The quotient each normal subgroup leaves

Counting subgroups by permutations gives a number and withholds the interesting half of it. The seventeen plane groups have eighty-two subgroups of index three between them and two hundred and eighty-one of index four, and a count like that says nothing about which of them a crystallographer would ever meet — because the subgroups that matter are the ones that are normal, and a normal subgroup is the only kind there is anything to divide by.

Dividing by one is what an ordering transition does. A crystal that loses half its translations keeps a group H inside its old group G, and the thing that indexes its antiphase domains is the quotient G/H — which exists as a group only when H is normal. So the question this essay asks is the one the counting left out: of the subgroups of small index, which are normal, and what is the quotient?

The column that cannot have an exception #

At index two, every subgroup is normal, in all seventeen groups and in every group there has ever been. The reason is one line and worth having: a subgroup H of index two splits G into H and everything else, and it splits it that way whether the cosets are written on the left or on the right. Left cosets and right cosets are the same two sets, which is the definition of normal.

That makes the seventy-four index-two subgroups of the plane groups seventy-four normal ones, and it is why the two-colourings work at all: a two-colouring of a pattern is a subgroup of index two together with the rule that everything outside it swaps the colours, and the rule is consistent precisely because the outside is a single coset.

At index three the picture inverts. Eighty-two subgroups, and thirteen of them normal — and the thirteen are concentrated: four in p1, four in p3, and one each in pm, pg, cm, p31m and p6. Ten of the seventeen have subgroups of index three and not one that is normal.

At index four, ninety-seven of the two hundred and eighty-one.

Why no enumeration is needed #

The three numbers above were computed twice, and the second computation does not enumerate anything.

A quotient G/N is abelian exactly when N contains every commutator, so the normal subgroups with abelian quotient correspond one for one with the subgroups of the abelianisation — the group with the order of multiplication forgotten, which this collection already computes by putting the presentation’s relator matrix into Smith normal form.

And every group of order two, three, four or five is abelian. There is no non-abelian group below order six.

Put those together and the whole question collapses. At index two, three or four, a normal subgroup is a subgroup of the abelianisation and nothing else, because its quotient has an order too small to be anything but abelian, so it contains the commutator subgroup automatically. The count is then arithmetic on the abelianisation’s invariant factors, and no permutation is ever written down.

The arithmetic is short. An abelian group is a free part of some rank together with a list of invariant factors, and the homomorphisms into a cyclic group of order m are counted factor by factor — each free generator contributes m, and an invariant factor d contributes the greatest common divisor of d and m — so |Hom(A, C_m)| is a product of greatest common divisors. Surjections are what is wanted, so the homomorphisms landing inside a proper subgroup are taken out by inclusion and exclusion; and each subgroup is hit once per automorphism of the quotient, so the surjections are divided by one for C₂, two for C₃ and C₄, and six for C₂ × C₂.

That the two agree is the whole reason to believe either. The permutation route is a search over tuples of permutations satisfying the relators — for four points that is a walk over twenty-four to the fourth power of candidates — and it knows nothing about abelianisations. The arithmetic route never looks at a group element. A hundred and eighty-four normal subgroups over three indices, and the columns match at every one of the fifty-one entries.

A disagreement would have meant one of exactly two things, and both are worth stating because they are what the agreement rules out. Either the abelianisation is wrong — a plausible failure, since it comes from a Smith normal form of a matrix built from a presentation — or some quotient of order two, three or four is not abelian, which is false and decidable.

Which quotient, and what decides it #

At index four there are two possible quotients, C₄ and C₂ × C₂, and which occurs is read straight off the invariant factors.

A cyclic quotient of order four requires a surjection onto C₄, which requires an element of order four somewhere in the abelianisation. Seven of the seventeen have abelianisations of exponent two — p2 is ℤ₂ ⊕ ℤ₂ ⊕ ℤ₂, pmm is four copies of ℤ₂ — and every element of those has order dividing two. So those seven have no cyclic quotient of order four at all, whatever else they have: pmm has sixty-seven subgroups of index four, thirty-five of them normal, and every one of the thirty-five leaves the Klein group.

