e-mail : 

Sequel to Group Theory

Table 9.1

Table 9.2

In order to turn this set of ordered pairs (i.e. all possible ordered pairs formed from the elements of the two groups) into a group, we must define a group operation for this new type of group, a group operation valid for all possible product groups. And with "valid" we mean of course that it always must turn the set into a group.
Well, on our set of ordered pairs we define the general operation * such that to every pair of pairs from our set of ordered pairs another pair, also from our set, is assigned (analogous to the fact that we, in ordinary multiplication assign to two numbers, say 5 and 7, a third number, namely 35, which we call the product of 5 and 7).
We will call a general element of the first group  x, while a general element of te second group wil be called  y. So let's define the operation * assigned to the above set of ordered pairs.

(x₁, y₁) * (x₂, y₂) = (x₁x₂, y₁y₂).

For example (now omitting * and using juxtaposition), we have :

jh = (1, r²) (-i, r) = (-i, 1) = c.
k² = (i, r²)² = (i, r²) (i, r²) = (-1, r) = g.

The resulting ordered pair P always consists of (1) a product of two elements of the first group, and that product is also is element of that first group, and (2) a product of two elements of the second group, and that product is also an element of that second group. So the resulting ordered pair P also consists of two members of which the first one comes from the first group, and the second one from the second group. THe product does not bring us outside our set of ordered pairs. So the set of ordered pairs as listed above is closed under the operation *.
Associativity, i.e.

[(x₁, y₁) * (x₂, y₂)] * (x₃, y₃)   =   (x₁, y₁) * [(x₂, y₂) * (x₃, y₃)]

is evident, because the definition of the operation * only contains ordinary multiplications which are themselves associative.
The identity element is (1, 1).
Finally each element of our new set has an inverse :
Let  x  be an element of the first group, that is the group {1, i, -1, -i} , and  y  be an element of the second group, that is the group {1, r, r²}, then  1 / x  is the inverse of  x  [ because the operation in that group is ordinary multiplication which implies that  x  times  1 / x  =  x / x = 1 ], and  1 / y  is the inverse of  y  for the same reason with respect to the second group. And because we have to do with groups, the inverses  1 / x  and  1 / y  belong to these respective groups, and thus the pair (1 / x, 1 / y) automatically belongs to the set of ordered pairs.
Now (x, y) (1 / x, 1 / y) = (x / x, y / y) = (1, 1) (which is indeed the identity element of our set of ordered pairs), so that the inverse of ( i, r²), for example, is (1 / i, 1 / r²) = (-i, r),
i.e. k^-1 = h.
Let's show that  1 / i  is indeed equal to -i.
1 / i  =  1/Sqr(-1)  =  Sqr(-1) / -1  =  -Sqr(-1)  =  -i.
Let's also show that 1 / r²  =  r.
1 / r²  =  r² / r⁴  =  r² / (r³r)  =  r² / r  =  r.
Thus the twelve elements form a group.
Note that in the array of elements (ordered pairs) as listed above, the top row is the subgroup C₄, and the first column is the subgroup C₃.

Now,
f = (i, r).
f ² = (-1, r²) = n.
f ³ = (-i, 1) = c.
f ⁴ = (1, r) = d.
f ⁵ = (i, r²) = k.
f ⁶ = (-1, 1) = b. And in the same way we can cintinue :
f ⁷ = h.
f ⁸ = j.
f ⁹ = a.
f ¹⁰ = g.
f ¹¹ = m.
f ¹² = e.

Hence element  f  is of period 12, it generates all the elements of the group, and thus the group must be C₁₂.
We are now ready to give a formal (and general) definition of the direct product of two groups :

(g_p, g_q') (g_r, g_s') = (g_p *  g_r  ,   g_q'  *'  g_s')

is a group of order  nn', and is called the Direct Product Group of G and G', and is abbreviated G x G'.

Sometimes instead of the term "Direct Product", the term "Direct Sum" is used.

Because  g_p  and  g_r  belong to the first group, their product  ( g_p* g_r )  also belongs to that group.
And because  g_q'  and  g_s'  belong to the second group, their product  ( g_q' *' g_s' )  also belongs to that group.
And thus the pair (g_p *  g_r  ,   g_q'  *'  g_s') belongs to the set of ordered pairs generally denoted by (g_p, g_q'), which means that in all cases the set of ordered pairs is closed under the defined operation.

