What is left when the order is forgotten

Two of the seventeen have been sitting side by side on this site since its first phase, and the way they are told apart has never quite been a proof.

The essay that introduced the pair makes the difference exactly: the mirrors of p3m1 pass through the three-fold centres, the mirrors of p31m run between them, and no rotation of the drawing carries one arrangement to the other. Another essay shows that the corresponding point groups are a single crystal class seen on two lattices, so the difference is not in the operations but in how they sit.

Every one of those statements is about the plane. And an isomorphism between two groups is not required to respect anybody’s plane: it is a bijection preserving multiplication, and it may do anything it likes to mirror lines. So the question are p3m1 and p31m the same group? has never actually been answered here.

It has a one-line answer, and the line contains no geometry at all.

Abelianising, and why it is an invariant #

Take a group and force every pair of elements to commute. Formally, quotient by the subgroup generated by all commutators; informally, throw away the order of the letters in every word, keeping only how many times each appears.

The result is an invariant. If two groups are isomorphic then their abelianisations are isomorphic, because an isomorphism carries commutators to commutators and so descends to the quotient. That is the whole of the logic, and it runs one way only: different abelianisations prove the groups are different; equal abelianisations prove nothing whatever.

The computation is short. A relator is a word; throwing away the order turns it into a row of exponent sums, one entry per letter. Stack the rows and the abelianised group is the cokernel of that integer matrix — generators, modulo the relations the rows impose. Reduce the matrix to Smith normal form and the invariant factors sit on the diagonal, with the generators that no row constrains left free.

p3m1 abelianises to ℤ2. p31m abelianises to ℤ6. A group of order two is not a group of order six, so the two plane groups are not isomorphic, and no drawing, relabelling, change of basis or choice of origin can make them so.

The difference between the two calculations is a single sign. In p3m1 the reflection sends the first translation to the inverse of the second; in p31m it sends it to the second. Abelianised, the first case forces the translations to square to nothing and then to vanish outright, and the second leaves a factor of three standing. One sign in one matrix column, and it decides a question that five essays of pictures could only illustrate.

Seventeen groups, ten answers #

Running the same computation across the whole classification gives a table that is interesting for both of the things it does.

What sits alone is settled. p1 is the only group abelianising to ℤ², pmm the only one giving four factors of two, p3 the only one giving ℤ3 ⊕ ℤ3. Each of those is now known to be non-isomorphic to every other plane group, from a computation that took a Smith normal form of an integer matrix with at most eight rows.

What shares a row is not settled either way. p2, pmg, cmm and p4m all abelianise to three factors of two. They are in fact pairwise non-isomorphic — they have different point group orders, for one thing, which an isomorphism would have to respect once the translation subgroup is pinned down — but this invariant does not see it, and saying that it does would be exactly the mistake the previous essay’s refusals exist to prevent.

The clusters are worth reading, because they are not random. pg and cm share an answer, and they are the two groups whose single reflection-like operation differs by whether it is a glide or a mirror — the very difference the centring essay argues is about the lattice rather than the pattern. p31m shares its answer with p6, which is a genuine coincidence and looks like one. And pgg, p4 and p4g agree on ℤ2 ⊕ ℤ4, which is three groups of two different point-group orders arriving at one small abelian group.

One more cheap invariant nearly finishes the job. The previous rung’s coset enumeration gives a second number for free — the index of the translation subgroup, which is the order of the point group — and it is an invariant of the group in the same sense, since the translations are the unique maximal abelian normal subgroup of finite index and an isomorphism has to carry them to each other. Sorting the seventeen by the pair takes ten classes to thirteen, leaving four tied pairs: pg with cm, pmg with cmm, pgg with p4, and p31m with p6. Two computations that never mention the plane get within four pairs of the whole classification, and neither of them can close the gap.

The free rank has a plain meaning. Four groups keep a free ℤ: p1, pm, pg and cm. Those are exactly the groups with a direction along which no operation reverses a translation — p1 has two such directions and the other three have one, the direction of the mirror or glide line. Everywhere else some operation sends a translation to its inverse, which in an abelian group forces it to have order two, and the free part dies. The rank is therefore a count of directions in which the group can wind without being sent back.

