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Sequel to Group Theory

In the previous documents we have discussed finite cyclic groups, i.e. cyclic groups of finite order, which means that they consist of a finite number of elements. We considered groups of prime order meaning that the number  n, indicating the number of elements is a prime number, i.e. a number that is divisible by itself and by 1 only. Groups of prime order are always cyclic, and they have no subgroups. We also considered cyclic groups of composite order (i.e. where the order  n is not prime). These groups have subgroups. The latter entail a lot of interesting features, like cosets, normal subgroups and homomorphisms. We have exemplified these with the cyclic group of order 12, i.e. C₁₂.

Having seen something of cyclic groups you will be in a position to appreciate the following theorems, which are given here without proof (BUDDEN, 1978, pp. 175) :

Our treatment of Group Theory is geared towards a deeper understanding of the symmetries of crystals. Especially for the symmetries of the internal order of crystals, cyclic groups of infinite order, and thus having infinitely many elements, are very important. Realized crystallographically the elements of such groups are symmetry operations, one type of them being translations. When we have some infinite pattern, and when we shift this pattern a certain distance along some direction, and when, after having done so, the pattern is precisely mapped onto itself, the pattern is said to possess translational symmetry. Such a translation can be indicated by a vector.

The set Z of integers, which is the set {..., -3, -2, -1, 0, 1, 2, 3, ...} forms a group under ordinary addition. We write (Z, +). The element 0 is the identity element, because 0 + x = x + 0 = x, for every element  x  of this group. And the group is infinite because repeated application of any element except the identity will never reach 0, which means that the period of every element except the identity is infinite.
This group, and isomorphic copies of it, is the cyclic group of infinite order, C_infinite. It is infinite, because and can be generated by a single element 1 and its inverse -1 (which can also be written as 1^-1).
Another way of generating C_infinite is by means of a single translation, as in the next Figure.

Figure 1.  The symmetry group C_infinite

Figure 2.  The symmetry group C₉

Infinite groups can also be generated by elements each of which has a finite period, for instance the reflections in two parallel mirrors, and indeed a reflection has period 2. An infinite sequence of images is generated (We are familiar with this phenomenon when we look into a mirror while having placed a second mirror behind us, parallel to the first. We then see an infinity of images of ourselves. The infinite group that is thus generated is the group D_infinite).

The symmetry of objects having in addition to rotational symmetries (direct symmetries) with respect to one center, also reflectional symmetries (opposite symmetries), like the regular polygons, can be described by one or another dihedral group.

The group elements of the symmetry of a regular polygon are rotations and reflections and all possible combinations of them, for instance a rotation followed by a reflection in some reflection line (mirror line). The problem is that we must decide whether such a reflection line is itself rotated by the mentioned rotation, or whether it should be considered a fixed reflection line. In order to avoid confusion we will distinguish between the vertices themselves of a polygon, and their respective positions. See next Figure, depicting the simplest regular polygon, the equilateral triangle.

Figure 3. The equilateral triangle in its fixed background. The vertices A, B and C of the triangle can move from one position to another. The positions themselves, (1), (2) and (3), are fixed. Also the three mirror lines, supported by the triangle, are not moved (one such mirror line is depicted). The positions of the vertices as they are in this Figure will be considered as the identity transformation.

The initial position of vertex A is position 1.
The initial position of vertex B is position 2.
The initial position of vertex C is position 3.

The triangle can be rotated by 0, 120 or 240 degrees about its mid-point effecting the triangle occupying the same patch of space again as before those operations had taken place. So the vertices -- which are parts of the triangle -- move with respect to the background. And now comes the important thing :  The symmetry lines, supported by the triangle are not part of the triangle, so they are not moved when the triangle is rotated.
The next Figure illustrates an anticlockwise rotation, named  p, of the triangle by 120⁰.

Figure 4. The equilateral triangle of Figure 3 is rotated anticlockwise through an angle of 120⁰. This symmetry transformation is here called  p.

One vertex of the triangle moves from position  1  to position  2.
One vertex of the triangle moves from position  2  to position  3.
One vertex of the triangle moves from position  3  to position  1.

So we describe the rotation  p  by the permutation of the positions :

Next we will illustrate a reflection  b. And that is a reflection of the triangle as it is in Figure 3, through the mirror line indicated in that figure. This mirror line has a fixed position, because it is not a part of the triangle.