In the example above, what we have done is to form the direct product of C₄ and C₃, and we have shown that the direct product group so obtained is in fact the familiar C₁₂.
If we were to rewrite the elements of C₄ and C₃ in terms of generators, thus C₄ { i⁰, i¹, i², i³ } and C₃ { r, r, r }, our ordered pairs would then appear as :

and in applying the product rule, for example,

jh = (i⁰, r²) (i³, r¹) = (i³, r³) = (i³, r⁰) = c, and
k² = (i¹, r²) (i¹, r²) = (i², r⁴) = (i², r¹) = g,

it is clear that we are adding the indices modulo 4 and modulo 3 respectively, so that, if we represent our ordered pairs by depicting only the indices, thus :

we may now write

jh = (0, 2)(3, 1) = ((0 + 3)mod 4, (2 + 1)mod 3) = (3, 0) = c, and
k² = (1, 2)(1, 2) = (1 + 1)mod 4, (2 + 2)mod 3) = (2, 1) = g,
and so on.

So the group C₄ x C₃ is seen in a new guise, as the set of oredered pairs (x, y) with  x  being an element of {0, 1, 2, 3}, and  y  being an element of {0, 1, 2}, combined according to the composition rule

(x₁, y₁) (x₂, y₂) = ((x₁ + x₂)mod 4, (y₁ + y₂)mod 3).

This adding of the corresponding members of ordered pairs also takes place in the case of vector addition in 2-dimensional space, albeit without applying a finite modulus. So then, with respect to the pairs (0, 2) and (3, 1) we get the vector addition

(0, 2)(3, 1) = (0, 2) + (3, 1) = ((0 + 3), (2 + 1)) = (3, 3), and
(3, 1)(2, 2) = (3, 1) + (2, 2) = ((3 + 2), (1 + 2)) = (5, 3).

But in the case being considered above, the x-coordinates have been reduced modulo 4, and the y-coordinates reduced modulo 3, and this implies working within a confined area, as the next Figure illustrates (after BUDDEN, F. J., 1978, The fascination of Groups, p. 256/7).

Figure 1.  Vector addition with modular arthmetic.
In non-modular arithmetic the vectors (3, 1) and (2, 2) would add to give the vector (5, 3), but in that modular arithmetic which is isomorphic with our product group C₄ x C₃, the two vectors add up to ((3 + 2)mod 4, (1 + 2)mod 3) = (1, 0).

h(x₁, y₁) = ((x₁)mod 4, (y₁)mod 3).
h(x₂, y₂) = ((x₂)mod 4, (y₂)mod 3).

h(x₁, y₁) + h(x₂, y₂) = ((x₁)mod 4, (y₁)mod 3) + ((x₂)mod 4, (y₂)mod 3) =
([(x₁)mod 4 + (x₂)mod 4]MOD 4, [(y₁)mod 3 + (y₂)mod 3]MOD 3) =
((x₁ + x₂)mod 4, (y₁ + y₂)mod 3).

h[(x₁, y₁) + (x₂, y₂)] = h[(x₁ + x₂, y₁ + y₂] =
((x₁ + x₂)mod4, (y₁ + y₂)mod 3).

So h[(x₁, y₁) + (x₂, y₂)] = h(x₁, y₁) + h(x₂, y₂), and thus  h  is a homomorphism from C_infinite x C_infinite  to  C₁₂, according to the definition of homomorphism.

Figure 2.  The cyclic permutations of the four symbols A, B, C, and D. They can be interpreted as the rotations of a square (regular tetragon).

Figure 3.  The cyclic permutations of three symbols P, Q, and R. They can be interpreted as the rotations of an equilateral triangle.

