Three of them, and they are equivalent

The subgroup tables are lists of counts. Under a group, a type and an index, an entry reads three subgroups, and the reader is expected to know what the three have to do with one another. Sometimes an operation of the group carries one onto another, so that nothing computed inside the group can tell them apart; sometimes nothing in the group moves any of them, and they are still exchanged by a motion that leaves the group as a whole exactly where it was. The first relation is conjugacy. The second is the one the same site under two names found acting on Wyckoff positions, and it acts on subgroups in the same way and for the same reason.

The distinction decides what a count means. Three conjugate subgroups are one subgroup seen from three places inside the group. Three subgroups in one normaliser set are three genuinely different subgroups that no description of the crystal can prefer between. This essay computes both partitions for all seventeen plane groups at the two smallest indices, and finds the second coarser than the first in nine of the thirty-four cases.

Over the seventeen there are 74 subgroups of index two, in 74 conjugacy classes and 56 sets, and 82 of index three, in 36 classes and 32 sets. The first row of numbers says something on its own: at index two, classes and subgroups agree, because a subgroup of index two is always normal and a normal subgroup is a class by itself. Everything that happens at index two, then, happens after conjugacy has already given up.

What a subgroup of small index cannot escape #

The computation needs every subgroup, not a chain of maximal ones, and the subgroups are infinite objects. What makes them finite is an observation about translations.

Let H have index n in G, and let T be the translations. The translations inside H are T ∩ H, and the cosets of T ∩ H in T inject into the cosets of H in G, so $[T : T \cap H]$ divides n. A sublattice of index m contains every vector multiplied by m, so it contains $n\Lambda$ whenever m divides n. Every subgroup of index n therefore contains n times the lattice, and so is a subgroup of the finite group $G/n\Lambda$, whose order is the point group’s order times $n^2$ — 108 elements at worst here, for p6m at index three.

So the enumeration is finite and complete: list the subgroups of that finite group, keep those of index n, and act on them twice. Conjugating by the group’s own elements gives the conjugacy classes. Conjugating by the Euclidean normaliser — every motion that carries the group’s operations onto the group’s operations, which is not a property of the group alone but of the group on its cell — gives the sets.

Two of the counts have already been made by another route entirely. How many subgroups of index three counts subgroups by their transitive actions on three points — permutations, with no matrix and no lattice anywhere in it — and counts the conjugacy classes as the orbits of those actions. Both numbers are checked against this enumeration for every group at both indices, and they agree in all thirty-four cases.

Three copies of p3, and the shift that exchanges them #

p3 has four subgroups of index three. One is its lattice of translations with every three-fold turn discarded, which is p1. The other three are copies of p3 itself on a cell three times as large — the kind of subgroup that grows the cell by a prime rather than dropping an operation — and they are the case the title is about.

A three-fold centre of p3 sits at the corner of the cell, at a third along both diagonals, and at two thirds — three centres to a cell, and no operation of p3 carries one kind onto another, which is why the classes of a plane group count three separate classes of turn rather than one. A copy of p3 on the tripled cell has to choose: it keeps every centre of one kind and abandons the other two.

Why the group cannot move one copy onto another is a one-line calculation. Conjugating a rotation about the point c by a translation v gives a rotation about c + v — the centre moves by exactly the translation. The translations of p3 are whole cells, so conjugating any of these three subgroups by any operation of p3 moves its centres by whole cells and returns the same subgroup. Each is normal, each is a conjugacy class of one, and the tables print three entries.

Now shift by two thirds along one edge and one third along the other. That is not a translation of p3. But it carries the corner centres onto the second kind, the second kind onto the third, and the third back onto the corners — so it carries the set of p3’s operations onto itself, which is what normalising means. Conjugation by it takes the first subgroup onto the second. The three subgroups are one set: three ways of writing one relation between a crystal and a superstructure three times its cell.

A pair in one class each, and the half-cell that joins them #

p4 gives the same shape of answer with two copies instead of three, and shows what the fusing motion has to be.

p4’s four-fold centres sit at the cell corners and at the cell centres, and they are two classes: a translation moves a corner centre to another corner centre and never to a cell centre. A copy of p4 on the doubled cell keeps one class and demotes the other to two-fold, and there are two such copies. Conjugating either by a translation of p4 moves its centres by a whole cell and gives it back.