Figure 5. The equilateral triangle of Figure 3 is reflected in the mirror line indicated (This mirror line is called  b).

Thus we will describe the reflection  b  as the following permutation of positions :

Next we will illustrate the rotation  p, immediately followed by the reflection  b, i.e. the symmetry transformation  bp :  first an anticlockwise rotation of the triangle by 120⁰ followed by a reflection in the fixed axis (mirror line)  b, i.e. we consider the mirror line  b NOT having been moved by the rotation  p,   because, so we could say, mirror lines do not, as a concrete part, belong to the triangle. Reflection takes place across its original position. See next Figure.

Figure 5. The equilateral triangle of Figure 3 is first rotated anticlockwise by 120⁰ and then reflected in the mirror line indicated (This mirror line is called  b). The mirror line has not been moved by the rotation. The total symmetry transformation is  bp.

And this is equivalent to a reflection  c , i.e. a reflection across a mirror line that goes through vertex C in its original position. We can indicate this symmetry transformation  bp = c  by the following permutation of positions :

So with respect to the symmetry transformations of the lateral triangle (trigon) we have some first results :

We still have to account for a rotation by 240⁰ ( =  p² ) ( p³ amounts to a rotation by 360⁰, which is equivalent to no rotation at all, i.e. to a rotation of 0⁰ ). Further there are two more reflections, one across the axis through vertex A-in-its-original-position, this reflection we will call  a, and another across the axis through the vertex C-in-its-original-position. This reflection we will call  c.

p² is indicated in the next Figure.

Figure 6. The equilateral triangle of Figure 3 is rotated anticlockwise by 240⁰. This symmetry transformation is here called  p² (i.e. applying  p  two times).

One vertex of the triangle has moved from position 1 to position 3,
One vertex of the triangle has moved from position 3 to position 2,
One vertex of the triangle has moved from position 2 to position 1.

The original positions of the vertices A, B and C (representing the identity transformation) are depicted in Figure 3, and explicitly stated below that Figure.

This rotation  p²  can now be expressed as the following permutation of positions :

Next we have the reflection  a, which is a reflection across the bisector through vertex A-in-its-original-position. See next Figure.

Figure 7.  Reflection of the equilateral triangle of Figure 3 across the bisector through the vertex A-in-its-original-position. This symmetry transformation is here called  a.

So transformation  a  can be expressed as a permutation of positions :

Finally, the transformation  c  is a reflection of the triangle across the bisector through the vertex C-in-its-original-position. See next Figure.

Figure 8.  Reflection of the equilateral triangle of Figure 3 across the bisector through vertex C-in-its-original-position. This transformation is here called  c.

So transformation  c  can be expressed as the following permutation of positions :

So we can now summarize all six permutations expressing the symmetry of the equilateral triangle :

These six permutations, representing the corresponding symmetry transformations, form the dihedral group D₃. When we want to construct the group table we must determine all possible binary combinations (which should turn out always to be one of the elements, 1, p, p², a, b, c), i.e. combinations (products) such as  ap  (which means transformation  p  followed by transformation  a).
We can determine these by means of the established permutations of positions. We could for instance determine  ap  by successively applying the corresponding permutations. In doing so we write the permutations below each other, starting with the identity, and writing down only the lower set of symbols of the relevant permutation. So the permutation  p  replaces 1 by 2, 2 by 3, and 3 by 1, while the permutation  a  replaces 1 by 1, 2 by 3, and 3 by 2.
Determining  ap  then goes as follows :

1 2 3
2 3 1  application of  p  to the identity.
3 2 1  application of  a. The result is equal to the permutation  b.

So  ap = b.

In the same way we determine all the products :

The product  pa.
1 2 3
1 3 2
2 1 3  = c

So pa = c.

We see that pa is different from ap, so the group (D₃) is not commutative (non-Abelian).

The product  pb.
1 2 3
3 2 1
1 3 2  = a

So pb = a.

The product  pc.
1 2 3
2 1 3
3 2 1  = b

So pc = b.

The product  p²a.
1 2 3
1 3 2
3 2 1  = b

So p²a = b.

The product  p²b.
1 2 3
3 2 1
2 1 3  = c

So p²b = c.

The product  p²c.
1 2 3
2 1 3
1 3 2  = a

So p²c = a.