The permutations for A, B, C, D, and for P, Q, R, as established above, can now be combined such that the first part, consisting of the letters A, B, C, and D, corresponds to the first member of the mentioned pairs, while the second part, consisting of the letters P, Q, and R, corresponds to the second member of those pairs. The result is 12 permutations of seven symbols, A, B, C, D, P, Q, R. In the list below we will only list the lower part of each permutation, for example the permutation

Let's now list the 12 permutations accordingly :

A B C D     P Q R
D A B C     P Q R
C D A B     P Q R
B C D A     P Q R
A B C D     R P Q
D A B C     R P Q
C D A B     R P Q
B C D A     R P Q
A B C D     Q R P
D A B C     Q R P
C D A B     Q R P
B C D A     Q R P

When we look to the part of the permutations played by the letters A, B, C, D, we see that the order of the letters is always according to the alphabetical sequence A, B, C, D, if we read them, starting with A and when arrived at the right-hand end, continue with the left-hand end, reading to the right.
The effect of the second permutation of those four letters is that they are shifted one place to the right.
The effect of the third permutation of those four letters is that they are shifted two places to the right.
The effect of the fourth permutation of those four letters is that they are shifted three places to the right.
After this the original places of the letters are restored again (A is again at the first place).
Well, the first members of the above pairs  e = (0, 0), a = (1, 0), b = (2, 0), etc. must be added modulo 4, and thus they correspond exactly with the cyclic permutations of the four letters A, B, C, and D. So we can specify the respective positions of A, as 0 (this is the case in the permutation A B C D, which is the identity permutation), 1 (this is the case in the permutation D A B C), 2 (this is the case in the permutation C D A B), etc.
Precisely the same we can do with the permutations of P Q R. They correspond to the second member of every pair, i.e. the pairs e = (0, 0), a = 1, 0), b = 2, 0), etc.
Also there we see that the alphabetical sequence of the letters P, Q, and R is preserved. When we start reading with P and go to the right, till the end, and then continue at the left-hand end, always reading to the right, we will always encounter P Q R, in this (alphabetical) order.
And also here we see that
the effect of the second permutation of those three letters is a shift of one place to the right,
the effect of the third permutation of those three letter is a shift of two places to the right,
while a fourth permutation would restore the letters in their original places.
Well, the second member of the pairs e = (0,0), a = (1, 0), b = (2, 0), etc. must be added modulo 3, and so they correspond exactly with the just mentioned cyclic permutations of the three letters P, Q, and R.
So we can specify the respective positions of P as 0 (this is the case in the permutation P Q R (the identity permutation)), 1 (this is the case in the permutation R P Q ), and 2 (this is the case in the permutation Q R P).
And indeed, when we accordingly denote any particular permutation of the above list by an ordered pair specifying the position of the letters A and P, calling the positions in the cycles respectively 0, 1, 2, 3 and 0, 1, 2, then, for example  h, which is (3, 1), comes to correspond with the permutation B  C  D  A    R  P  Q, since A occurs in the final position (of 0 1 2 3) and P in the middle position (of 0 1 2).
In this way we have set up a full correspondence between the 12 permutations and the 12 ordered pairs.

We know how to write a permutation as one or more cycles, for example the permutation BCDARPQ means that

A becomes B
B becomes C
C becomes D
D becomes A

P becomes R
Q becomes P
R becomes Q

In the cycle notation we then write (ABCD)(PRQ), we thus have a 4-cycle and a 3-cycle.
What is the overall period of this set of two cycles?
After four shifts (A becomes B, B becomes C, etc.) we are back at A again, but then the second cycle has also performed four shifts, which implies that it effectively has performed one shift : after it has returned to P (P becomes R, R becomes Q, and Q becomes P) (three shifts), it goes one step further (a total of four shifts) : P becomes R.
After a second round of four shifts (totalling eight shifts) it is clear that the cycle (PRQ) still has not returned to P, because this only will happen after three, or a multiple of three shifts. And indeed a third round of four shifts (totalling 12 shifts) will not only bring the cycle (ABCD) back to its starting position, but also the cycle (PRQ). In (ABCD) this happened for the third time, while in (PRQ) this happened for the fourth time.
So in this way we can write down the above permutations as cycles and determine their overall period (from the periods of the individual cycles which are contained in such a permutation). This determination of the overall period from the periods of the individual cycles can now be stated as follows :

The period of a combination of cycles is equal to the Lowest Common Multiple (L.C.M) of the periods of the individual cycles.

