Two operations make rock salt's group, and Pmmm needs six

Seven space groups cannot make their own translations sorted the 219 affine types of space group by which of their own operations generate them. It counted the lattice as something the operations must produce, not something handed to them. Two hundred and twelve are generated by their own rotations, screws, mirrors, glides, inversions and rotoinversions. Seven, the primitive polar groups with no glide such as P2 and P4, need a translation given outright. That census handed each group every operation of each kind at once. It asked whether a kind of operation suffices, not how many operations do.

The question it left is the one how few operations make a pattern answered for the seventeen plane groups: how few of a group’s operations are enough to rebuild the whole infinite group? In the plane the answer was two for most groups, three for p2 and p3m1, and four for pmm. The plane answer rested on a floor from the abelianisation and a ceiling from a search through the operations within one cell. For three groups the floor sat below the ceiling and the search alone decided, with the caveat that a pool had been exhausted rather than a possibility ruled out.

This essay answers the same question for the 212 space groups, and settles every one of them exactly. Sixty-four space groups are made by two of their own operations, ninety-three by three, forty-six by four, eight by five, and one, Pmmm, needs six. The group of rock salt, copper and aluminium, $Fm\bar{3}m$ with its forty-eight operations per lattice point, is made by one screw and one glide. The other forty-six operations per lattice point, and every translation, are products of those two, operations nobody put in that the closure supplies. For 198 of the 212 the count meets a lower bound read off a quotient of the group. For the other fourteen that bound is one short, and a second argument, a finite quotient in which no smaller set of elements of any kind generates, proves the extra operation is needed. One group, P312, needs two more than its bound.

A quotient that every generating set must generate #

The lower bound comes from a principle simple enough to state in one line. If a set generates a group, its image generates every quotient of the group. So a group needs at least as many generating elements as any of its quotients does, and the useful quotients are those whose generating number is easy to read.

The easiest are the elementary abelian ones. Divide a group $G$ by its commutators and by all $p$-th powers of its elements, for a prime $p$. What is left is abelian with every element of order $p$, so it is a vector space over the field of $p$ elements, of some dimension $d_p$. A vector space of dimension $d_p$ needs $d_p$ generators, so $G$ needs at least $d_p$ for every prime. What is left when order is forgotten computed the abelianisations of the seventeen plane groups, and $d_p$ is the number of their invariant factors divisible by $p$.

For a space group the quotient can be computed exactly on a finite group. Every homomorphism from $G$ onto a group of exponent $p$ kills $p$ times every lattice vector, so it factors through $G/pL$, where $L$ is the lattice. That group is finite, of order $p^3$ times the number of cosets of the lattice, at most $27 \times 48 = 1{,}296$. In it, the subgroup made of commutators and $p$-th powers is the normal closure of the commutators and $p$-th powers of any generating set. It is found by closing a handful of elements under products and conjugation, and $d_p$ is the base-$p$ logarithm of its index.

Two primes suffice. A space group’s point group has order divisible only by 2 and 3. A prime that divides no point-group order sees only the free part of the abelianised lattice, and the prime 2 sees that part too. The bound is therefore the larger of $d_2$ and $d_3$, and never less than two. A group made by one element is cyclic and so abelian, and the only abelian space group is P1, which is $\mathbb{Z}^3$ and needs three.

The table shows the bound at work. $P\bar{1}$ is $\mathbb{Z}^3$ with a centre of inversion, and its quotient by squares and commutators is $(\mathbb{Z}/2)^4$: one factor for the inversion and three for the lattice, since the inversion sends every translation to its inverse and a translation is congruent to its inverse modulo 2. So $P\bar{1}$ needs four operations of some kind, and four inversions do it. Two inversions through centres $c_1$ and $c_2$ compose to a translation by $2(c_2 - c_1)$, and four centres placed half a lattice vector apart along three directions make all three lattice vectors. Pmmm is three copies of the infinite dihedral group, one along each axis, since its three mirror families act on the three coordinates separately. Each copy contributes $(\mathbb{Z}/2)^2$, so Pmmm needs six. P222 needs five, P2/m five, $P6/mmm$ four and $Pm\bar{3}m$ three.

At the other end, $P2_12_12_1$ and $Fm\bar{3}m$ have quotients of rank two, so the bound is two, and two operations suffice. $P2_12_12_1$ is made by two of its three screws. Each squared climbs its own axis, and their product is the third screw, which supplies the third direction.

A search that stops at the bound #

The upper bound is a search. For each group it draws sets of the group’s own operations, a size at a time starting from the bound, and tests each set by Schreier’s lemma as the earlier census did. A set generates the group when its products reach every coset of the lattice and the translations among those products are the whole lattice, and the lemma computes those translations exactly from a finite list.

