CHEM 548: Materials Chemistry

Point Symmetry, Point Groups, and Crystal Systems

Meeting 2 · 08/26/2026

Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Ch. 1 §1.1.1, §§1.1.2–1.1.3

Learning Objectives

Five elementsdefine and describe identity, inversion, mirror, rotation, and rotoinversion

Point groupsay what makes a set of operations one, and assemble it by closing the set

Seven crystal systemsidentify each by the minimum symmetry it requires

Exactly 32account for why there are 32 point groups, and connect them to their crystal system

Directions and planesname a crystallographic direction and a lattice plane inside a unit cell

Oblique: 1 ↔ 2

a ≠ b, γ arbitrary. In lecture 1 we chose the cell that shows each lattice's symmetry; now we write that symmetry down as a Hermann–Mauguin symbol, a list of the point symmetry elements the pattern keeps. The bare lattice carries each shape's maximum symmetry; a motif can only lower it. Each title reads minimum ↔ maximum.

Lattice alone: the maximum

Every lattice keeps a twofold rotation

Hermann–Mauguin: 2

With a motif: down to the minimum

An asymmetric molecule removes it

Hermann–Mauguin: 1

Per Solid State Materials Chemistry, §§1.1.1–1.1.2.

In lecture 1 we chose the cell that shows the symmetry of its lattice.

We write that symmetry down as a Hermann–Mauguin symbol: a list of the point symmetry elements the pattern keeps
Oblique is the general case: a and b unequal, γ arbitrary

[advance: apply the twofold rotation]

Rotate 180° about any lattice point: the cell copy lands on the cell across the axis
Every lattice has this twofold, so we write down a 2 — the maximum this shape allows

[advance: attach a molecule to every lattice point]

A crystal is lattice plus motif: attach the same molecule to every point
This molecule has no symmetry: the twofold is gone, the symbol drops to 1, the minimum

The lattice by itself always has the maximum point symmetry allowed by its metric (e.g., the triangular lattice has 6mm).

But when you attach a motif, you can reduce the symmetry—never increase it—because the motif may not respect all the lattice operations.

Next is the rectangular lattice, where the right angle makes mirrors possible.

Rectangular: m ↔ 2mm

a ≠ b, γ = 90°

Lattice alone: the maximum

Twofold plus two perpendicular mirror families

Hermann–Mauguin: 2mm

The fold flips the flag: mirrors reverse handedness, rotations never do.

With a motif: down to the minimum

A bent molecule keeps one mirror family

Hermann–Mauguin: m

Per Solid State Materials Chemistry, §§1.1.1–1.1.2.

In the rectangular lattice, γ is 90 degrees and a and b are still unequal.

[advance: apply the twofold rotation]

The twofold comes first, as for every lattice: we write down a 2

[advance: apply the mirror family along b]

The right angle makes a mirror along b possible: the cell copy folds across the red line
The flag in the cell copy flips left-handed: mirrors reverse handedness, rotations never do
One line stands for a whole family of parallel mirrors: we add an m

[advance: apply the mirror family along a]

A second family along a, perpendicular to the first, completes the symbol: 2mm — the rectangular maximum

[advance: attach a molecule to every lattice point]

A bent molecule keeps one mirror, breaks the twofold and the other family: the symbol drops to m, the minimum

Next is the square lattice, where equal edges allow a fourfold rotation.

Square: 4 ↔ 4mm

a = b, γ = 90°

Lattice alone: the maximum

Fourfold plus edge and diagonal mirror families

Hermann–Mauguin: 4mm

The turned flags stay right-handed; each m makes a copy no rotation can.

With a motif: down to the minimum

A pinwheel molecule keeps only the fourfold

Hermann–Mauguin: 4

Per Solid State Materials Chemistry, §§1.1.1–1.1.2.

The square lattice has equal edges and a right angle.

[advance: apply the fourfold rotation]

A 90° turn lands the cell copy on the next cell over: a fourfold axis, so we write down a 4

[advance: apply the edge mirror family]

One mirror family along the cell edges: we add an m
Same landing cell as the rotation, but the flag flips: turned and solid under the fourfold, flipped and hollow under the mirror
Each m makes a copy no rotation can

[advance: apply the diagonal mirror family]

The second family along the diagonals folds the cell onto itself: 4mm — the square maximum

[advance: attach a molecule to every lattice point]

A pinwheel molecule: its blades have no mirror image, only the fourfold survives, the symbol drops to 4, the minimum

The last shape is the hexagonal lattice.

