The four groups with a centre

Composing two symmetries lands on a third, and the order matters: a rotation followed by a translation is not the translation followed by the rotation. That is the first thing that separates a group from a list, and it is why conjugation is the right notion of sameness — two operations are the same operation seen from two places exactly when one is a conjugate of the other.

The centre is what is left when that fails to be interesting. An element is central when it commutes with everything, so a central element is one whose conjugates are all itself: it looks the same from every place. The centre is a subgroup, it is automatically normal, and the quotient by it is how much of the group is visible in the way it acts on itself.

For plane groups the answer is short, and the argument is shorter than the answer.

A central element has to be a translation #

Write an operation as a matrix and a translation, (M, t), composing by (M, t)(N, s) = (MN, Ms + t). That is the whole of the arithmetic and it is the composition this collection has used since the four motions.

Take a central element (M, t) and commute it with a pure translation (I, v). One way round gives (M, Mv + t); the other gives (M, t + v). Equal for every v in the lattice means Mv = v for every lattice vector, and a matrix fixing two independent vectors is the identity.

So a central element is a translation, and that is settled before anything about a particular group is known. No rotation, no mirror, no glide is ever central in a plane group, whatever the group.

The second half is the same calculation read the other way. A translation (I, t) commutes with (N, s) when Nt = t, so the centre is exactly the set of lattice vectors fixed by every operation of the point group. That is a fixed-subspace calculation over the integers, and in the plane it has three possible answers: the whole lattice, a line’s worth, or nothing.

What a rotation does to it #

A rotation of the plane fixes only the origin. So a point group containing any rotation fixes no non-zero vector, and the centre of such a group is trivial.

Thirteen of the seventeen contain a rotation. Four do not — p1, whose point group is trivial; and pm, pg and cm, whose point group is a single mirror.

And those four are exactly the four with a centre. The equivalence is the result rather than a coincidence about which four they happen to be: a group has a centre if and only if its point group has no rotation, and the reason is that a rotation’s only fixed vector is zero.

For p1 the point group is trivial, so every translation is central and the centre is the whole group — which is to say p1 is abelian, and is the only plane group that is.

For pm, pg and cm the point group is one mirror, whose fixed vectors are the lattice vectors along it. The centre is a rank-one subgroup: an infinite cyclic group of translations, running along the mirror.

The three mirror groups, and how their centres differ #

pm, pg and cm all have a centre of rank one, and the three vectors are worth comparing because they say something about the three groups that their symbols do not.

In pm the mirror lies along a lattice vector and the centre is generated by that vector: the shortest central translation is the shortest lattice vector in the mirror.

In pg the operation is a glide rather than a mirror, and its linear part is the same matrix — the reflection — so the fixed direction is the same. But the group’s own shortest operation along that direction is the glide, whose square is the translation by a full lattice vector. So the centre of pg is generated by the square of a glide, and the glide itself is not central: it is a reflection, and no reflection commutes with a translation across it.

In cm the lattice is centred and the mirror lies along a diagonal of the rectangle, so the shortest central translation is a diagonal vector rather than an edge one — the centre is generated by (1, 1) in the primitive basis rather than by (1, 0).

The three centres are three different subgroups of three different lattices, and they are the same statement in each case: the translations along the mirror. That the answers differ is a fact about the lattices; that the calculation is the same is a fact about the point groups.

What a trivial centre means about the group #

A group acts on itself by conjugation, and the kernel of that action is precisely the centre. So the thirteen groups with a trivial centre act on themselves faithfully: no operation of any of them is invisible to conjugation, and the group of inner automorphisms is the whole group again.

That is worth putting beside a fact this collection already has. The abelianisation of a plane group is finite for thirteen of the seventeen — the same thirteen? No: the abelianisations with a free part are p1, pm, pg and cm, and those are the same four. The two calculations agree, and they are not the same calculation.

The abelianisation asks what survives when the order of composition is forgotten; the centre asks what never depended on it. For a plane group both answers are decided by the same thing — whether the point group has a rotation in it — because a rotation is what makes a translation fail to commute and is also what kills a translation in the abelianisation, by identifying v with Mv.

