Seventeen and No Moreevery way a pattern can repeat across a plane — the whole list, and why it ends

Take any pattern that repeats forever in two independent directions — a real wallpaper, a tiled floor, the atoms in a slab of salt. Ask only what rigid motions leave it looking unchanged: slides, turns, mirror-flips, and glide-reflections (flip-and-slide, the way footprints repeat). The complete answer was settled by 1891, by the Russian crystallographer Evgraf Fedorov, and again by George Pólya in 1924: there are precisely 17 such symmetry groups in the plane. This page is that catalogue, made to play with — and, more to the point, the two reasons the list can never grow.

The first reason forbids most of what you might imagine. The second counts what remains, and lands on seventeen.

INSTRUMENT I — THE FORBIDDEN PENTAGONThe crystallographic restriction

Here is a fact that surprises everyone the first time: a repeating planar pattern can have centres of 2-, 3-, 4-, or 6-fold rotation, and nothing else. No 5-fold centre. No 7-fold, no 8-fold. You can draw a single beautiful pentagon, but you cannot tile the plane periodically so that a fifth-turn leaves it unchanged. (Quasicrystals get five-fold symmetry precisely by giving up periodicity — that is another story entirely.)

The proof is two lines. A rotation that maps a lattice onto itself, written in the lattice's own basis, is a matrix of whole numbers — it sends grid points to grid points. The trace of a rotation by angle θ is 2 cos θ. A whole-number matrix has a whole-number trace. So 2 cos(2π/n) must be an integer — and that happens only for n = 1, 2, 3, 4, 6. The pentagon's value, 2 cos 72° = 0.618…, is the reciprocal golden ratio: a root of x² + x − 1, stubbornly irrational, never a whole number.

The restriction constrains the rotational symmetry of periodic crystals. It does not choose hexagons for a honeycomb. Regular pentagons cannot tile a flat floor because their 108° corners cannot fill 360° around a vertex. The lattice restriction cuts the infinite zoo of imaginable symmetries down to a handful of allowed rotation centres. What it does not yet tell us is how those centres combine with mirrors and glides, and that is where the number seventeen actually comes from.

INSTRUMENT II — THE WHOLE CATALOGUEAll seventeen, built live

Pick a group below. Its pattern builds itself from a single asymmetric motif, repeated by exactly the rigid motions that group allows — and nothing more. Drag inside the canvas to move the motif; every copy moves in lockstep, which is what "symmetry" really means. cyan copies are mirror-images (left-handed); gold copies are direct (right-handed). Toggle the skeleton to see the rotation centres (◆ 2-fold, ▲ 3-fold, ■ 4-fold, ⬢ 6-fold), mirror lines (solid), and glide axes (dashed).

Notice how the four "families" of rotation — 2-fold, 3-fold, 4-fold, 6-fold — each branch by what mirrors they admit. p4 has four-fold turns and no mirror; add mirrors through the centres and you get p4m; offset them and you get p4g. The catalogue is not a random list of seventeen things; it is a tree with a very particular number of leaves. The next instrument says exactly why that number is seventeen.

INSTRUMENT III — THE MAGIC THEOREMConway's cost ledger

The cleanest proof that the count is seventeen, and not sixteen or eighteen, is John Conway's, from The Symmetries of Things (2008). Fold the repeating plane down by its own symmetry and you get a small patched surface called an orbifold; its features carry a "cost," and the orbifold of a flat, repeating pattern must cost exactly 2. For an admissible signature, less than 2 gives spherical symmetry; more than 2 gives hyperbolic symmetry (infinitely many groups). Below 2 there are exceptions: a lone cone or two unequal cones, and the corresponding mirror-boundary signatures, are not spherical symmetry groups. Cost exactly 2 is the knife-edge of flatness, and the symbols that balance there are seventeen, no more.

Run the enumeration over every finite orbifold symbol and ask which cost exactly 2, and you get these seventeen and not one more. Allow an infinite-order symbol (an ∞ in the signature) and you also pick up the seven frieze groups — the symmetries of a border or a railing, repeating in one direction only. Everything is the same machine; only the total changes the world it lives in.

THE HONEST FOOTNOTEAre all seventeen really in the Alhambra?