Hexagonal: 3 ↔ 6mm

a = b, γ = 60°. The 120° cell from lecture 1 describes the same net.

Lattice alone: the maximum

Sixfold plus six mirrors in two alternating families

Hermann–Mauguin: 6mm

With a motif: down to the minimum

A three-bladed molecule keeps only a threefold

Hermann–Mauguin: 3

Per Solid State Materials Chemistry, §§1.1.1–1.1.2.

The hexagonal lattice has equal edges and a 60-degree angle.

In lecture 1 we drew the same triangular net with the 120-degree cell: both cells describe one lattice

[advance: apply the sixfold rotation]

Turn 60°: the cell copy stands on its corner and every corner lands back on the net, so we write down a 6

[advance: apply the first mirror family]

Three mirrors through the nearest-neighbor rows: the cell copy folds across the horizontal one, and we add an m

[advance: apply the second mirror family]

Three more bisect them: six mirrors in two alternating families, the full symbol is 6mm — the hexagonal maximum

[advance: attach a molecule to every lattice point]

A molecule with three blades repeats every 120° and nothing else: the symbol drops to 3, the minimum

Next we collect all four shapes and their symbols in one table.

Summary: the four 2D crystal systems

In 4mm the number is a rotation axis, each m a family of mirrors containing it. A system is defined by the point symmetry it can carry: minimum is what membership requires, maximum what the bare lattice has.

2‑D crystal systems: minimum ↔ maximum point-group symmetry

Crystal system	Unit-cell metric (a,b,γ)	Minimum symmetry (H–M)	Maximum symmetry (H–M)
Oblique	$a \neq b,\; \gamma \neq 90^\circ, 60^\circ$	1 (no symmetry)	2 (one twofold rotation)
Rectangular	$a \neq b,\; \gamma = 90^\circ$	m (one mirror)	2mm (twofold + two perpendicular mirror families)
Square	$a = b,\; \gamma = 90^\circ$	4 (four-fold only)	4mm (fourfold + mirrors along edges and diagonals)
Hexagonal (triangular net)	$a = b,\; \gamma = 60^\circ$	3 (three-fold only)	6mm (sixfold + six mirrors)

A motif can lower the symmetry of a lattice while the lattice retains its own maximum symmetry.

This table collects the minimum and maximum point symmetry of the four lattice shapes.

This range is the definition of a crystal system: a system is defined by the point symmetry it can carry
The minimum is the least symmetry a motif can leave; the maximum is the symmetry of the bare lattice
Oblique: the lattice has only a twofold, and a generic motif breaks even that
Rectangular: a motif keeps one mirror at most, the lattice has both families and the twofold
Square: a motif can drop the mirrors and keep the fourfold
Hexagonal: a motif can drop the lattice to a threefold, the lattice keeps the sixfold and all six mirrors

Next we count the two-dimensional Bravais lattices, where the centered rectangular cell from lecture 1 makes five out of four.

2-D Bravais lattices (5 total)

Bravais lattice	Lattice type	Metric constraints	Notes / equivalences
Oblique (p)	primitive	$a \neq b,\; \gamma \neq 90^\circ, 60^\circ$	General parallelogram.
Rectangular (p)	primitive	$a \neq b,\; \gamma = 90^\circ$	Edges orthogonal, unequal lengths.
Rectangular (c)	centered	$a \neq b,\; \gamma = 90^\circ$	Distinct Bravais type. Conventional rectangle has a lattice point at the center; primitive basis can be taken as $\mathbf{u} = \tfrac{1}{2}(\mathbf{a}+\mathbf{b})$, $\mathbf{v} = \tfrac{1}{2}(-\mathbf{a}+\mathbf{b})$ (a rhombus).
Square (p)	primitive	$a = b,\; \gamma = 90^\circ$	Any “centered square” reduces to a primitive square via a $45^\circ$ rotation and rescale (not a new Bravais type).
Hexagonal / Triangular (p)	primitive	$a = b,\; \gamma = 60^\circ$	Often called “triangular” in 2-D. Any centering is equivalent to primitive.

Five Bravais lattices cover every two-dimensional crystal, and this table is the whole count.