The draws are made in two stages, because most sets fail for a cheap reason. First the search draws linear parts and keeps only sets whose linear parts generate the point group. A set whose products miss a coset cannot generate anything. Then it draws a translate of each operation from a box of neighbouring cells, since the same coset holds a mirror on one plane and a glide on the next, and tests the set. The search stops at the first success. When that first success is at the bound, the answer is exact, because no search can beat an argument that no smaller set exists.

The picture at the head of this essay is the result, and the bars are exact. For 198 of the 212 types the search finds a generating set at the bound on its first attempt at that size. For the remaining fourteen it finds none at the bound and one at the next size, and the next section proves that the gap is real.

The distribution has a clear shape. Sixty-four types need two operations, ninety-three three, forty-six four, eight five, and one six. The bulk sits at three. The groups that need many are the ones whose operations act on few directions at a time. Pmmm’s operations act on one axis each, P222’s rotations on two axes each, and P2/m’s on one axis and the plane across it, so each operation contributes to few factors of the quotient.

Fourteen groups need more than their quotient says #

A lower bound that is not met could mean the search was not thorough enough, or that the bound is too weak. For the fourteen, it is the bound.

The proof uses the same principle as the bound, with a quotient that is not abelian. If $k$ elements generate $G$, their images generate $G/mL$ for every whole number $m$. For each of the fourteen, one value of $m$, 2 or 3, gives a finite group of order at most 324 in which no set of $k - 1$ elements of any kind generates. That is checked over every such set, not sampled, and so no $k - 1$ elements of $G$ generate $G$.

The fourteen fall into two families. Ten are trigonal or hexagonal and are caught modulo 3: P321 and P312, P3m1 and its non-symmorphic partner P3c1, $P\bar{3}$, R32, $P6_322$, $P\bar{3}1c$, $P\bar{6}c2$ and $P3_112$, which stands with its mirror image $P3_212$ as one affine type. Four are cubic and are caught modulo 2: P23, F23, $Pm\bar{3}$ and $Fm\bar{3}$. Every one of the fourteen has a three-fold axis. The four cubic ones are the symmorphic groups of the two cubic classes with no four-fold axis, 23 and $m\bar{3}$, on the primitive and the face-centred lattice. The body-centred I23 and $Im\bar{3}$, and every non-symmorphic group of those two classes, $P2_13$ and $Pa\bar{3}$ among them, meet their bound of two.

Nine of the fourteen are symmorphic. That runs against the pattern of the census as a whole, where screws and glides do more work than rotations and mirrors: a symmorphic group’s operations include fewer screws and glides, and here that costs an operation.

What a pair of operations reaches #

The quotient says that no pair generates. Counting which translations each pair does reach shows why.

The picture takes every pair of own operations whose linear parts generate the point group, for five of the fourteen, and records which of the lattice’s translations modulo $m$ the pair’s products reach. For P23, 2,304 pairs qualify. 576 reach no translation at all modulo 2, and 1,728 reach exactly two: the zero vector and $(1, 1, 1)$, the body diagonal. No pair reaches more. The translations modulo 2 form the three-dimensional space over two elements whose seven non-zero points only two and three can centre a lattice drew as the Fano plane. A generating set must reach all of that space. A pair of P23’s operations is confined to the one point of the Fano plane that the cubic group’s three-fold axes fix, the body centring I. The obstruction to generating P23 with two operations is the same invariant subspace that makes body-centred cubic a lattice.

The trigonal groups are confined differently. Modulo 3, P321’s pairs reach at most the nine translations of the basal plane and never the step up the axis. P312’s reach at most three: the zero vector and the line through $(1, 2, 0)$, which is the one line of the hexagonal net that the three-fold rotation fixes modulo 3. Only two and three can centre a lattice met the same line as the centring that turns the hexagonal net into itself rotated by thirty degrees. P3m1’s pairs reach at most nine, that line together with the axis. A third operation lifts P321 and P3m1 out of their sublattice, and P312 needs a fourth. Even with three elements of any kind, the quotient modulo 3 of P312 is never generated. The exhaustive check runs over every triple of its 162 elements.

P312 and P321 have the same point group on the same lattice, differently oriented. In P321 the two-fold axes lie along the lattice vectors of the hexagonal net. In P312 they lie across them, and one family of them lies along the line $(1, 2, 0)$ that the three-fold rotation fixes modulo 3. The orientation is the whole difference between the two groups, and the counts show what it costs. A pair of P321’s operations can reach the whole basal plane modulo 3, and a pair of P312’s reaches only that fixed line. Both groups have the same abelian quotients, of rank two, and one needs three operations where the other needs four. An abelian quotient cannot see which way an axis points.

Own operations cost nothing #

A generating set here was restricted to the group’s own operations, with every lattice translation left out. It is natural to ask whether that costs an operation. A translation handed over freely might replace two operations that are needed only to produce it.

It never does. The abelian bound applies to every generating set, of any elements at all, so for the 198 groups whose own-operation count meets the bound, no set of anything is smaller. For the fourteen, the quotient test was run over every element of $G/mL$, translations included, and still found no smaller generating set. So for every one of the 212 groups, the fewest own operations is the group’s minimal number of generators as an abstract group. Leaving the lattice out of the offer costs nothing anywhere, which is a quantitative form of what the earlier census found qualitatively: outside the seven, a space group’s lattice is always a by-product of its other operations.