You will read, in a hundred popular accounts, that the Moorish artisans of the Alhambra in Granada discovered all seventeen groups by hand, centuries before mathematicians named them. It is a wonderful story, and it may be too good. In their 1986 paper reporting the survey made in 1983, Branko Grünbaum, Zdenka Grünbaum and G. C. Shephard wrote that they could find clear examples of at most thirteen of the seventeen, and argued that p2, pg, pgg and p3m1 are simply absent. Grünbaum returned to the question in the Notices of the AMS in 2006 under a pointed title: "What Symmetry Groups Are Present in the Alhambra?". The answer was: fewer than the legend claims, and the exact count depends on which colours, interlacing and extent of a pattern you count.

So the honest version is the better one. The seventeen groups are a theorem, complete and exact; the claim that one building contains them all is an empirical assertion that careful counting does not support. The mathematics is certain. The tilework is human, and partial, and still argued over — which is exactly why it is worth telling the truth about.

INSTRUMENT I, MADE CERTAIN · added 2026-06-29 · full restriction 2026-08-20All five orders, checked by the kernel

The argument in Instrument I runs through the real numbers: it leans on the trace being 2 cos θ, and on 2 cos 72° being irrational. That is true and it is enough, but it asks you to trust a fact about cosines. The restriction can be pinned to something smaller and harder: a statement with no cosines, no reals, no geometry at all, just integers, and then handed to a proof assistant, which certifies it for every integer matrix at once.

An orientation-preserving lattice rotation, in the lattice basis, is a 2×2 integer matrix of determinant 1 (it is area- and orientation-preserving). A finite-order rotation is one whose kth power is the identity for some k. So the full restriction ("the only orders are 1, 2, 3, 4, 6") is exactly:

theorem crystallographic_restriction (M : M2)
    (hdet : M.det = 1) (h : M.npow (k+1) = I) :
    M = I ∨ M.npow 2 = I ∨ M.npow 3 = I
         ∨ M.npow 4 = I ∨ M.npow 6 = I

The proof is pure integer algebra, and the extra reach past n=5 costs almost nothing over the pentagon argument. Cayley–Hamilton gives M² = t·M − I with t = trace M. Iterating it packages every power as a two-term combination: M^k+1 = U_k+1(t)·M − U_k(t)·I, where U is the Chebyshev-like integer sequence U₀=0, U₁=1, U_k+2 = t·U_k+1 − U_k. The load-bearing arithmetic fact is that for every integer |t| ≥ 3 the sequence |U_k| is strictly increasing from 1, so it never returns to zero. That rules the trace out of {…,−4,−3,3,4,…} without any real-number machinery. What is left is five allowed traces: t = −2 gives −I (order 2), t = 2 gives I (order 1), and the three middle values give M³=I, M⁴=I, M⁶=I unconditionally (no finite-order hypothesis needed for the middle three; they hold for every det-1 integer matrix of that trace).

The pentagon case that shipped first, no_order_five, is now a two-line corollary via a Euclidean-descent lemma that needs no matrix inverses: if M^p=I and M^q=I with p ≥ q, then M^p−q = M^p after cancelling I on the right, so M^p−q=I. Applied to the umbrella's five disjuncts with p=5, each one descends to M¹=I.

THE 4D COUNTERPOINT · same file, two dimensions higherThe pentagon does come back, in four dimensions

The 2D restriction is sharp in dimension and only in dimension. What breaks the pentagon is not that five is impossible for a lattice; it is that five is impossible for a lattice in two dimensions. Raise the ambient rank by two and the argument stops binding: at dimension 4, an integer matrix of order exactly 5 exists, and one can write it down.

The witness is the companion matrix of the fifth cyclotomic polynomial Φ₅(x) = x⁴ + x³ + x² + x + 1. Written out, it is

C =  ⎡ 0  0  0  −1 ⎤
     ⎢ 1  0  0  −1 ⎥
     ⎢ 0  1  0  −1 ⎥
     ⎣ 0  0  1  −1 ⎦

Two things about this. First, four is not arbitrary; it is the smallest integer dimension the pentagon can occupy. Second, this is what quasicrystals are doing. Three-dimensional quasicrystals are described by cut-and-project models: five-dimensional lattices for decagonal quasicrystals, six-dimensional lattices for icosahedral ones, including Shechtman's rapidly-quenched Al–Mn alloy (1982; Nobel Prize in chemistry 2011). The Nobel scientific background, p. 9, distinguishes these two dimensions. Periodicity is what forbids the pentagon at rank 2 and 3, and giving up periodicity in 3-space (by living in the shadow of a higher-dimensional lattice) is what returns it.