Oblique P: the general parallelogram, a and b unequal, γ neither 90 nor 60 degrees
Rectangular P: orthogonal edges of unequal length
Rectangular C is the fourth type: a point at the center of the conventional rectangle, primitive basis u = ½(a + b), v = ½(b − a)
Square P: equal edges and a right angle; a centered square is a primitive square after a 45-degree rotation and a rescale
Hexagonal P, called triangular in two dimensions: equal edges, 60 degrees, and any centering of it reduces to the primitive cell

Four shapes and five lattices: next we separate what a crystal system classifies from what a Bravais lattice classifies.

Crystal systems and Bravais lattices

Aspect	Crystal system	Bravais lattice
What is classified?	Sets of point groups (rotations/mirrors about a fixed point)	Sets of vector lattices (translations) distinguished up to centering
Includes translations?	No	Yes (pure translations only; point symmetry comes from the lattice metric)
What it fixes	Cell metric/shape constraints (e.g., $a = b$, $\gamma = 90^\circ$ for square)	Metric and whether a distinct centering exists (e.g., primitive vs centered)
2-D inventory	4 systems: oblique, rectangular, square, hexagonal	5 Bravais types: p-oblique, p-rectangular, c-rectangular, p-square, p-hexagonal
3-D inventory	7 systems: triclinic, monoclinic, orthorhombic, tetragonal, trigonal, hexagonal, cubic	14 Bravais types (e.g., cubic-P, cubic-I, cubic-F; orthorhombic-P/C/I/F; etc.)
Example mapping	“Square system” (requires $a = b,\; \gamma = 90^\circ$)	Within square: only p-square (any ‘centered square’ reduces to p-square). Rectangular admits p and c.

A crystal system and a Bravais lattice answer two different questions.

A crystal system classifies point symmetry and constrains the cell shape: a = b, γ = 90° for square
A Bravais lattice classifies the translations: it fixes the metric and whether a distinct centering exists
That is why the counts differ: 4 systems against 5 Bravais types, because only rectangular carries both a primitive and a centered lattice
In three dimensions the same split gives 7 systems against 14 Bravais types, cubic P, I, and F among them
The square system makes the point in one line: only the primitive square, since any centered square reduces to it

Next we take the point symmetry these systems classify and define its operations one at a time.

The Five Point Symmetry Elements

This morning rotations and mirrors sorted every 2D lattice into four systems; every symbol we wrote, like 2mm, was a list of them. Today's second half sorts the 3D lattices the same way: five elements, two of them new.

Point symmetry element	Symbol	Symmetry operation	vs. 2D
Identity	1	Every point stays in place; every object has it	same
Proper rotation axis	N	Rotate by 360/N° about the axis; only N = 1, 2, 3, 4, 6 fit a lattice	a point becomes an axis
Mirror plane	m	Reflect through the plane	a line becomes a plane
Inversion center	$\bar{1}$	Send every point through the center to the opposite side	new in 3D
Rotoinversion axis	$\bar{N}$	Rotate by 360/N° about the axis, then invert through a point on it	new in 3D

"Point symmetry" means each operation leaves some location unmoved: a rotation moves every point except those on its axis, a mirror leaves its plane, inversion leaves one center. A translation moves every point; the two elements that carry one arrive next meeting.

5 elements → groups → 7 crystal systems → 32 point groups

This morning two kinds of element, rotations and mirrors, were all we needed: every symbol we wrote, 2mm, 4mm, was a list of them, and that list sorted the four two-dimensional systems.

Second half of today sorts the 3D lattices the same way; the strip at the bottom of the slide lists the steps
Define the five elements, close them into point groups, let required symmetry sort the seven crystal systems, until all 32 groups are on the table; we are at step one
Three of the five carry over from 2D. Identity leaves every point in place, and every object has it
Rotations keep the orders 1, 2, 3, 4, and 6, taken now about an axis rather than a point
A mirror line becomes a mirror plane, and an axis or a plane now has a direction inside the cell, the second half of today

[advance: reveal the two elements new in three dimensions]

Inversion is new in three dimensions: every point goes through a center to the opposite side
Rotoinversion is new as well: rotate about an axis, then invert
Each operation leaves some location unmoved: a rotation its axis, a mirror its plane, inversion its center; that is what point symmetry means
Next meeting two more elements add a translation, the operation that moves every point

Next is the inversion center.