We can, in the same way determine the periods of the ordered pairs e = (0, 0), a = (1, 0), b = (2, 0), etc. The first members of which must be added modulo 4, the second members must be added modulo 3.
The periods of the first members are :
0 = 0, Period(0) = 1
1 + 1 + 1 + 1 = 0, Period(1) = 4
2 + 2 = 0, Period(2) = 2
3 + 3 + 3 + 3 = 0, Period(3) = 4

The periods of the second members are :
0 = 0, Period(0) = 1
1 + 1 + 1 = 0, Period(1) = 3
2 + 2 + 2 = 0, Period(2) = 3

The period of an ordered pair is now equal to the L.C.M. of the periods of the individual members :

e = (0, 0),  (Period 1, Period 1), implying overall period of 1.
a = (1, 0),  (Period 4, Period 1), implying overall period of 4.
b = (2, 0),  (Period 2, Period 1), implying overall period of 2.
c = (3, 0),  (Period 4, Period 1), implying overall period of 4.
d = (0, 1),  (Period 1, Period 3), implying overall period of 3.
f = (1, 1),  (Period 4, Period 3), implying overall period of 12.
g = (2, 1),  (Period 2, Period 3), implying overall period of 6.
h = (3, 1),  (Period 4, Period 3), implying overall period of 12.
j = (0, 2),  (Period 1, Period 3), implying overall period of 3.
k = (1, 2),  (Period 4, Period 3), implying overall period of 12.
n = (2, 2),  (Period 2, Period 3), implying overall period of 6.
m = (3, 2),  (Period 4, Period 3), implying overall period of 12.

Now we will determine the periods of the 12 permutations by considering their cycle notation :

A B C D     P Q R  = (A) (B) (C) (D) (P) (Q) (R).   Period = 1
D A B C     P Q R  = (ADCB) (P) (Q) (R).   Period = 4
C D A B     P Q R  = (AC) (BD) (P) (Q) (R).   Period = 2
B C D A     P Q R  = (ABCD) (P) (Q) (R).   Period = 4
A B C D     R P Q  = (A) (B) (C) (D) (PRQ).   Period = 3
D A B C     R P Q  = (ADCB) (PRQ).   Period = 12
C D A B     R P Q  = (AC) (BD) (PRQ).   Period = 6
B C D A     R P Q  = (ABCD) (PRQ).  Period = 12
A B C D     Q R P  = (A) (B) (C) (D) (PQR).  Period = 3
D A B C     Q R P  = (ADCB) (PQR).  Period = 12
C D A B     Q R P  = (AC) (BD) (PQR).  Period = 6
B C D A     Q R P  = (ABCD)(PQR).  Period = 12

And now we can set up the correspondence between the 12 ordered pairs and the 12 permutations :

e --- ABCD PQR
a --- DABC PQR
b --- CDAB PQR
c --- BCDA PQR
d --- ABCD RPQ
f --- DABC RPQ
g --- CDAB RPQ
h --- BCDA RPQ
j --- ABCD QRP
k --- DABC QRP
n --- CDAB QRP
m --- BCDA QRP

Because there is at least one element with period 12 (and the group is of order 12) we can identify our direct product group as (isomorphic with) C₁₂.
Thus we can say that C₄ x C₃ is isomorphic with C₁₂.

A B C D E F  P Q  (A) (B) (C) (D) (E) (F) (P) (Q),  Period = 1
F A B C D E  P Q  (AFEDCB)  (P) (Q),  Period = 6
E F A B C D  P Q  (AEC) (BFD)  (P) (Q),  Period = 3
D E F A B C  P Q  (AD) (BE)  (CF)  (P) (Q),  Period = 2
C D E F A B  P Q  (ACE) (BDF)  (P) (Q),  Period = 3
B C D E F A  P Q  (ABCDEF)  (P) (Q),  Period = 6
A B C D E F  Q P  (A) (B) (C) (D) (E) (F)  (PQ),  Period = 2
F A B C D E  Q P  (AFEDCB)  (PQ),  Period = 6
E F A B C D  Q P  (AEC)  (BFD)  (PQ),  Period = 6
D E F A B C  Q P  (AD)  (BE)  (CF)  (PQ),  Period = 2
C D E F A B  Q P  (ACE)  (BDF)  (PQ),  Period = 6
B C D E F A  Q P  (ABCDEF)  (PQ),  Period = 6

We see that there is no element of period 12, so our new group is definitely not C₁₂. The underlying reason for this is, of course, that 2 is a factor of 6, so that when either of the twelve permutations has been performed six times, the letter P and Q are the right way round again, and the order of the whole set is restored. So no element can be of period greater than 6, and thus we do certainly not have to do with C₁₂. In fact, C₆ x C₂ (or C₂ x C₆) is a new group of order 12 which is not isomorphic with either C₁₂, or D₆ or Q₆ (This last one is the quaterion group of order 12). In D₆, which can be realized in the symmetries of the Regular Hexagon, there are 7 elements of period 2, while in our present group there are only 3 of them. In the group A₄, which is also of order 12, and which can be realized as the direct symmetries of the Regular Tetrahedron, has no elements of order 6, so our present group cannot be isomorphic with A₄.
Of the five groups of order 12, D₆, Q₆, and A₄ are all non-Abelian, whereas C₁₂ and C₆ x C₂ are Abelian.