The shift by half a cell along each edge is the missing motion. It carries corner centres onto cell centres and cell centres onto corner centres, so it normalises p4 and swaps the two subgroups. p4’s third subgroup of index two is p2 — every turn halved, the lattice untouched — and nothing exchanges it with anything, because nothing else has its type. Three subgroups, three classes, two sets, and the set with two members in it is the one the tables would mark equivalent.

Normal, and still exchangeable #

It is tempting to read normality as the end of the question: a normal subgroup is fixed by every conjugation the group can perform, so surely nothing more can be said. pmm shows that the two properties are about different things.

pmm has fifteen subgroups of index two. Three keep the whole lattice and drop operations — two of type pm, one of type p2 — and twelve sit on a halved lattice: four of type pmm, four of pmg and four of cmm. All fifteen are normal — each leaves a quotient — so the group’s own conjugation sorts them into fifteen classes of one. The normaliser sorts them into eight sets: the four cmm subgroups become a single set, the four pmm subgroups become two sets of two, the four pmg subgroups likewise, and the three on the full lattice stay apart because nothing shares their type.

The four cmm copies differ only in where their mirror lines fall — at the cell’s edges, or half a cell along one direction, or the other, or both. pmm’s own translations move each of those patterns onto itself. The half-cell shifts, which normalise pmm because its mirrors already come at every half cell, move each onto the next. Normality says the group cannot move a subgroup at all; it says nothing about whether two different subgroups are interchangeable, and here they are.

Twelve subgroups, four classes, two sets #

p3m1 has the richest row in the table, and it makes both partitions do work at once.

Three of the twelve keep the whole lattice: they are the three copies of cm got by keeping one of p3m1’s three mirror directions and throwing away the three-fold turns. Those three are conjugate — the three-fold turn of p3m1 carries one mirror direction onto the next — so they are one class of three, and the tables print them as one entry. Conjugacy has done its work there, and the normaliser adds nothing.

The other nine are copies of p31m on the tripled cell, and they need both partitions. They fall into three conjugacy classes of three, so the tables print three entries of three conjugate subgroups; and the normaliser’s third-of-a-cell shifts carry any of the three classes onto any other, so all nine are one set, and the three entries are one relation written three times.

That p31m is what sits inside p3m1 is the pairing the two ways down found: the tripled cell is turned through thirty degrees against the original, and a turn of thirty degrees exchanges the two ways three mirrors can sit on a hexagonal lattice, which is why neither of the pair contains a copy of itself at index three. The two are not symmetric in the counting, either. p31m has four subgroups of index three — three of type cm in one class, and exactly one of type p3m1 — against p3m1’s twelve. The tripled cells available to one are not available to the other, and the count records it.

Seven subgroups of p2, counted by hand #

The table’s numbers are small enough to derive, and p2’s row is worth doing in full because every ingredient of the general answer appears in it.

A subgroup of index two either keeps the whole lattice and halves the point group, or keeps the point group and halves the lattice. p2’s point group is the half-turn and the identity, so the first kind drops the half-turn altogether: one subgroup, of type p1. For the second kind, a lattice has three sublattices of index two — the one that doubles along each edge, and the centred one that keeps the sum of the edges — and on each of them a copy of p2 must choose which half-turn centres to keep. p2 has four classes of centre to a cell; a copy on a doubled cell keeps two of them and demotes the other two to nothing, and there are exactly two ways to do it. Three sublattices, two choices each, and six copies of p2, which with the p1 makes the seven the table prints.

Conjugating by a translation of p2 moves a centre by a whole cell and so returns each of the six to itself: six classes, all normal. The normaliser’s half-cell shifts move the kept centres onto the dropped ones, and they do it one sublattice at a time — a shift along an edge cannot turn a lattice doubled along that edge into the centred one. So the six pair up into three sets, one for each sublattice, and with p1 alone that is four sets, which is what the computation reports.

The same reasoning read backwards says where fusing is impossible. If a group’s copies on one sublattice are told apart by something other than which coset of centres they keep — a direction, say, rather than an origin — then no shift will exchange them, because a shift cannot turn one direction into another.