Inversion Center

Inversion through a point, called the center of symmetry
The element is written $\bar{1}$, read “one bar”

Sixfold Axis of Rotation

A hexagonal prism crystal with a vertical rotation axis drawn through it.

When rotated about its axis, the crystal repeats itself every 60° (six times in a 360° rotation).

In three dimensions a rotation turns about an axis, a line through the crystal, not a point.

Rotation Axes: Names and Symbols

An $n$ fold rotational symmetry operation rotates an object by $360^\circ/n$. Only $n$ = 1, 2, 3, 4, and 6 are permitted in a periodic lattice.

Two dog motifs related by a twofold rotation about a central lens marker.

2 fold
‘Diad’

Three dog motifs related by a threefold rotation about a central triangle marker.

3 fold
‘Triad’

Four dog motifs related by a fourfold rotation about a central square marker.

4 fold
‘Tetrad’

Six dog motifs related by a sixfold rotation about a central hexagon marker.

6 fold
‘Hexad’

The blue marker at each center is the axis’s graphical symbol: lens, triangle, square, hexagon.

An N-fold axis turns the object by 360 degrees over N, and these diagrams run through the four orders a lattice actually uses: twofold at 180 degrees, threefold at 120, fourfold at 90, and sixfold at 60.

Each order has a spoken name: diad, triad, tetrad, hexad
And a graphical symbol, the blue marker at each center: a lens for the twofold, a triangle for the threefold, a square for the fourfold, a hexagon for the sixfold
These marks label axes on every crystallographic diagram from here on: the symbol table later today collects them, and the space-group diagrams of the coming lectures are drawn with them
Only the orders 1, 2, 3, 4, and 6 occur in a periodic lattice; we proved that in lecture 1, so today we take it as given

Next is the element that combines a rotation with a second motion, rotoinversion.

RotoInversion $\bar{N}$

$\bar{4}$: rotate $90^\circ$, then invert

$\bar{N}$: rotate by $360^\circ/N$ about the axis, then invert through a point on that axis.

Rotate $90^\circ$, invert: $0 \to 1$

Again: $1 \to 2$

Again: $2 \to 3$, then the set closes: four points

The $\bar{N}$ family

$\bar{1}$: inversion center

$\bar{2}$: $m$, a mirror plane

$\bar{3}$: rotate $120^\circ$, invert; six points

$\bar{4}$: rotate $90^\circ$, invert; four points

$\bar{6}$: rotate $60^\circ$, invert; six points

Rotoinversion is a compound element: rotate by 360 degrees over N about the axis, then invert the result through a point on that axis.

Neither step alone has to be a symmetry of the object: only the pair together brings it back onto itself
The vertical line is the axis, the two circles hold the positions the operation visits, and the dot between them is the inversion center
Point 0 starts on the upper circle, and we run 4 bar on it: N is four, so the turn is 90 degrees

[advance: turn point 0 by 90 degrees and invert it]

The hollow circle is where the 90-degree turn leaves the point, a quarter of the way around the upper circle
Straight through the center, the same distance out the far side, and we land on point 1 on the lower circle
Two motions, one operation, and point 1 is the first equivalent position

[advance: run the operation again]

Point 1 turns another 90 degrees and inverts back up to point 2 on the upper circle
Every application flips the point from one circle to the other and walks it three quarters of the way around

[advance: run it a third time]

Point 3 lands on the lower circle and the set closes: one more turn and inversion carries it straight back onto point 0
So 4 bar generates four equivalent points

[advance: bring up the N bar family]

1 bar is a 360-degree turn followed by inversion, which is just an inversion center
2 bar is a 180-degree turn followed by inversion, and that pair is a mirror plane perpendicular to the axis
That leaves 3 bar, 4 bar, and 6 bar as the rotoinversion axes with a character of their own
3 bar turns 120 degrees and generates six equivalent points, 4 bar is the one we just walked with four, and 6 bar turns 60 degrees and also closes at six

Now let's take that family one member at a time, starting with 1 bar.

$\bar{1}$: A Full Turn, Then Invert

$\bar{1}$: turn by $360^\circ/1$ about the axis, then invert through the center.

Rotate $360^\circ$ (back to start), invert: straight through the center

$\bar{1}$ is the inversion center

$\bar{1} \equiv i$: rotating a full turn changes nothing, so all that is left is the inversion

One bar is the first member of the family, and it is a onefold rotation followed by an inversion.