A group can have a large abelianisation and a trivial centre, and pmm is the example: its abelianisation is ℤ₂ ⊕ ℤ₂ ⊕ ℤ₂ ⊕ ℤ₂, sixteen elements, and its centre is the identity alone. Nothing about a group commuting modulo commutators implies anything commutes.

The same question in three dimensions, and in one #

The argument above uses nothing about two dimensions except that a rotation of the plane fixes only the origin, and that is the step that changes.

In three dimensions a rotation fixes its own axis, so a space group whose point group is a single rotation of order n has a centre: the lattice translations along the axis. The screw axes make it sharper — the centre of a group generated by a screw is generated by the screw’s nth power, which is the shortest lattice translation along the axis, exactly as pg’s centre is generated by the square of its glide. So space groups with a centre are common where plane groups with one are rare, and the difference is one dimension of freedom for a rotation to leave alone.

In one dimension — the seven friezes — the point group can contain the half-turn, the mirror across the strip, the mirror along it, or nothing. The translations along the strip are fixed by everything except the half-turn and the transverse mirror, so four of the seven friezes have the whole translation group as their centre and three have nothing. The count is different and the calculation is identical, which is what makes it a calculation rather than a table.

The pattern across the three dimensions is one sentence. The centre of a crystallographic group is the translations its point group fixes, so it is large exactly when the point group has a common invariant direction — never in the plane unless the point group is a mirror or trivial, often in space where an axis is a direction, and usually on a strip where there is only one direction to fix.

The conjugacy classes the centre is made of #

There is a second way to see the same answer, and it is the way a reader who has met conjugacy classes will find natural.

An element’s conjugacy class is the set of everything it is carried to by conjugation, and an element is central exactly when its class has one member. So counting central elements is counting classes of size one, and the classes of a plane group are already an object this collection computes: two operations are in one class when some operation of the group carries the first to the second.

Conjugating the translation by v gives the translation by Mv, so the class of a translation is the orbit of v under the point group. A vector fixed by the whole point group has an orbit of size one and is central; a vector moved by anything has an orbit of at least two and is not. That is the same condition arrived at without composing anything, and it makes the answer visible in a picture: the centre is the set of lattice points the point group leaves alone, which for a mirror is the points on the mirror and for a rotation is the origin.

It also says how far from central the rest is. In p2 a translation’s class has two members, v and −v, so no translation is central and every one is exactly two-fold non-central. In p4 a class has four members and in p6 it has six, so the higher the rotation the larger the classes and the further the group is from commuting. The centre is the extreme case of a quantity that varies, rather than a yes-or-no property, and the class sizes are the quantity.

Why the centre is a normal subgroup nobody chose #

The centre is normal for a reason worth stating, because it is the only normal subgroup a group has without anybody constructing one. Conjugating a central element gives g z g⁻¹ = z g g⁻¹ = z, so every conjugate of a central element is itself, so the centre is carried to itself by everything — normality for free.

That makes it the first entry in the census of normal subgroups, and the entry that census cannot see. Its indices are two, three and four, and the centre of pm has infinite index — it is an infinite subgroup of an infinite group, and the quotient by it is not finite at all. So a permutation census of small index misses the one normal subgroup every group has.

The quotient G/Z is the group of inner automorphisms, and for pm it is the group of everything the mirror and the across-mirror translations do, with the along-mirror translations divided out. That is not one of the seventeen: it is an infinite group with a mirror and one direction of translation, which is a frieze group — the seven of them are what happens when a plane group is cut down to one direction, and here one arrives by dividing rather than by cutting.

The centre and the origin, which are not the same freedom #

There is a confusion this essay should head off, because the two things it is between are both called “choosing a place”.

The origin of a plane group is a choice, and moving it changes every translation part in the description while changing no group. That freedom is the coboundaries of the extension arithmetic, and it is why two descriptions of one group can look completely different.