Where the two-generated groups are #

By crystal family the counts follow how many directions each operation moves. Twenty-eight of the thirty-five cubic types are made by two operations. A cubic group’s three-fold axes along the body diagonals turn every direction into two others, so one translation made anywhere is made along all three axes, and a pair can cover the lattice. The seven cubic types that need three are the four caught modulo 2 above and three whose bound is three: $Pm\bar{3}m$, $Im\bar{3}m$ and one non-symmorphic group in the class of $Pm\bar{3}m$. The hexagonal family has twenty of forty-three types two-generated. The tetragonal family has twelve of sixty-four, and the orthorhombic family only three of fifty-eight, $P2_12_12_1$ and two others made by a screw and a glide.

The triclinic and monoclinic families sit at the high end. $P\bar{1}$ needs four and P2/m five. No operation of either turns one lattice direction into another, so no translation made in one direction is copied into a second, and each direction has to be produced separately. Pmmm, needing six, is the extreme case of that. It is generated by mirrors in three perpendicular families, each mirror acting on one coordinate only, so each of its three directions needs a pair of parallel mirrors to itself.

The rock-salt group is at the other extreme. $Fm\bar{3}m$ is made by a four-fold screw and a glide plane. It is made just as well by two rotoinversions, one of order four and one of order six, and by a two-fold screw and a four-fold rotoinversion. Runs of the search from different starting points find all three. Whichever pair is taken, its products reach all 48 cosets of the face-centred lattice and every lattice translation, so a crystal with 48 operations at every lattice point is fixed by knowing two of them.

What the numbers rest on #

The quotient proof is exhaustive and the search is not. For the 198 groups at the bound, the bound is a theorem and the generating set found is checked by Schreier’s lemma, so the count is exact regardless of how the search ran. For the fourteen, the proof that one fewer is impossible runs over every set of that size in a finite group, and a set of the larger size is exhibited and checked. Nothing in the reported counts depends on a search having been thorough.

The operations are taken with translates in a box. An operation’s translates within one cell of its representative are offered to the search. A generating set found this way is a generating set whatever box it came from, so the box can make the search fail to find a small set but cannot make it report a false one. For the counts reported, it never failed at a size where a set exists.

The groups are types, not settings. Each of the 212 is one affine type, taken on a primitive basis of its lattice as built in two hundred and nineteen over seventy-three. A change of basis or of origin carries generating sets to generating sets, so the count is a property of the type. Enantiomorphic pairs such as $P3_112$ and $P3_212$ are one affine type and have one count.

One correction to the older census was needed. The routine that tests a set by Schreier’s lemma computed each translation as the product on the wrong side. It took the chosen element of a coset times the generator, where the lemma requires the inverse order. The result is still a translation of the subgroup, conjugated by that element, so the routine never accepted a set that fails to generate. It could reject a set that does: two glides of Cc, one lattice step apart, generate it, and the old routine found translations of rank two. The earlier census handed over every operation of a kind at once, and its conjugates are already in such a set. Rerun after the correction, it gives exactly the same 219 verdicts.

Who worked the pieces out #

The minimal number of generators of a finite group, and the gap between it and the abelian bound, is a subject of its own in group theory. Its lower bounds come from the Frattini quotient, the same quotient used here, and the reason a group can need more than its abelianisation lets on is its non-abelian quotients. The computational side belongs to the programs Schreier’s lemma underlies, from Sims’s work on permutation groups in the 1970s onward. For space groups the International Tables list a set of generating operations for each group, chosen to follow the structure of its symbol rather than to be few, and every such list begins with the three lattice translations.

What the census must refuse #

The refusal is the one the bound makes. Six operations of Pmmm generate it, and the census must decline to accept five of them as generating it. Pmmm’s quotient has rank six, so no five elements of it generate, and the five tested make a lattice of rank below three. The test that guards the whole census is the count settled for every group. It holds only if each of the 212 either meets its bound or has its excess proved, and an unproved gap anywhere would fail it.

Still open: the kinds that can do it #

The count says how many operations suffice, not which kinds. $Fm\bar{3}m$ turned out to be made by a screw and a glide and by two rotoinversions, but whether two rotations, or two mirrors, could also do it was never asked, since the search stops at its first success. Asking it for every group would turn one number per group into a table of which pairs of kinds are enough. In the plane, six groups are made of glides found that the kind matters a great deal: glides make six of the seventeen and mirrors four. In space the pools are larger, 48 cosets times the translates of each, and an exhaustive test of every pair of kinds would need the quotient method to prune rather than the search to find. The same quotient test would also settle the plane’s three unproved cases, p3m1, p31m and p6, where the plane count rests on an exhausted pool rather than a proof. Neither has been run.