A onefold rotation is a full 360-degree turn, so it carries every point right back to where it started
Point 0 sits on the upper circle, the axis runs up the middle, and the dot at the middle of the axis is the inversion center

[advance: turn point 0 all the way around and invert it]

The sweep runs the whole way around the upper circle and comes back to where it began, so the hollow ring sits right on top of point 0
Nothing has happened yet, and now we invert: straight through the center and the same distance out the far side
We land on point 1 on the lower circle

[advance: name what 1 bar really is]

The turn contributed nothing, so all that is left is the inversion, and that is why we write 1 bar equals i
When we listed the family a moment ago, 1 bar was written as an inversion center: one element with two names

Next is 2 bar, where the turn is half a circle.

$\bar{2}$: Half a Turn, Then Invert

$\bar{2}$: turn by $360^\circ/2$ about the axis, then invert through the center.

Rotate $180^\circ$, invert: the point lands straight below

$\bar{2} \equiv m$: a mirror plane perpendicular to the axis

Two bar is a twofold rotation followed by an inversion, so the turn is 180 degrees, half a circle.

Point 0 starts in the same place on the upper circle

[advance: turn point 0 by 180 degrees and invert it]

Half a turn carries the point to the far side of the upper circle, where the hollow ring marks it
Inverting through the center brings it down to point 1, directly below point 0 and the same distance out from the axis

[advance: draw in the plane those two points define]

Point 0 and point 1 sit one above the other, the same distance above and below the level of the center
That is exactly what a mirror plane at right angles to the axis does to a point, and the shaded disc is that plane
So two bar is m: turning by half and inverting is the same operation as reflecting through the horizontal plane

Next is 3 bar, the first member with a character of its own.

$\bar{3}$: Rotate $120^\circ$, Then Invert

$\bar{3}$: turn by $360^\circ/3$ about the axis, then invert through the center.

Rotate $120^\circ$, invert: $0 \to 1$

Six equivalent points before it closes

$\bar{3}$ contains both a threefold and an inversion center

Three bar is a threefold rotation followed by an inversion, so the turn is 120 degrees, a third of a circle.

Point 0 starts on the upper circle again

[advance: run 3 bar once]

The turn carries point 0 a third of the way around the upper circle to the hollow ring
Inverting through the center drops it to point 1 on the lower circle: one application, 0 goes to 1

[advance: fill in the rest of the set]

Running it over and over walks the point around, flipping circles each time, and it takes six applications before we come back to where we started
Three points end up on the upper circle and three on the lower: six equivalent points in all
The hollow ring turns out to be point 4, because the three points on top are 120 degrees apart, which is a threefold axis in its own right
Every point on top also has a partner straight through the center below it, which is an inversion center
So three bar carries both a threefold axis and an inversion center inside it

Next is 6 bar, the last member of the family.

$\bar{6}$: Rotate $60^\circ$, Then Invert

$\bar{6}$: turn by $360^\circ/6$ about the axis, then invert through the center.

Rotate $60^\circ$, invert: $0 \to 1$

Two eclipsed triangles, one per circle

$\bar{6} \equiv 3/m$: a threefold axis with a mirror plane perpendicular to that axis

$N/m$: the slash always makes the mirror perpendicular to the $N$ axis it follows; an $m$ beside the number, as in $4mm$, contains the axis instead

Six bar is a sixfold rotation followed by an inversion, so the turn is 60 degrees.

Point 0 starts on the upper circle

[advance: run 6 bar once]

Sixty degrees around the upper circle takes us to the hollow ring
Inverting through the center lands point 1 on the lower circle: one application, 0 goes to 1

[advance: fill in the rest of the set]

Six applications close the set here too, the same count as 3 bar, but the points land in a different arrangement
The three on the upper circle are 120 degrees apart, the three on the lower circle are 120 degrees apart, and each lower point sits directly beneath an upper one: point 3 is straight below point 0
That is two triangles, one on each circle, eclipsing each other rather than staggered
The shaded plane through the center reflects the top triangle onto the bottom one, so this set carries a threefold axis and a perpendicular mirror plane
We write that 3 over m: the slash always means the mirror plane is perpendicular to the axis it follows, so six bar is the same element as 3/m
When the m sits beside the number with no slash, as in 4mm from the two-dimensional symbols, the mirror contains the axis instead

That completes the point symmetry elements, so next we put them all in one table.