The centre is not that. It is a set of operations of the group, and moving the origin does not change which operations commute with which. A translation is central or it is not, and the answer is the same in every description.

The two do meet at one point, and the meeting is instructive. An origin shift by s changes a translation part by (M − I)s, and it changes nothing at all when M fixes s — so the origin shifts that do nothing are exactly the central translations, at least when s is taken in the lattice. That is not a coincidence: an origin shift acts on the group’s descriptions by conjugation, and the shifts acting trivially are the ones commuting with everything, which is the centre again.

So the centre is the amount of the origin’s freedom that is not freedom. For p1 every origin shift does nothing to the description, which is another way of seeing that p1 has one class in its cohomology; for p4 no shift is wasted, which is why its origin has to be chosen carefully and why the choice has a convention attached.

What the calculation refuses #

The last is the control and it is the one that would be easy to leave out. A calculation that always returned “trivial” would agree with thirteen of the seventeen and be wrong about four, and the wrongness would be invisible because the thirteen are the majority and the answer looks plausible. Requiring that p1 come back with a rank-two centre is what makes the thirteen zeros evidence rather than a default.

The second is the one that carries the argument. Four groups having a centre is a list; the four being exactly those with no rotation is a statement, and it is checked in both directions — no group with a rotation has a centre, and no group without one lacks it.

What the centre is good for #

A quantity computed for its own sake is worth less than one that decides something, so it is worth saying where the centre is used.

It bounds how badly a group can be described. Two descriptions of a group differ by an automorphism, and the automorphisms coming from the group itself — conjugation by one of its own elements — form a group isomorphic to G/Z. A large centre means few inner automorphisms and a group whose descriptions differ mostly in ways that come from outside; a trivial centre means the group’s own elements supply as many relabellings as they can.

It decides whether a projective representation can be made ordinary. A representation of a group up to a phase is an ordinary representation of a central extension of it, and the central extensions of a group with trivial centre are constrained differently from those of a group with a large one. This collection has met that already, from the other end: the projective representations at a zone boundary are exactly what a factor system produces when the phases cannot be removed.

And it is the first thing to compute about a group nobody has met. The centre is cheap — one linear condition per generator — and its answer immediately says whether the group is abelian, whether it acts on itself faithfully, and whether there is a normal subgroup to divide by before anything harder is attempted. For the seventeen the answer is unexciting, which is itself the useful fact: thirteen of them have nothing there, so every normal subgroup of those thirteen has to be constructed rather than found.

Where the exactness stops #

Computed here: the distinct linear parts of each of the seventeen; the rank of the stacked matrix M − I over all of them, by elimination over the rationals; a primitive integer vector in the kernel where the rank permits one; the commutator of every operation with each of three lattice vectors; and the verdict for each group.

The centre is computed as a subgroup of the translations, which is where it lives. The argument that no operation with a non-trivial linear part can be central is general and short, and it is made once rather than checked group by group — so the census enumerates fixed vectors and does not enumerate operations. A reader wanting the second can have it from the commutator table, which is the same fact laid out one operation at a time.

Rank one means rank one, and the generator is a choice of sign. The centre of pm is generated by (1, 0) or equally by (−1, 0); the search reports the positive one, and nothing depends on which.

This is the centre of the group, not the centre of a pattern. A pattern with pm symmetry has translations along its mirror that commute with everything in its symmetry group, and that says nothing about any point of the pattern being special. The word “centre” here is group theory’s and not geometry’s, and the two meet nowhere on this page.

Where the ladder goes next #

Back, to the composition the whole calculation is: where the product is, where composing two operations is worked out in coordinates, and conjugation and sameness, where the conjugates of an operation are what the centre is defined by having none of.

Sideways, to the other calculation that separates the same four groups: what is left when the order is forgotten, where the abelianisation’s free part appears for p1, pm, pg and cm and for nothing else.

Onward, to the normal subgroups a census can reach: the quotient each normal subgroup leaves, where the subgroups of index two, three and four are counted and the centre is the one they cannot see.