For us it is important that C₆ x C₂ has a crystallographic realization, namely in crystals of the Hexagonal-bipyramidal Class (6/m) of the Hexagonal Crystal System. Such crystals have one mirror plane, coinciding with the equatorial plane. No other mirror planes are present. Perpendicular to the equatorial plane is a 6-fold rotation axis. In this realization the group C₆ x C₂ is called C_6i .
Indeed we can interpret the above 12 permutations as the symmetry transformations of the crystals of this Class (6/m). These symmetries are rotations about the main axis and reflection across the equatorial plane.
Let us first list and name those symmetry transformations :

1 = identity (0⁰ rotation).
r = rotation by 60⁰.
r² = rotation by 120⁰.
r³ = rotation by 180⁰.
r⁴ = rotation by 240⁰.
r⁵ = rotation by 300⁰.
a = reflection in equatorial plane.
ar = ra = reflection in equatorial plane followed by a rotation of 60⁰.
ar² = r²a = reflection in equatorial plane followed by a rotation of 120⁰.
ar³ = r³a = reflection in equatorial plane followed by a rotation of 180⁰.
ar⁴ = r⁴a = reflection in equatorial plane followed by a rotation of 240⁰.
ar⁵ = r⁵a = reflection in equatorial plane followed by a rotation of 300⁰.

We can now let these symmetries to be corresponded to the above twelve permutations :

A B C D E F  P Q =  1.   Period = 1
F A B C D E  P Q =   r.   Period = 6
E F A B C D  P Q =   r².   Period = 3
D E F A B C  P Q =   r³.   Period = 2
C D E F A B  P Q =   r⁴.   Period = 3
B C D E F A  P Q =   r⁵.   Period = 6
A B C D E F  Q P =   a.   Period = 2
F A B C D E  Q P =   ar = ra.   Period = 6
E F A B C D  Q P =   ar² = r²a.   Period = 6
D E F A B C  Q P =   ar³ = r³a.   Period = 2
C D E F A B  Q P =   ar⁴ = r⁴a.   Period = 6
B C D E F A  Q P =   ar⁵ = r⁵a.   Period = 6

We can now construct the group table for the group C₆ x C₂ with the notation and interpretation of its twelve elements geared to the description of the group of symmetries of crystals of the Class 6/m :

Table 9.3

As has been said, the group C₆ x C₂ is crystallographically realized in crystals of the Hexagonal-bipyramidal Class (6/m) of the Hexagonal Crystal System. The promorph or stereometric basic form of all the single non-twinned crystals of this Class (6/m) is the Regular Hexagonal Gyroid Bipyramid, and it belongs to the Isosigmostaura sextamera (Stauraxonia homopola sigmostaura) of the Promorphological System of Basic Forms. See next Figure.

Figure 4.   Regular Hexagonal (i.e. six-fold) Gyroid Bipyramid, Stereometric Basic Form of the (single) crystals of the Hexagonal-bipyramidal Class of the Hexagonal Crystal System. Because of the view direction the lower pyramid is not visible.

We will demonstrate that the direct product of groups is commutative as well as associative. We will derive these properties from the definition of the direct product. Let's give this definition again :

(g_p, g_q') (g_r, g_s') = (g_p *  g_r  ,   g_q'  *'  g_s')

is a group of order  nn', and is called the Direct Product Group of G and G', and is abbreviated G x G'.

When we reverse the order of the two (general) groups G and G' of the definition, the order within the ordered pairs are reversed. We get for G' x G :

(g_q', g_p) (g_s', g_r) = (g_q'  *'  g_s' ,  g_p  *  g_r).