The check, which is the interesting half #

The calculation above is a small matrix reduced by hand-written integer arithmetic. It is exactly the kind of computation that can be quietly wrong: nothing in it is long enough to look suspicious, and its output is a short string that reads plausibly whatever it says.

So it is checked against a route sharing none of its code.

Take a finite quotient. Counting translations modulo N cells in each direction gives a finite group of order |P|·N² — finite because the sublattice of N-fold translations is invariant under the point group and therefore normal. That group can be enumerated element by element, its commutator subgroup generated from the commutators of its generators, the quotient formed as a set of cosets with a multiplication table, and the invariant factors peeled off it one at a time by repeatedly splitting off a cyclic factor of maximal order.

Then predict the same thing from the presentation. Adding the relators x^N and y^N to the presentation is a presentation of that same finite quotient, so the Smith normal form of the enlarged matrix must give the same answer.

The second figure is the one that earns its place. If both routes were wrong in the same way — reading the same matrix, sharing a helper — they would agree at every N. Agreeing on different answers at N = 2 and N = 3, where the finite groups have different orders and different structures, is much harder to do by accident.

The subgroup that was quotiented out #

The abelianisation is a quotient, so there is a subgroup underneath it, and for a plane group that subgroup is another plane group — which makes the invariant say something a reader can go and look at.

The index of the commutator subgroup is the order of the abelianisation. For p3m1 that is two, so the commutator subgroup has index two in p3m1, and there is only one candidate: the operations of determinant one. The commutator subgroup of p3m1 is p3, sitting inside it exactly as the two ways down describes a subgroup that keeps every translation and loses half the point operations.

For p31m the index is six, and the arithmetic forces a different shape. An index-six subgroup containing no reflection and only a third of the translations is a p3 on a sublattice three times as coarse — one of the sublattices that a three-fold rotation permits — which is a klassengleiche descent stacked on a translationengleiche one.

Four groups have a commutator subgroup of infinite index, which is what a free ℤ in the abelianisation means. p1’s commutator subgroup is trivial, since the group is already abelian; pm, pg and cm have one of index infinity, generated by the translations perpendicular to the mirror direction. An infinite index is not a failure of the computation — it is the statement that the group has a direction in which it winds freely, and it is exactly the free rank read backwards.

So the small abelian group on the right of the table is a measurement of how far a plane group is from commuting, and the subgroup it measures against is a plane group a reader has already met.

What no convention can move #

The reason this invariant is worth the trouble is that nothing a describer chooses can affect it.

A group’s description on this site is full of choices, and the essays say so repeatedly: the cell is a choice and the lattice is not, the origin can be moved to a second inequivalent position, the axes can be permuted so that one group answers to six symbols, and the normaliser is precisely the group of redescriptions that change nothing real.

Every one of those moves the matrices. A change of origin adds a translation to each operation’s translation part; a change of basis conjugates every matrix; a change of setting permutes the axes and renames the symbol. The abelianisation is untouched by all of it, because it is computed from a presentation and a change of description merely renames the letters.

That is the sense in which it is a stronger kind of statement than the ones the pictures make. p3m1’s mirrors run through its rotation centres is true, useful, and stated about a drawing; a reader who chose different axes would have to restate it. p3m1 abelianises to ℤ2 is true of the group, and there is no description in which it is false.

What this does not decide #

The seventeen are pairwise non-isomorphic, and the abelianisation does not prove it. The full result is Bieberbach’s: two crystallographic groups of the same dimension are isomorphic if and only if they are conjugate in the affine group. Since the seventeen are seventeen distinct affine classes — which is what the classification proves — they are seventeen distinct abstract groups. That is a considerably heavier theorem than a Smith normal form, and it is the one that settles the eleven cases the invariant leaves open.

What the abelianisation buys is a cheap, exact and completely convincing answer in the cases where it answers at all, obtained without appeal to any classification. For p3m1 and p31m — the pair a reader is most likely to suspect of being the same group described twice — it answers.

A finite quotient is not the group. The check above verifies the presentation against enumerations of G/(N·L), which are different groups from G and from each other. It tests the machinery, not the limit.

Nothing here is about the plane, which is the point and also the limitation. The abelianisation cannot tell where a mirror line runs, cannot distinguish two settings of one group, and has nothing to say about a pattern. It answers one question, exactly.