The four groups whose abelianisation carries a ℤ₄ are pgg, p4, p4g and p1, and that ℤ₄ is the same object in three of them: the quarter-turn, whose fourth power is a translation and which therefore survives abelianisation as an element of order four. p1’s comes from its free part instead — ℤ² surjects onto C₄ in six ways.

The distribution is worth reading as a fact about the subject rather than about the arithmetic. Seventy-seven of the ninety-seven quotients are the Klein group, which is the group of “two independent things each of which can be done or not”. That is what an ordering transition usually is: two translations lost independently, or a translation and a rotation, and the domains that result are indexed by a pair of signs rather than by a cycle.

What a subgroup that is not normal looks like #

Sixty-nine of the eighty-two index-three subgroups are not normal, and it is worth having a picture of what that means rather than only the definition, because the picture is the same in every case.

A subgroup that is not normal has conjugates that are not itself. Conjugating H by an operation g gives gHg⁻¹, which is another subgroup of the same index, and the number of distinct ones is the index of H’s normaliser. For a subgroup of index three that is not normal, the conjugates come in a set of three, permuted cyclically by whatever operation of G fails to fix H.

Take p2, which has twelve subgroups of index three and none normal. Its abelianisation is ℤ₂ ⊕ ℤ₂ ⊕ ℤ₂, which has no element of order three at all, so it has no quotient of order three — and yet it has twelve subgroups of index three. Those twelve are the sublattices of index three of its translation lattice, four of them, each carried along with the half-turns in three ways. The half-turn does not fix a sublattice of index three: multiplying by minus one carries the sublattice generated by (3, 0) and (0, 1) to itself, but the half-turn’s centres move, and the subgroup containing a particular set of centres goes to the subgroup containing another set.

That is what the difference between eighty-two and thirteen is made of. A subgroup of index three that is not normal is one whose cosets cannot be given a consistent group law, and the failure shows up physically: a crystal that lost those particular operations would have three inequivalent ways of having lost them, related by an operation of the parent, and the resulting domains do not form a group under any composition.

The seven groups that do have a normal subgroup of index three are the ones whose abelianisation has a factor of three in it — p1 and p3 from their lattices, and pm, pg, cm, p31m and p6 from a single ℤ or ℤ₆ factor. Nothing else can have one, and the arithmetic says so before any subgroup is written down.

The quotient neither count can reach #

Every plane group has a normal subgroup that no census here finds, and it is the most important one it has.

The translations of a plane group form a subgroup, and that subgroup is normal — conjugating a translation by any operation gives another translation, because the linear part of a conjugate is the conjugate of the linear part and the linear part of a translation is the identity. The quotient by it is the point group, which is where the thirty-two classes and their plane analogues come from. So the point group is a quotient of the plane group by a normal subgroup, always, and its index is the point group’s own order.

For twelve of the seventeen that quotient is abelian: C₂ for p2, pm, pg and cm; C₂ × C₂ for pmm, pmg, pgg and cmm; C₃, C₄ and C₆ for p3, p4 and p6; and the trivial group for p1. For five it is not. p4m and p4g have point group 4mm, which is the dihedral group of order eight; p3m1 and p31m have 3m, which is S₃; p6m has 6mm, of order twelve.

All five sit at index six, eight or twelve, and the permutation census stops at four because the search over tuples of permutations grows as the factorial. So the honest statement of everything above is bounded: at index two, three and four every quotient is abelian and the abelianisation decides everything. It is not a statement about normal subgroups in general, and the five groups in that table are the standing counterexample to the version of the claim without its bound.

The two kinds of descent, seen from here #

This essay’s census cuts across a distinction the collection already has, and putting the two together says something neither says alone.

Two ways down from a group separates the subgroups of a plane group into the translationengleiche — same translations, fewer point operations — and the klassengleiche, which keep the class and lose translations. That split is about what is given up. Normality is about whether what is left can be divided by, and the two are independent: there are normal subgroups of both kinds, and non-normal ones of both kinds.