The product  ap².
1 2 3
3 1 2
2 1 3  = c

So ap² = c.

The product  aa.
1 2 3
1 3 2
1 2 3  = 1 (identity). Indeed, a  is a reflection, and if we perform it two times we end up with what things were, before we applied the reflections.

So aa = 1.

The product  ab.
1 2 3
3 2 1
2 3 1  = p

So ab = p.

The product  ac.
1 2 3
2 1 3
3 1 2  = p²

So ac = p².

The product  bp.
1 2 3
2 3 1
2 1 3  = c

So bp = c.

The product  bp².
1 2 3
3 1 2
1 3 2  = a

So bp² = a.

The product  ba.
1 2 3
1 3 2
3 1 2  = p²

So ba = p².

bb = 1 (because  b is a reflection).

The product  bc.
1 2 3
2 1 3
2 3 1  = p

So bc = p.

The product  cp.
1 2 3
2 3 1
1 3 2  = a

So cp = a.

The product  cp².
1 2 3
3 1 2
3 2 1  = b

So cp² = b.

The product  ca.
1 2 3
1 3 2
2 3 1  = p

So ca = p.

The product  cb.
1 2 3
3 2 1
3 1 2  = p²

So cb = p².

cc = 1, because  c  is a reflection.

pp² = p²p = p³ = 1, because 120⁰ + 240⁰ = 360⁰ = 0⁰ = 1 (identity).

p²p² = p⁴ = pp³ = p1 = p.

On the basis of these results we can set up a group table for our group D₃ :

Table 7.1

We can see that the group admits of a homomorphic mapping to the group C₂.

The group also has some subgroups, namely {1, p, p²}, {1, a}, {1, b} and {1, c}.

Table 7.2
The group table of D₃ with the subgroup {1, p, p²} ( = C₃ ) highlighted (red).

Table 7.3

The table of the group D₃ with the subgroup {1, a} ( = C₂ ) highlighted (red).

Table 7.4

The table of the group D₃ with the subgroup {1, b} ( = C₂ ) highlighted (red).

Table 7.5

The table of the group D₃ with the subgroup {1, c} ( = C₂ ) highlighted (red).

Figure 8a.  The trigonal pyramid representing the promorph of all the (single non-twinned) crystals of the Ditrigonal-pyramidal Class of the Hexagonal Crystal System.

Now because the reflections  a, b and  c  can also be interpreted as half-turns (180⁰ rotations) of the equilateral triangle through the third dimension about the respective reflection lines, this group of six symmetry transformations is at the same time the group of direct symmetries (i.e. the rotational symmetries only) of the trigonal bipyramid, and also of the trigonal prism. The three reflections  a, b and  c  are then half-turns about axes coinciding with the bisectors of the triangular equatorial plane of the trigonal bipyramid (or trigonal prism). See next Figure.

Figure 9.  The trigonal bipyramid.
Its direct symmetries are according to the group D₃. It has one 3-fold rotation axis, allowing for rotations of 0⁰ (identity), 120⁰ (p), and 240⁰ (p²). Further it has three 2-fold rotation axes (one is depicted in the Figure), coinciding with the three bisectors of the triangular equatorial plane. These are equivalent to the three mirror lines of tyhe equilateral triangle.

When we want to describe the full symmetry of the trigonal bipyramid we must add the enantiomorphs, i.e. its reflectional symmetries. These are :

A reflection across the equatorial plane. This leaves the positions of the vertices of the equilateral triangle (= the equatorial plane of the bipyramid) unchanged, but interchanges the positions of X and Y (See Figure 9).
A reflection across the equatorial plane followed by an anticlockwise rotation about the main axis of 120⁰.
A reflection across the equatorial plane followed by an anticlockwise rotation about the main axis of 240⁰.
A reflection across the (vertical) meridian plane through vertex A-in-its-original-position.
A reflection across the (vertical) meridian plane through vertex B-in-its-original-position.
A reflection across the (vertical) meridian plane through vertex C-in-its-original-position.

This effects a total of 12 symmetry transformations forming the group D₆.
So the full symmetry of the trigonal bipyramid (and also of the trigonal prism) is according to the group D₆.

And this group D₆ is also the group of all symmetries of the regular hexagon.