Symmetry Element Symbols

Table 1.28, point symmetry elements, from West, Solid State Chemistry and its Applications.

The written symbol goes into a point-group or space-group symbol; the graphical symbol marks the element on a crystallographic diagram.
The filled shape counts the order: a lens has two sides, a triangle three, a square four, a hexagon six.

This is the symbol key for the elements, Table 1.28 from West, Solid State Chemistry and its Applications.

Each element has a written symbol, the one that goes into a Hermann–Mauguin string
And a graphical symbol, the one printed on a crystallographic diagram: filled lens for the twofold, triangle for the threefold, square for the fourfold, hexagon for the sixfold, solid line for a mirror — the shape counts the order
The operation column is the lecture so far in five rows: identity does nothing, inversion carries x, y, z to minus x, minus y, minus z, a mirror reflects across its plane, a rotation axis turns by 360 over N, a rotoinversion axis turns and then inverts
Orders on offer still only 1, 2, 3, 4, and 6: a fivefold or an eightfold axis cannot fill space without leaving gaps or overlaps
The written column builds point-group symbols later today; the graphical column is what next lecture's space-group diagrams print
Two entries have no graphical symbol of their own: 1 bar, the center of symmetry, and 2 bar, the mirror plane written another way

That is every element that holds a point fixed. Next we ask why we bundle them into groups.

Why Group the Symmetry Elements?

Lecture 1 sorted cells into seven crystal systems by shape. Shape is the consequence: each system is defined by the symmetry its lattice must carry.

One element at a time is not enough. A crystal's operations act together, and the set must close on itself: a group.

7 crystal systems → 32 point groups → × 14 Bravais lattices + screws and glides → 230 space groups → any crystal structure

A published structure is one space-group symbol plus a short list of atoms; symmetry generates the rest.

5 elements → groups → 7 crystal systems → 32 point groups

So far today we have collected symmetry elements one at a time. Here is why we now bundle them into groups.

In the first lecture we sorted unit cells into seven crystal systems by the shape of the cell
Shape is really a consequence: a cube is cubic because of its threefold axes along the body diagonals, not because its edges happen to come out equal

[advance: say why one symmetry element is not enough]

Operations combine: do a rotation, then a reflection, and you have performed some third operation
Whatever set of operations a crystal has must close on itself, and a closed set like that is called a group

[advance: show the flow out to space groups]

There are exactly 32 of these groups compatible with a lattice
Combine the 32 with the compatible Bravais lattices, then let rotations become screws and mirrors become glides, and the list fills out to exactly 230
That enumeration was completed in the 1890s; next lecture we learn to read the 230, not to re-derive them
Every published crystal structure is one space-group symbol plus a handful of atomic positions; the symmetry generates the rest of the cell

Before we take the requirements one at a time, it is worth seeing where these thirty-two packages actually came from.

Why Point Groups? Faces, Not Atoms

The 32 classes were found before atoms could be seen, from two laws of faces, measured with a goniometer.

Steno 1669the angles between a mineral's faces are constant

Haüy 1784faces sit at orientations given by small whole numbers

Purple fluorite crystals grown as interpenetrating cubes with flat square faces. — Fluorite — cubic

Brassy pyrite cubes whose square faces carry parallel striations, each face striated at right angles to its neighbors. — Pyrite — cubic, striations break the fourfold

A smoky quartz crystal: a six-sided prism capped by inclined terminal faces. — Quartz — trigonal, left- and right-handed forms

A pale gypsum swallowtail twin: two monoclinic blades joined along a plane so the pair opens like a V. — Gypsum — monoclinic

Faces come in symmetry-related sets: the class is readable from the shape.

Symmetries never come one at a time: two operations in a row are some third operation, so a crystal's operations form a closed set.

Hessel (1830) counted the closed sets the face laws allow: exactly 32. Unnoticed; rederived by Gadolin (1867), 80 years before X-rays.

5 elements → groups → 7 crystal systems → 32 point groups

So far we have said that a crystal's operations must form a group. Before we write the mathematical requirements down, it is worth asking where these groups came from, because it was not from atoms.