To establish the isomorphy between G x G' and G' x G we first show that between the elements of G and G' there is a 1,1 correspondence (blue arrow) :

What we now have to do (in order to prove the isomorphism) is to show that products are conserved (which it self is a prof for homomorphy) :

Call the 1,1 mapping from G to G' the mapping  h. Then we have :

h [(g_p *  g_r  ,   g_q'  *'  g_s')  (g_a *  g_c  ,   g_b'  *'  g_d')] =
h [(g_p *  g_r ) *  (g_a *  g_c)  ,  (g_q'  *'  g_s') *' (g_b'  *'  g_d')]  =

((g_q'  *'  g_s') *' (g_b'  *'  g_d')  ,  (g_p *  g_r ) *  (g_a *  g_c))

In words :  The image of the product [i.e. (the image under  h  of) the product of two elements (both are ordered pairs) of the group G x G'] is a new ordered pair, and is calculated as follows :  first we determine the product, and this results in a pair of which the first member consists of the product (*) of the first members of the initial two ordered pairs to be multiplied, and of which the second member consists of the product (* ') of the second members of the initial two ordered pairs to be multiplied. Then, to obtain the image under  h, we reverse the members of the obtained ordered pair.

h [(g_p *  g_r  ,   g_q'  *'  g_s')]  =  (g_q'  *'  g_s'  ,   g_p *  g_r ).

h [(g_a *  g_c  ,   g_b'  *'  g_d')]  = (g_b'  *'  g_d'  ,  g_a *  g_c).

In words :  The image under  h  of the first element of G x G' is obtained by the reversal of the two members of that pair that constituted that first element. The image of that element is an element of G' x G.
The image under  h  of the second element of G x G' is obtained by the reversal of the two members of that pair that constituted that second element. The image of that element is also an element of G' x G.

h [(g_p *  g_r  ,   g_q'  *'  g_s')]  h [(g_a *  g_c  ,   g_b'  *'  g_d')]  =

(g_q'  *'  g_s'  ,   g_p *  g_r )   (g_b'  *'  g_d'  ,  g_a *  g_c)  =

((g_q'  *'  g_s')  *'  (g_b'  *'  g_d')  ,  (g_p *  g_r ) * ( g_a *  g_c)).

Thus we have demonstrated that

h [(g_p *  g_r  ,   g_q'  *'  g_s')  (g_a *  g_c  ,   g_b'  *'  g_d')] =
h [(g_p *  g_r  ,   g_q'  *'  g_s')]  h [(g_a *  g_c  ,   g_b'  *'  g_d')].

Or in words :  The image (under h ) of the product is equal to the product of the images, which means that products are preserved in the mapping  h  from G x G' to G' x G. So the mapping  h  is a homomorphism. And because it is a 1,1 homomorphism (bijective homomorphism) it is an isomorphism. It is an isomorphism between the groups G x G' and G' x G, so the commutativity of the direct product is demonstrated.

Thus we may write A x B x C without ambiguity, and, moreover, the letters may be permuted in any manner A x B x C is isomorphic with B x C x A, etc. For example,

C₂ x C₆ is isomorphic with C₆ x C₂, and since C₆ is isomorphic with C₃ x C₂ (because the L.C.M. of 3 and 2, i.e. elements of period 3 to be combined with elements of period 2, is 6), we may also write C₂ x C₆ is isomorphic with C₂ x (C₂ x C₃) is isomorphic with C₂ x C₂ x C₃ is isomorphic with C₃ x C₂ x C₂, etc.

We will shortly show that C₂ x C₂ is isomorphic with D₂, and then we can say :
C₂ x C₆ is isomorphic with D₂ x C₃.

This means that the group C₂ x C₆ may be realized by the product of two disjoint sets of permutations one being those permutations of four letters  A B C D, which give the group D₂, the other being the cyclic permutations of  P Q R.

D₂ can be represented by the permutations

1 = A B C D = (A)(B)(C)(D), implying this element to be of period 1.
a = B A D C = (AB) (CD), implying this element to be of period 2.
b = C D A B = (AC) (BD), implying this element to be of period 2.
ab (ba) = D C B A = (AD) (BC), implying this element to be of period 2.

and C₃ by the permutations

1 = P Q R = (P) (Q) (R), implying this element to be of period 1.
p = R P Q = (PRQ), implying this element to be of period 3.
p² = Q R P = (PQR), implying this element to be of period 3.