Where the fusing happens, and where it cannot #

The nine rows that fuse are p2, pm, pg, pmm, pmg, p4 and p4m at index two, and p3 and p3m1 at index three. What they have in common is the motion doing the fusing: a translation by half a cell, or by a third of one, that is not a translation of the group but leaves the group’s operations as a set unmoved. Those are exactly the groups with equivalent origins that are not already their own translations — the count the normaliser essay prints as the number of descriptions of one arrangement.

The rows that do not fuse fail for two different reasons, and the difference is worth keeping. cm, cmm, pgg, p4g, p6 and p6m have subgroups of these indices and no fusion, because their normalisers add no translation that moves a subgroup: the half-cell shifts that would do it are already operations of the group, or would not normalise it. p1 has the largest normaliser of all — every motion of the plane normalises it — and still fuses nothing at index two, because conjugating a sublattice by a translation gives that sublattice back, and the only linear parts available on a general oblique cell are plus and minus the identity, which fix each of the three sublattices. A large normaliser is not the point; a normaliser with the right kind of element is.

What the classes look like at index three #

At index two conjugacy has nothing to do, because every subgroup of index two is normal. At index three it does: the 82 subgroups fall into 36 classes, so most classes hold more than one subgroup, and a class that is not a single subgroup holds exactly three — a subgroup of index three is either normal or has exactly three conjugates, since the group permutes its conjugates transitively and there are only three cosets to permute.

p2 is the plain case. It has twelve subgroups of index three in four classes, each class holding three subgroups that the group’s own half-turn shuffles, and its normaliser fuses none of them: four classes, four sets. The tables print four entries, each reading three conjugate subgroups, and every word of that is the group’s own doing. Set p2 beside p3m1, which also has twelve subgroups of index three in four classes, and the difference is entirely in the second partition: p3m1’s normaliser fuses three of its four classes and p2’s fuses nothing.

What a fused set means for a report #

A structure report names a subgroup whenever it describes a superstructure, a phase transition or an ordering: the low-temperature phase has the symmetry of this subgroup of that group, at index three. If the subgroup named belongs to a set with three members, then two reports of the same material may name different members and both be right, and a comparison that matches on the name will call them different.

That is the same error one crystal and sixteen coordinate lists is about, moved one level up, and it is the error Wyckoff sets fix for positions. The repair is the same too: compare sets, not members, unless the cell as measured says the members are genuinely different. And it is the cell that decides, because the Euclidean normaliser is the normaliser on that cell — the affine version, which allows any change of basis that preserves the group, fuses more and is the right unit for classifying kinds of descent rather than for identifying one crystal.

The tables carry both pieces of information, and this is what the second one is for. An entry that says three subgroups under one type, with the three marked conjugate, is one relation. Three separate entries whose subgroups a normaliser exchanges are also one relation, and nothing in the entry itself says so.

The checks, and what they refuse #

The quarter-cell refusal is the one that keeps the whole procedure honest. A shift by a quarter of a cell moves pmm’s mirrors onto lines where pmm has none, so it is not in the normaliser, and it is not allowed to identify two of pmm’s subgroups even though it would happily carry one set of operations onto another set of operations. The fusing motion has to normalise the parent, not merely move the child, and a computation that forgot the difference would report far fewer sets than there are.

Who counted them #

The subgroup tables of the International Tables list, for every space group, its maximal subgroups with their indices and — since the volume on subgroups was published in 2004 — which of them are conjugate. Euclidean normalisers were tabulated by Erwin Koch and Werner Fischer in the 1970s and appear in the Tables’ volume A; using them to sort subgroups into equivalence classes is the work of Hans Wondratschek and, at the Bilbao crystallographic server, of Mois Aroyo and his colleagues, whose subgroup tools report conjugacy and normaliser equivalence side by side for the space groups. The plane groups are small enough to recompute from nothing, which is what the numbers above are.

Where this goes: the same question one dimension up #

In space the normaliser of a monoclinic or triclinic group contains changes of basis that mix the axes, and whether every identification it makes is realised by an isometry of some cell is a question the plane answers only for itself. The subgroup counts grow too: 230 groups, indices two, three and four, and normalisers that are sometimes continuous. The interesting case would be a space group with subgroups in more conjugacy classes than the plane can show, fused by a normaliser whose extra motion is a shift along an axis rather than across a cell — and whether the pattern here, that fusing needs a fractional translation the group lacks, survives the screw axes that already carry fractional translations of their own.