Nobody had seen an atom, and X-rays were eighty years away; it came from faces
Crystals of one substance grow with different face sizes, but the angles between the faces are always the same — Steno, 1669
The angle was the one thing an early goniometer could measure precisely
Haüy added in 1784 that the faces sit at orientations described by small whole numbers

[advance: reveal the mineral strip]

Fluorite grows cubes
Pyrite grows cubes too, but its faces carry striations that break the fourfold symmetry, so the faces tell us not just the crystal system but the class within it
Quartz grows small extra faces that come in left-handed and right-handed versions
Gypsum shows the single mirror plane of a monoclinic crystal

[advance: reveal why the answer is a list of groups]

Symmetry operations never come one at a time: a rotation then a reflection is some third operation, so a crystal's set of operations must close on itself
Johann Hessel counted the closed sets the face laws allow in 1830 and found exactly thirty-two
Nobody noticed; Axel Gadolin found the same thirty-two independently in 1867
So this classification was read off mineral faces, half a century before anyone could check a single atom

It starts with what makes a set of operations a group.

Point Groups

A point group is a set of symmetry operations, acting on an isolated object, that fulfils the four requirements of a group.

A group must have

closuretwo combined give one of the set

the associative lawany pair in a chain may go first

an identitythe do-nothing operation, written 1

an inverseevery operation has one that undoes it

A point group is a set of operations, the point symmetry elements, that fulfil the mathematical requirements of being a group and act on an isolated object.

Closure: combining two operations of the set gives an operation of the set
The associative law: in a chain of operations it does not matter which pair we combine first
An identity: the do-nothing operation, written 1, which leaves every operation unchanged when we combine it with them
An inverse for every operation: a partner operation that undoes it, so that combined, the two give the identity
Closure is the one that does the real work today

We take each requirement in turn on one shape, an equilateral triangle with a threefold axis through its center, starting with closure.

Closure

This triangle has exactly three operations: do nothing (1), rotate 120°, and rotate 240°.

120° then 120° is the same net move as a single 240° rotation, and 240° is one of the three operations above. Combining two of them gave another one: closure.

Each operation lands the triangle back on its own outline. The red dot is not part of the triangle; it tells the three identical-looking positions apart.

Here is the triangle: three equal sides, and a threefold rotation axis through its center.

Rotate it 120 degrees, or 240 degrees, or leave it alone, and it lands exactly on top of its own outline, every corner on a corner position
Those three rotations, and only those, are the operations of the triangle: the set we test
A plain triangle looks identical in all three positions, so one corner is colored red
The red dot is not part of the triangle; it is a label we add so we can tell the three identical-looking positions apart

[advance: rotate the triangle 120 degrees]

120 degrees about the axis: the outline lands back on itself, and without the red dot nothing seems to have happened
The dot shows the corner really moved one place around

[advance: rotate another 120 degrees]

Another 120 degrees: the marked corner is now in the third position

[advance: name the combined operation]

Two operations one after the other turned the triangle 240 degrees
A single 240 degree rotation is one of the three operations we started with
Combining two operations of the set gave an operation of the set, and that is closure

Next we ask what happens when we chain three operations together and combine them in different orders.

The Associative Law

Both groupings land the marked corner in the same place.

The three rotations happen in the same order both times. The only choice is which two neighbors to merge into a single move first, and either choice gives the same result.

Now chain three operations together: rotate 120 degrees, then 120 degrees again, then 240 degrees.

The associative law says the chain stays in that order, and we choose which two neighboring rotations to merge into a single move first: the first two, or the last two

[advance: combine the first two rotations first]

Top row: merge the first two rotations of the chain, the 120 and the 120, into one move: 120 and 120 make 240
Then the chain's last rotation, 240 degrees, still has to happen, and it carries the marked corner around to the lower right

[advance: combine the last two rotations first]

Bottom row: merge the last two rotations instead, the 120 and the 240: 120 and 240 make a full turn, which is the identity
The chain's first rotation, 120 degrees, still has to happen, and it is now the only real move left

[advance: compare where the corner landed]

The marked corner lands in the same place both ways
That is what associativity gives us: we never have to write brackets in a chain of symmetry operations

Next is the operation that does nothing at all, the identity.

Identity

Combining 1 with any operation leaves that operation unchanged.

Inverse

120° and 240° undo each other: combined, they give the identity, so each is the inverse of the other.

An inverse is a partner operation, and every operation in a group has one. The inversion center $\bar{1}$ is a different thing: one particular symmetry operation, which happens to be its own inverse.