Hence the group D₂ x C₃ is the group of the set of twelve permutations obtained by combining these (i.e. by combining the just listed permutations), such as

pab  is :  first permutation  b, then permutation  a, and then permutation  p.
The permutation  b  is then

C D A B P Q R

While the permutation  a  is

B A D C P Q R

So the permutation  ab  is

D C B A P Q R

The permutation  p  is

A B C D R P Q

So the permutation  pab  is

D C B A R P Q.

The group C₂ is a cyclic group consisting of two elements, the identity (1), and one element of period 2 (let us call it  h, and thus hh = 1) (for example a rotation by 180⁰). The group can thus be denoted by {1, h}. So in order to form the product C₂ xC₂, we must form all the possible ordered pairs (i.e. the cartesian product) of the two sets {1, h} and {1, h}. These pairs are :

(1, 1), (1, h), (h, 1), (h, h).

We will now investigate how these pairs behave under multiplication according to the definition of a direct product.

(1, 1) (1, 1) = (1.1, 1.1) = (1, 1)
(1, 1) (1, h) = (1.1, 1.h) = (1, h)
(1, 1) (h, 1) = (1.h, 1.1) = (h, 1)
(1, 1) (h, h) = (1.h, 1.h) = (h, h)
(1, h) (1, 1) = (1.1, h.1) = (1, h)
(1, h) (1, h) = (1.1, h.h) = (1, 1)
(1, h) (h, 1) = (1.h, h.1) = (h, h)
(1, h) (h, h) = (1.h, h.h) = (h, 1)
(h, 1) (1, 1) = (h.1, 1.1) = (h, 1)
(h, 1) (1, h) = (h.1, 1.h) = (h, h)
(h, 1) (h, 1) = (h.h, 1.1) = (1, 1)
(h, 1) (h, h) = (h.h, 1.h) = (1, h)
(h, h) (1, 1) = (h.1, h.1) = (h, h)
(h, h) (1, h) = (h.1, h.h) = (h, 1)
(h, h) (h, 1) = (h.h, h.1) = (1, h)
(h, h) (h, h) = (h.h, h.h) = (1, 1)

We clearly see that the set {(1, 1), (1, h), (h, 1), (h, h)} is closed under the operation of the multiplication of ordered pairs according to the definition of the direct product.

In order to show that these four pairs, (1, 1), (1, h), (h, 1), (h, h), i.e. the elements of the group C₂ x C₂, behave according to the group D₂, we will rename these elements as repectively 1, X, Y, and Z. Thus :
(1, 1) = 1
(1, h) = X
(h, 1) = Y
(h, h) = Z

So we can translate the above multiplications accordingly :

(1, 1) (1, 1) = (1, 1) is equivalent to 1.1 = 1
(1, 1) (1, h) = (1, h) is eqquivalent to 1.X = X
(1, 1) (h, 1) = (h, 1) is equivalent to 1.Y = Y
(1, 1) (h, h) = (h, h) is equivalent to 1.Z = Z
(1, h) (1, 1) = (1, h) is equivalent to X.1 = X
(1, h) (1, h) = (1, 1) is equivalent to X.X = 1
(1, h) (h, 1) = (h, h) is equivalent to X.Y = Z
(1, h) (h, h) = (h, 1) is equivalent to X.Z = Y
(h, 1) (1, 1) = (h, 1) is equivalent to Y.1 = Y
(h, 1) (1, h) = (h, h) is equivalent to Y.X = Z
(h, 1) (h, 1) = (1, 1) is equivalent to Y.Y = 1
(h, 1) (h, h) = (1, h) is equivalent to Y.Z = X
(h, h) (1, 1) = (h, h) is equivalent to Z.1 = Z
(h, h) (1, h) = (h, 1) is equivalent to Z.X = Y
(h, h) (h, 1) = (1, h) is equivalent to Z.Y = X
(h, h) (h, h) = (1, 1) is equivalent to Z.Z = 1

We can now construct a group table of these products :

D₃ expresses the symmetry of the equilateral triangle, and at the same time the symmetry of the regular trigonal pyramid (i.e. a single pyramid with an equilateral triangle as its base). If we now add a second trigonal pyramid, congruent with the first one, such that the two pyramids are attached to each other by their bases, then we have the regular trigonal bipyramid. In fact we now have combined the D₃ symmetry with a 'flip-flop' symmetry, or, in other words, we have combined the elements of D₃ with a new element of period 2. And this combination yields six more elements (i.e. six more symmetry transformations).
Further down we will list all these elements as named permutations of the letters A, B, C, P, Q.
In the mentioned trigonal bipyramid
The upper apex of this pyramid is called P,
The lower apex is called Q,
The vertices at the triangular equatorial plane are called A, B, and C.
See next Figure.