In order to verify this we must examine the 12 symmetry transformations especially with respect to their periods. In order to do so we must see by what permutations they can be represented. With respect to the trigonal bipyramid we must add two more vertices X and Y, to represent the apices of the bipyramid (See Figure 9). So we now have to do with permutations of five positions that the five vertices A, B, C, X and Y could visit. We will give the permutations by their lower part only. The rotations  p  and  p²  do not let X and Y change position. But the reflections  a, b, and  c  can be interpreted as 'flipping-overs' of the original equilateral triangle, and thus as half-turns about the bisectors lying in the equatorial plane of the bipyramid (One such axis is indicated in Figure 9). So these half-turns (180⁰) do interchange the positions of the vertices X and Y. Let the initial position of vertex X be U (= up), and that of Y be D (= down). So for the direct symmetries of the trigonal bipyramid we have the following permutations and their periods :

1 =   1 2 3 U D.  Period = 1
p =   2 3 1 U D.  Period = 3
p² = 3 1 2 U D.  Period = 3
a =   1 3 2 D U.  Period = 2
b =   3 2 1 D U.  Period = 2
c =   2 1 3 D U.  Period = 2

Now we have to add the above mentioned enantiomorphs (reflectional symmetries, opposite symmetries).

I.  A reflection across the equatorial plane. This leaves the positions of the vertices of the equilateral triangle (= the equatorial plane of the bipyramid) unchanged, but interchanges the positions of X and Y (See Figure 9). This corresponds to the permutation 1 2 3 D U.

II.  A reflection across the equatorial plane followed by an anticlockwise rotation about the main axis of 120⁰. This corresponds to the permutation 2 3 1 D U.

III.  A reflection across the equatorial plane followed by an anticlockwise rotation about the main axis of 240⁰. This corresponds to the permutation 3 1 2 D U.

IV.  A reflection across the (vertical) meridian plane through vertex A-in-its-original-position. This is just a reflection across a vertical plane of the trigonal bipyramid, so the positions of X and Y are not interchanged. Thuso the corresponding permutation is 1 3 2 U D.

V.  A reflection across the (vertical) meridian plane through vertex B-in-its-original-position. This also is just a reflection across a vertical plane of the trigonal bipyramid, so the positions of X and Y are not interchanged. Thus the corresponding permutation is 3 2 1 U D.

VI.  A reflection across the (vertical) meridian plane through vertex C-in-its-original-position. Also this, finally, is just a reflection across a vertical plane of the trigonal bipyramid, so the positions of X and Y are not interchanged.
Thus the corresponding permutation is 2 1 3 U D.

We will now determine the periods of these latter six permutations, by reatedly apply each permutation till the identity is reached.

For I :

1 2 3 U D
1 2 3 D U
1 2 3 U D

So the period of element  I  is  2.

For II :

1 2 3 U D
2 3 1 D U
3 1 2 U D
1 2 3 D U
2 3 1 U D
3 1 2 D U
1 2 3 U D

So the period of element  II  is  6.

For III :

1 2 3 U D
3 1 2 D U
2 3 1 U D
1 2 3 D U
3 1 2 U D
2 3 1 D U
1 2 3 U D

So the period of element  III  is  6.

For IV :

1 2 3 U D
1 3 2 U D
1 2 3 U D

So the period of element  IV  is  2.

For V :

1 2 3 U D
3 2 1 U D
1 2 3 U D

So the period of element  V  is  2.

For VI :

1 2 3 U D
2 1 3 U D
1 2 3 U D

So the period of element  VI  is  2.

Let us now summarize all these twelve permutations representing the symmetry transformations of the trigonal bipyramid :

1 =   1 2 3 U D.  Period = 1
p =   2 3 1 U D.  Period = 3
p² = 3 1 2 U D.  Period = 3
a =   1 3 2 D U.  Period = 2
b =   3 2 1 D U.  Period = 2
c =   2 1 3 D U.  Period = 2
I =   1 2 3 D U.  Period = 2
II =  2 3 1 D U .  Period = 6
III = 3 1 2 D U .  Period = 6
IV = 1 3 2 U D .  Period = 2
V =   3 2 1 U D .  Period = 2
VI = 2 1 3 U D .  Period = 2

This period distribution is indeed the finger print of the dihedral group D₆.

The Trigonal Bipyramid which has -- as we found out -- a D₆ symmetry, is the promorph or stereometric basic form of all the (single, non-twinned) crystals of the Ditrigonal-bipyramidal Class, 6* m 2, of the Hexagonal Crystal System. This promorph belongs to the Isostaura polypleura (hexapleura) (Stauraxonia homopola) in the Promorphological System of Basic Forms .

