 2×2 torus group has four non-normal subgroups. Each defines a colouring with four colours but a colour group of order eight — non-regular. Two of them produce the same Conway symbol and yet are different colourings.

Regular versus non-regular

A colouring is regular when the number of colours equals the order of the colour group, which happens exactly when H = K. Compare:

ABBA
Regular. H = K = {e, XY, s, XYs}, a normal subgroup of index 2. Two colours, colour group C₂ of order 2. Colours = group order, so H is recoverable from K and Conway writes a single slash.
ABCD
Non-regular. H has order 2 and K = core(H) = {e}. Four colours, colour group of order 8. Eight ≠ four, so H is not recoverable — double slash. The picture is trivial; all the structure is in the action.

The action is where the content is

For the rainbow pattern every element of Γ permutes the four colours. Rows shaded green lie in H (they fix colour A); the darker row is K.

elementordercolour permutation fixes Ain K
e1identity
X2(A C) (B D)
XY2(A D) (B C)
XYs2(A D)
Y2(A B) (C D)
s2(B C)
Xs4(A C D B)
Ys4(A B D C)

Only the identity fixes every colour, so K = {e}. Two elements fix A, so H has order 2 and index 4 — the four colours. The map Γ → S₄ is injective but lands in a subgroup of index 3, which is exactly the gap between 8 and 4.

The two cases

H = {e, s}

This one is realized by the grid itself: the four cells are the coset space Γ/⟨s⟩, and the rainbow pattern is its colouring.

esYsYXYXYsXsX

Each thick coloured edge is one coset of H — one colour. Thin grey edges change the colour.

the four cosets
A{e, s}sends AA
B{Y, Ys}sends AB
C{X, Xs}sends AC
D{XY, XYs}sends AD
colour stabilizers
A{e, s}
B{XYs, e}
C{XYs, e}
D{e, s}

2 distinct stabilizers for 4 colours, since |N(H)|/|H| = 4 / 2.

G = Γ (8)  ·  H = order 2, index 4  ·  K = core(H) = {e}  ·  4 colours, colour group of order 8 → non-regular

H = {X, e}

This one is not realized by the cells: ⟨X⟩ has only two orbits on them, so the colouring needs a different four-element Γ-set. Drawn abstractly, as the four cosets.

esYsYXYXYsXsX

Each thick coloured edge is one coset of H — one colour. Thin grey edges change the colour.

the four cosets
A{X, e}sends AC
B{Ys, s}sends AB
C{XY, Y}sends AD
D{XYs, Xs}sends AD
colour stabilizers
A{e, s}
B{XYs, e}
C{XYs, e}
D{e, s}

2 distinct stabilizers for 4 colours, since |N(H)|/|H| = 4 / 2.

G = Γ (8)  ·  H = order 2, index 4  ·  K = core(H) = {e}  ·  4 colours, colour group of order 8 → non-regular

Why the symbol is not enough

Both cases give identical G, identical K, and the same number of colours — therefore the same Conway symbol Γ⁴//{e}:

H = ⟨s⟩ H = ⟨X⟩
GΓ, order 8Γ, order 8
K = core(H){e}{e}
colours44
symbolΓ⁴//{e}Γ⁴//{e}
realized on the cells?yesno
the colourings aredifferent

They differ because ⟨s⟩ and ⟨X⟩ are not conjugate — they lie in the two separate classes of non-normal subgroups — so the Γ-sets Γ/⟨s⟩ and Γ/⟨X⟩ are not isomorphic. Same symbol, two colourings.

That is the whole reason Conway's notation carries three groups rather than two. The double slash in Gⁿ//K is a warning that H has been suppressed and cannot be reconstructed; the full form Gⁿ/H/K names it. On an abelian ambient this never arises — every subgroup is normal, H = K always, and a single slash suffices.

Generated by scripts/generate_torus_nonregular.py. Companion to the Cayley graph page and the torus tutorial.