The fourth requirement is that every operation has an inverse, a partner operation that undoes it.

We test it on the same triangle, with the marked corner at the top

[advance: rotate 120 degrees]

Rotate 120 degrees and the marked corner moves one place around

[advance: rotate 240 degrees on top of it]

Rotate 240 degrees on top of that and the marked corner is back where it started
The two rotations together give the identity, so each one is the inverse of the other

[advance: separate the inverse from the inversion center]

Watch the words here: an inverse is a partner operation, and every operation in a group has one
The inversion center, 1 bar, is a different thing: one particular symmetry operation, the one that sends every point through the center to the far side
It happens to be its own inverse, but that is a fact about one operation, not the requirement we just wrote down

That was the last of the four requirements; next we collect the results and name what we have built.

The Triangle's Point Group

The set: 1, rotate 120°, rotate 240°.

✓ closure: two rotations combined always gave a rotation of the set
✓ associative law: whichever neighboring pair we merged first, the corner landed in the same place
✓ identity: the do-nothing operation, 1, is in the set
✓ inverse: 120° and 240° undo each other, and 1 undoes itself

All four requirements hold, so these three rotations are a point group. Its symbol is 3: a single threefold axis.

The triangle also has three mirror planes through the axis; adding their operations gives a larger point group, 3m.

Look back at what the triangle just did.

Closure passed: two rotations combined always gave a rotation of the set
The associative law held: whichever neighboring pair we merged first, the corner landed in the same place
The set contains the do-nothing operation, 1
Every rotation has its undoing partner inside the set

[advance: name the point group]

A set that meets all four requirements is a point group, so we have just built one
Its symbol is 3: a single threefold axis
The triangle has more symmetry still, three mirror planes through the axis; add their operations and the set closes again, on a larger point group written 3m

Now we take a twofold rotation axis with a mirror plane perpendicular to it and test that pair against the four requirements.

Question

Let's take a twofold rotation axis. And let's have a mirror plane perpendicular to it.

Is that a valid point group, just those two symmetry operations?

If not, what other symmetry operations do we need to add to those two to get a valid point group?

Point Group $2/m$

Four operations, four copies of the motif: 1 left it in place, 2 rotated it, m reflected it, and the new diagonal jump is $\bar{1}$, inversion through the center. Closure adds that inversion — the group is $\{1,\ 2,\ m,\ \bar{1}\}$, written 2/m: a twofold axis with a mirror plane perpendicular to it.

Here is the setup again: a twofold axis through the motif and a mirror plane perpendicular to it, and we test those two symmetry elements against closure, one operation at a time.

Start with the identity, which every group has and which leaves the motif exactly where it is

[advance: apply the twofold rotation]

Apply the twofold: the motif turns 180 degrees about the axis and lands on the far side of it.

[advance: apply the mirror plane]

Apply the mirror: reflection across the horizontal plane sends the motif to a partner on the other side of the plane.

[advance: combine the two operations]

Now do the twofold first and the mirror second: the motif lands diagonally opposite, straight through the point where the axis meets the plane, and that is a third operation, inversion.

The mirror copy from the last pane is still there in the upper left, and each copy now carries the label of the operation that put it there: 1 lower left, 2 lower right, m upper left, 1 bar upper right.

The dashed lines join the two pairs of copies that the inversion center relates
Two symmetry elements were not enough: the set does not close until the inversion center is in it

[advance: reveal the completed group]

With the identity the group has four operations: 1, 2, m, and 1 bar
We write it 2/m; the slash says the mirror plane is perpendicular to the twofold axis

That set closes on itself, so it is a valid point group. Before we count how many such groups a crystal allows, we sort the crystals themselves by the symmetry their lattices have to carry.

The Seven Crystal Systems

A crystal system is defined by the point symmetry it can carry; the cell shape follows.

This morning · two dimensions · four systems

obliqueup to 2

rectangularup to 2mm

squareup to 4mm

hexagonalup to 6mm

Now · three dimensions · seven systems, least required symmetry to most

triclinicnothing required

monoclinicone twofold

orthorhombicthree twofolds

tetragonala fourfold

trigonala threefold

hexagonala sixfold

cubicfour threefolds

Each slide that follows names one system's requirement and collects its point groups; the running count lands on 32.

5 elements → groups → 7 crystal systems → 32 point groups