The connection is at the extremes. A klassengleiche subgroup of index n whose lost translations are a sublattice invariant under the whole point group is normal, and its quotient is the group of that sublattice’s cosets — which is where the antiphase domains come from and why they are counted by an index. A translationengleiche subgroup is normal exactly when the corresponding subgroup of the point group is, which reduces a question about an infinite group to one about a group of order at most twelve.

And the translation subgroup itself is the extreme case of the second kind, the largest translationengleiche subgroup there is, whose quotient is the whole point group. Everything in the table above is somewhere between that and the identity.

Why the boundary is at six and not somewhere convenient #

The index at which the argument stops is not an artefact of what could be computed. It is exactly the first order at which a group can fail to commute, and that is a fact about the integers rather than about crystallography.

A group of prime order is cyclic, which disposes of two, three and five. A group of order p² is abelian, which disposes of four and nine. Six is the first order that is neither prime nor a prime square, and S₃ duly exists there. So the arithmetic route works up to index five and fails at six — and it fails at exactly the place where the interesting quotients of plane groups begin, which is not a coincidence: the point groups with a mirror and a rotation are non-abelian, and those are most of crystallography.

There is a second reason the boundary sits where it does, and it is about how much a quotient can carry. A quotient of order four has at most two independent generators and no room for a relation between them beyond commuting; a quotient of order six has room for a rotation and a reflection that do not commute, which is the smallest arrangement in which “the order of operations matters” can be expressed at all. So the first non-abelian quotient is also the first quotient that can record which of two lost operations was lost first — and that is exactly the information an abelianisation throws away.

It is worth noticing what this does to the shape of the subject. The abelianisation is a coarse instrument — it cannot tell a translation from the identity when a commutator produces it, which is the reason the word problem needs more than exponent sums — and here that coarseness is exactly right, because the objects it is being asked about are too small to notice.

What the counts refuse #

The third is the one to keep. If normality turned out to be no restriction, the whole essay would be a re-run of the subgroup count with a longer name — and at index two it is exactly that, which is why the index-three column matters: twelve of the seventeen have a subgroup of index three that is not normal, and those subgroups are carried onto different subgroups by operations of the group they sit in.

The fourth is the bound made into a test rather than a caveat. A claim that holds “at every index reached” is worth very little if nothing exists beyond that index; here something does, five somethings, and the assertion fails if any of them ever comes into range.

Where the exactness stops #

Computed here: the abelianisation of each of the seventeen from a Smith normal form of its relator matrix; the number of surjections onto each abelian group of order two, three and four by a product of greatest common divisors with inclusion and exclusion; the resulting subgroup counts; the same counts by enumerating homomorphisms of each presentation into S₂, S₃ and S₄ and keeping the transitive ones with regular image; and the point group of each of the seventeen with its order and whether it commutes.

The indices are two, three and four, and nothing here is a statement about any other. The permutation route stops because S₅ has a hundred and twenty elements and the search is over tuples of them; the arithmetic route stops because at index six a quotient may be S₃ and the abelianisation cannot see it.

A count of normal subgroups is not a count of conjugacy classes of them. A normal subgroup is its own conjugacy class, so the two agree here by definition — but for the subgroups that are not normal the counts in the first table are counts of subgroups rather than of classes, and the class count is a different column.

And a quotient is an abstract group, not an action. C₂ × C₂ appears as seventy-seven quotients and those seventy-seven are seventy-seven different subgroups; knowing the quotient fixes the domain structure of an ordering transition and settles nothing at all about which translations were lost.

Where the ladder goes next #

Back, to the count this one splits: how many subgroups of index three, where the permutation correspondence is set up, and to the abelianisation, which turns out to hold the whole answer.

Sideways, to what a normal subgroup does in the physical subject: the domains a lost translation makes, where the quotient is what indexes the domains, and two colours and a symmetry that swaps them, which is the index-two case with the colouring attached.

Onward, to the quotients this census cannot reach: the two ways down from a group, where the translationengleiche and klassengleiche split is the same distinction seen through the point group, and the descent of symmetry, which is the graph of those quotients drawn out.