And from here it is possible to lay the connection with the symmetry of the regular hexagon :, i.e. it is possible to bring the symmetry transformations of the trigonal bipyramid into a 1,1 correspondence and thereby preserving products. Said differently, the group of all symmetry transformations of the trigonal bipyramid is isomorph to the group of all symmetry transformations of the regular hexagon.

We will now consider the symmetry transformations of the regular hexagon and see whether it is indeed possible to set up the mentioned isomorphism.

The regular hexagon possesses the following symmetry transformations :

Rotations of 0⁰, 60⁰, 120⁰, 180⁰, 240⁰, and 300⁰ about the hexagon's mid-point.
Two types of reflections : reflections across lines connecting opposite vertices, and reflection across lines connecting centers of opposite sides.

The next Figure indicates those symmetries.

Figure 10.  The Regular Hexagon and its symmetries.
There are six mirror lines, indicated as  a, b, c, d, f, g,  and six rotations of which one is indicated :  r = 60⁰. The other rotations are 1 (= 0⁰, identity), r² = 120⁰, r³ = 180⁰, r⁴ = 240⁰, r⁵ = 300⁰. The vertices A, B, C, D, E and F can move to the fixed positions 1, 2, 3, 4, 5 and 6.

Let's determine the periods of these elements :

The period of  1 (identity) is 1, because it is directly 1, without any powering.
The period of  r (60⁰ rotation) is 6, because 60 + 60 + 60 + 60 + 60 + 60 = r⁶ = 360⁰ = (effectively) 0⁰ = 1 (identity).
The period of  r² (120⁰) is 3, because 120 + 120 + 120 = (r²)³ = 360 = 1 (identity).
The period of  r³ (180⁰) is 2, because 180 + 180 = (r³)² = 360⁰ = 1 (identity).
The period of  r⁴ (240⁰) is 3, because 240 + 240 + 240 = 480 + 240 = 120 + 240 = 360⁰ = 1 (identity).
The period of  r⁵ (300⁰) is 6, because 300 + 300 + 300 + 300 + 300 + 300 = 240 + 300 + 300 + 300 + 300 = 180 + 300 + 300 + 300 = 120 + 300 + 300 = 60 + 300 = 360⁰ = 1 (identity).
Because a, b, c, d, f, and g are reflections, their period is 2. So we can say
The period of  a is 2.
The period of  b is 2.
The period of  c is 2.
The period of  d is 2.
The period of  f is 2.
The period of  g is 2.

So the pattern of periods of the symmetry transformations (group elements) of the regular hexagon is :

1, 6, 3, 2, 3, 6, 2, 2, 2, 2, 2, 2

And this is indeed the finger print of the dihedral group D₆.

We can now set up our isomorphism from the group of all symmetries of the trigonal bipyramid to the group of all symmetries of the regular hexagon. In this isomorphism the periods of the corresponding elements must match :

1 ----------  Period = 1  ---------- 1 (identity)
p ----------  Period = 3  ---------- r²
p² ----------  Period = 3  ---------- r⁴
a ----------  Period = 2  ---------- a
b ----------  Period = 2  ---------- b
c ----------  Period = 2  ---------- c
I ----------  Period = 2  ---------- d
II ----------  Period = 6  ---------- r
III ----------  Period = 6  ---------- r⁵
IV ----------  Period = 2  ---------- f
V ----------  Period = 2  ---------- g
VI ----------  Period = 2  ---------- r³

By this isomorphism there are some elements related to each other which are different symmetries, for example
the element III of the group of symmetries of the trigonal bipyramid is a reflection across the equatorial plane followed by an anticlockwise rotation about the main axis of 240⁰. So this is an opposite symmetry. It is -- by the above isomorphism -- related to the element r⁵ of the group of symmetries of the regular hexagon. And this is a direct symmetry (a rotation of 300⁰). This relating is fully legitimate, because the respective periods are the same.

It is perhaps instructive to illustrate this with respect to the group C₂. In addition to having a realization in finite arithmetics this group has three crystallographic realizations, realizations where different symmetries are involved, but whose groups are fully isomorph (i.e. group theoretically equivalent). See next group tables which are all expressing the group C₂.