Figure 5.  The Trigonal Bipyramid, its vertices named A, B, C, P, Q.
Equatorial plane indicated (green). The three cross axes lying in the equatorial plane are indicated (red).

• Identity.
• Rotation by 120⁰ about main axis of bipyramid.
• Rotation by 240⁰ about main axis of bipyramid.
• Reflection across the (vertical) meridian plane through vertex A.
• Reflection across the (vertical) meridian plane through vertex B.
• Reflection across the (vertical) meridian plane through vertex C.
• Reflection across the equatorial plane (The apices P and Q of the bipyramid are interchanged.
• Rotation by 120⁰ about main axis of bipyramid followed by a reflection across the equatorial plane.
• Rotation by 240⁰ about main axis of bipyramid followed by a reflection across the equatorial plane.
• Reflection across the (vertical) meridian plane through vertex A followed by a reflection in the equatorial plane. This whole operation is equivalent to a half-turn about the cross axis through vertex A. This cross axis is horizontal and lies in the equatorial plane.
• Reflection across the (vertical) meridian plane through vertex B followed by a reflection in the equatorial plane. This whole operation is equivalent to a half-turn about the cross axis through vertex B.
• Reflection across the (vertical) meridian plane through vertex C followed by a reflection in the equatorial plane. This whole operation is equivalent to a half-turn about the cross axis through vertex C.

1

A

B

C

P

Q

Identity

d

A

B

C

Q

P

eq. refl.

p

C

A

B

P

Q

rot. 1200

dp

C

A

B

Q

P

1200+eq.refl.

p2

B

C

A

P

Q

rot. 2400

dp2

B

C

A

Q

P

2400+eq.refl.

a

A

C

B

P

Q

refl. in a

da

A

C

B

Q

P

refl. in a + eq.refl.

b

C

B

A

P

Q

refl. in b

db

C

B

A

Q

P

refl. in b + eq.refl.

c

B

A

C

P

Q

refl. in c

dc

B

A

C

Q

P

refl. in c + eq.refl.

We can now determine the periods of these 12 elements :

ABC PQ =  (A) (B) (C) (P) (Q).  Period = 1
CAB PQ =  (ACB) (P) (Q).  Period = 3
BCA PQ =  (ABC) (P) (Q).  Period = 3
ACB PQ =  (A) (BC) (P) (Q).  Period = 2
CBA PQ =  (AC) (B) (P) (Q).  Period = 2
BAC PQ =  (AB) (C) (P) (Q).  Period = 2
ABC QP =  (A) (B) (C) (PQ).  Period = 2
CAB QP =  (ACB) (PQ).  Period = 6
BCA QP =  (ABC) (PQ).  Period = 6
ACB QP =  (A) (BC) (PQ).  Period = 2
CBA QP =  (AC) (B) (PQ).  Period = 2
BAC QP =  (AB) (C) (PQ).  Period = 2

The distribution of periods among the 12 elements is the finger print of the group D₆, thus we have demonstrated that D₃ x C₂ is isomorphic with D₆. So the full symmetry of the trigonal bipyramid and also that of the regular hexagon as well as that of the regular hexagonal pyramid (single pyramid), is according to the group D₆, which is isomorphic with D₃ x C₂.

e-mail : 

back to homepage

back to retrospect and continuation page

back to Part I of the theoretical preparation to the study of 3-D crystals

back to Part II of the theoretical preparation to the study of 3-D crystals

back to Part III of the theoretical preparation to the study of 3-D crystals

back to Part IV of the theoretical preparation to the study of 3-D crystals

back to Part V of the theoretical preparation to the study of 3-D crystals

back to Part VI of the theoretical preparation to the study of 3-D crystals

back to Part VII of the theoretical preparation to the study of 3-D crystals

back to Part VIII of the theoretical preparation to the study of 3-D crystals