Who found it, and when #

Abelianisation is as old as the commutator subgroup, which appears in work of Dedekind and of Frobenius in the 1880s and 1890s; the name derived group is the older one. Its use as an invariant — compute it for two groups, and if the answers differ, stop — is standard from the beginning of combinatorial group theory.

The Smith normal form is H. J. S. Smith’s, from an 1861 paper on systems of linear congruences, twenty years before anyone was abelianising groups. The theorem that any integer matrix reduces to a diagonal one whose entries successively divide each other, by integer row and column operations, is exactly what is needed to read the invariant factors off a relation matrix, and it was available long before the question was asked.

Bieberbach’s theorems are of 1911 and 1912, answering the part of Hilbert’s eighteenth problem that asks whether there are finitely many crystallographic groups in each dimension. The second of them — isomorphic implies affinely conjugate — is what makes the classification of the seventeen a classification of groups rather than only of patterns, and it is the reason a reader can be told that the seventeen are seventeen and believe it.

The connection worth carrying away is that the abelianisation of a plane group is the first homology of its orbifold, in the sense the orbifold essays use the word. A group’s quotient of the plane is a surface with cone points and mirror boundaries; its first homology is the abelianisation of the group; and Conway’s magic theorem, which counts the seventeen by spending exactly two dollars on features, is a statement about the Euler characteristic of the same object. The three computations — the cost, the homology, and the relation matrix — are looking at one thing from three sides.

The same answer, read off a folded surface #

The connection to orbifolds is named above as a remark, and it is a third route to the same numbers — one with no matrix in it at all, and with the pieces of the answer coming from visibly different features.

Fold a plane group’s pattern along its own symmetries and what remains is an orbifold with a symbol. The abelianisation of the group is the first homology of that orbifold, and homology of a surface with marked points is read off the symbol directly.

The free part comes from the handles and the boundary — a torus contributes ℤ², which is why p1 is the only group with two free factors. The torsion comes from the cone points: a cone of order n contributes a factor of order n, subject to one relation among all of them.

Run it on the pair. p3m1 folds to *333 — a disc with three corners of order three — and its homology comes out with a single factor of two. p31m folds to 3*3, a disc with one cone point of order three and one corner of order three, and the extra cone point contributes an extra factor of three. ℤ₂ against ℤ₆, from counting features of two surfaces.

That is a genuinely independent calculation. The matrix route reduces a presentation over the integers; the orbifold route reads a symbol. Neither mentions the plane, and they agree, which is the arrangement this collection puts under every number it reports.

Why counting subgroups adds nothing #

There is an obvious way to strengthen a weak invariant — count something else — and the natural candidates in this collection turn out to be the same invariant in disguise, which is worth knowing before anybody spends effort on them.

The two-colour census counts subgroups of index two, and the three-colour one counts index three. Both are counts of surjections onto a cyclic group of prime order, and a surjection onto an abelian group factors through the abelianisation — so both counts are determined by the table on this page and neither can separate two groups the table does not.

The arithmetic is explicit. The count at prime p is (pʳ − 1)/(p − 1), with r the rank of the abelianisation modulo p. So p2, pmg, cmm and p4m — which share a row here — necessarily share their two-colour counts and their three-colour counts as well, and they do.

So the four groups in that row cannot be separated by any count of abelian quotients whatever, and separating them needs an invariant that sees the order of the letters rather than one that throws it away. The growth of the group is one such; the arithmetic class is another; and the classification itself is the complete one.

That is a useful thing to know about a cheap invariant: strengthening it by counting more abelian things is not strengthening it at all.

Where this ladder goes #

The presentation has now been used to count cosets and to compute an invariant. The third use is a lower bound: the number of invariant factors of the abelianisation is the fewest generators the group can possibly have, because making a group abelian cannot make it harder to generate.

That number is a floor, and an exhaustive search over the operations within one cell of the origin supplies a ceiling. For fourteen of the seventeen the two meet and the fewest generators is settled exactly — including the fact that p2, a group whose every operation is a half-turn, cannot be generated by two of them. The next rung is that computation, and the three groups where the floor is not tight.