Table 7.6

Realization of the group C₂ in finite arithmetics. The group elements are numbers.

Table 7.7

Realization of the group C₂ in crystallography. The group elements are the identity (1) and a reflection (m). As such it represents the symmetry of crystals of the Domatic Class of the Monoclinic Crystal System.

Table 7.8

Another realization of the group C₂ in crystallography. The group elements are the identity (1) and a half-turn (rotation of 180⁰) (r). As such it represents the symmetry of crystals of the Sphenoidic Class of the Monoclinic Crystal System.

Table 7.9

Yet another realization of the group C₂ in crystallography. The group elements are the identity (1) and an inversion ( = reflection across a point, the mid-point of the crystal) ( i ). As such it represents the symmetry of crystals of the Pinacoidal Class of the Triclinic Crystal System.

Figure 10 (replay).  The Regular Hexagon and its symmetries.
There are six mirror lines, indicated as  a, b, c, d, f, g,  and six rotations of which one is indicated :  r = 60⁰. The other rotations are 1 (= 0⁰, identity), r² = 120⁰, r³ = 180⁰, r⁴ = 240⁰, r⁵ = 300⁰. The vertices A, B, C, D, E and F can move to the fixed positions 1, 2, 3, 4, 5 and 6.

See next Figure :

Figure 11.  The transformation r⁴ performed on the regular hexagon.

and by the cycle (153)(264).

In order to complete the transformation ar⁴ (i.e. the product of a and r⁴) we must now apply to the just obtained result the transformation a which is a reflection in the fixed axis a (Figure 10). The next Figure illustrates this completion.

Figure 12.  The complete transformation ar⁴ on the regular hexagon.

So the permutation representing the transformation a is :

and is the cycle (12)(36)(45).

The movements corresponding to the whole transformation ar⁴ are then as follows :
vertex A going from position 1 to position 4,
vertex B going from position 2 to position 3,
vertex C going from position 3 to position 2,
vertex D going from position 4 to position 1,
vertex E going from position 5 to position 6,
vertex F going from position 6 to position 5.

So the permutation representing the transformation ar⁴ is

which is the cycle (14)(23)(56), which corresponds to the transformation c, which is a reflection in the axis labelled c (Figure 10). So we can say that ar⁴ = c (i.e. the product of the elements a and r⁴ is equal to the element c).

In the same way we can determine the product fb :  first we perform the reflection in the axis b (Figure 10), and then the reflection in the axis f (Figure 10) :

Figure 13.  The transformation fb applied to the regular hexagon.

Further we can determine the product ba. First we apply reflection in the a axis, followed by a reflection in the b axis :

Figure 14.  The transformation ba applied to the regular hexagon.

Thus we have combined several pairs of symmetries of the regular hexagon, and this is a start towards drawing up a complete table of the group of symmetries.
The method presented, though it is instructive, is very laborious for us to obtain all the 12 x 12 = 144 possible combinations of symmetries. Instead we should, in each case, follow one particular vertex, see its starting position, and then determine its final position with respect to the relevant product. In the last example, concerning ba we know that the transformation a takes the vertex A from position 1 to position 2. Having arrived there, it will not suffer any further movement, because anything in that position remains where it was when a reflection in the axis b is applied. So, in terms of the permutation of positions, we get :

1 -- 2 and 2 -- 2, thus (as total result) 1 -- 2

Two reflections (a and b) effect a non-reflectional result, so the latter must be a rotation, and here this evidently is a rotation of 60⁰ anticlockwise and corresponds to the transformation r.

Indeed when we have followed the fate of just one vertex we must determine whether the result is a reflectional or a rotational transformation before assessing the final result as to what it is. Let's illustrate this with the example of the product r⁴a. First a, then r⁴. By the transformation a the vertex A is taken from position 1 to position 2. Then the transformation r⁴ takes it to position 6. So all in all we get a shift from position 1 to position 6. This looks like a rotation of 300⁰, i.e. r⁵, but, we know that the final result must be a reflection, because there has taken place one reflection, and not a second one to make the reflectional character undone. And thus the final result is a reflection in the axis f, and so we have
r⁴a = f.

In this way we can determine all 144 symmetry combinations resulting in the following group table for our dihedral group D₆ :

Table 7.10

Figure 15.  The hexagonal pyramid as the promorph of all the (single non-twinned) crystals of the Dihexagonal-pyramidal Class (6mm) of the Hexagonal Crystal System.

The group D₆ also describes the direct symmetries of the regular hexagonal bipyramid (and also of the regular hexagonal prism). When we now include the opposite symmetries as well, we get the full symmetry of the hexagonal bipyramid, according to the product group D₆ x C₂ of order 24 (in crystallography also known as D_6i ) (It is not the group D₁₂ :  If we have a dihedral group D_n representing the direct symmetries of an n-fold bipyramid, and if we then add the enantiomorphs (i.e. reflections), then we get the group D_n x C₂, and this group is only D_2n if  n  is odd (BUDDEN, p. 293)).  [Product groups are explained further down (Part IX and Part XII). As such the regular hexagonal bipyramid is the promorph or stereometric basic form of all the single non-twinned crystals of the Dihexagonal-bipyramidal Class (6/m 2/m 2/m) of the Hexagonal Crystal System.

Figure 16.  Regular Hexagonal Bipyramid as the promorph of all the crystals of the Dihexagonal-bipyramidal Class of the Hexagonal Crystal System.

The 24 symmetry transformations -- forming the group D₆ x C₂ -- of the regualar hexagonal bipyramid are the following :

1. Identity.   Period 1.

2. 60⁰ rotation about the main axis.    Period 6.

3. 120⁰ rotation about the main axis.   Period 3.

4. 180⁰ rotation about the main axis.   Period 2.

5. 240⁰ rotation about the main axis.   Period 3.

6. 300⁰ rotation about the main axis.   Period 6.

7. 180⁰ rotation about radial cross axis (See Figure 10)  b.   Period 2.

8. 180⁰ rotation about radial cross axis  d.   Period 2.

9. 180⁰ rotation about radial cross axis  g.   Period 2.

10. 180⁰ rotation about interradial cross axis  a.   Period 2.

11. 180⁰ rotation about interradial cross axis  c.   Period 2.

12. 180⁰ rotation about interradial cross axis  f.   Period 2.

13. Reflection across equatorial plane.   Period 2

14. Reflection across equatorial plane followed by rotation of 60⁰ about main axis.  Period 6.

15. Reflection across equatorial plane followed by rotation of 120⁰ about main axis.  Period 6.

16. Reflection across equatorial plane followed by rotation of 180⁰ about main axis.  Period 2.

17. Reflection across equatorial plane followed by rotation of 240⁰ about main axis.  Period 6.

18. Reflection across equatorial plane followed by rotation of 300⁰ about main axis.  Period 6.

19. Reflection in the (vertical) radial meridian plane containing cross axis  b.  Period 2.

20. Reflection in the (vertical) radial meridian plane containing cross axis  d.  Period 2.

21. Reflection in the (vertical) radial meridian plane containing cross axis  g.  Period 2.

22. Reflection in the (vertical) interradial meridian plane containing cross axis  a. Period 2.

23. Reflection in the (vertical) interradial meridian plane containing cross axis  c. Period 2.

24. Reflection in the (vertical) interradial meridian plane containing cross axis  f.  Period 2.

In the group table of the group D₆  (Table 7.10)  the order (arrangement) of the elements is chosen such that the cyclic subgroup H (C₆) = {1, r, r², r³, r⁴, r⁵} and its cosets stand out clearly. Let's check these cosets.

Left and right cosets of H with respect to the element 1 :
1{1, r, r², r³, r⁴, r⁵} = {1, r, r², r³, r⁴, r⁵}1 = {1, r, r², r³, r⁴, r⁵}.

Left and right cosets of H with respect to the element r :
r{1, r, r², r³, r⁴, r⁵} = {1, r, r², r³, r⁴, r⁵}r = (r1, rr, rr², rr³, rr⁴, rr⁵} = {r, r², r³, r⁴, r⁵, 1}.

Because the subgroup H is cyclic, multiplication with one of its elements does not lead us outside this subgroup, so all cosets of this subgroup with respect to elements of this subgroup are identical to the subgroup itself.

Left coset of H with respect to the element  a (This coset we can easily read off from the group table).
a{1, r, r², r³, r⁴, r⁵} = {a1, ar, ar², ar³, ar⁴, ar⁵} = {a, g, f, d, c, b}.

Right coset of H with respect to the element a.
{1, r, r², r³, r⁴, r⁵}a = {1a, ra, r²a, r³a, r⁴a, r⁵a} = {a, b, c, d, f, g}.

So aH = Ha.

In the same way we see that for every element  x  of the group (D₆) the following relation holds : xH = Hx. And this means that the subgroup H = {1, r, r², r³, r⁴, r⁵} is a normal subgroup in D₆. And this in turn means that this subgroup supports a homomorphism of the whole group to the group C₂, i.e. C₂ is the homomorphic image group of the group D₆ with respect to the subgroup H = {1, r, r², r³, r⁴, r⁵}.
Let's demonstrate this.
In the table we can see this homomorphism clearly : The red squares contain direct symmetries only, while the green squares contain opposite symmetries only. So the two elements of the homomorphic image group of D₆ could be DIRECT SYMMETRY, and OPPOSITE SYMMETRY, which we could indicate as DIRECT and OPPOSITE. Now we know that an opposite symmetry followed by another equivalent opposite symmetry results in a direct symmetry. For example, first a reflection in axis a, followed by a reflection in axis b  (which is a symmetry transformation that is equivalent to a reflection in  a) results in  r  (ba = r). And two direct symmetries -- two rotations -- being performed successively, result in a direct symmetry again. Further the successive application of a direct symmetry and an opposite symmetry always results in an opposite symmetry. So we have the group {DIRECT, OPPOSITE}, and its table looks like this :

Table 7.11

So the above table depicts the homomorphic image of the dihedral group D₆ with respect to the subgroup H = {1, r, r², r³, r⁴, r⁵}, and the structure of this image is that of the cyclic group of order 2,  C₂.

Table 7.12

Figure 17.   A Rhombic Bisphenoid, the Stereometric Basic Form of the crystals of the Rhombic-bisphenoidic Class of the Orthorhombic Crystal System.
The promorphological main axis is indicated by a red line.

The four symmetry transformations are in this case, the identity (1), and three half-turns (2_a, 2_b, 2_c) about three perpendicular, but nevertheless equivalent, axes, a, b and c. Two half-turns about the same axis results in the identity transformation, while two half-turns about to different axes result in a half-turn about the third axis :

Figure 18.  A half-turn about one axis, followed by a half-turn about the other axis, sends the motif to the same place as a half-turn about the third axis.

Figure 19.   A Rhombic Pyramid, the Stereometric Basic Form of the crystals of the Rhombic-pyramidal Class of the Orthorhombic Crystal System.
The base is a rhombus. The three crystallographic axes are given as reddish solid lines. The one that goes to the pyramid's tip is the promorphological main axis, the other two reddish solid lines are the two promorphological radial cross axes, while the dashed lines are the two interradial cross axes.

The group D₂ in its guise of the symmetries of the just mentioned Orthorhombic Crystal Class mm2, consists accordingly of four elements, the identity (1), two mirror planes m₁ and m₂, perpendicular to each other, and a half-turn about a 2-fold rotation axis coinciding with the line of intersection of the two mirror planes. This half-turn will be denoted 2.
The group table then looks like this :

Figure 20.   Slightly oblique top view of an Amphitect Gyroid Bipyramid, the Stereometric Basic Form of the crystals of the Prismatic Class of the Monoclinic Crystal System. The depicted bipyramid allows for four antimers to be recognized. The lower pyramid is not visible, but nevertheless present.

The group D₂ in its guise of representing the symmetries of the just mentioned Monoclinic Class 2/m, accordingly consists of the identity (1), a half-turn (r), a mirror plane (m) perpendicular to the axis of the half-turn, and a center of symmetry (i) (The latter is a reflection in a point, namely the mid-point of the 2-fold rotation axis). These symmetry transformations are depicted in the next Figure :

Figure 21.  The symmetry transformations in crystals of the Class 2/m.
The vertical blue line is the 2-fold rotation axis.
The center of symmetry (inversion center) is indicated by  i. The mirror plane goes through the center of inversion and is perpendicular to the 2-fold rotation axis.
The movement of the starting motif is indicated for the transformations 1 (identity), i (inversion in the point i), m (reflection) and r (half-turn).

ir = m
ri = m
im = r
mi = r
mr = i
rm = i
i² = 1
r² = 1
m² = 1

On the basis of this we can draw the group table of this crystallographic realization (2/m) of the group D₂ :

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