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The Total Symmetry of Three-dimensional Crystals

Part V
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e-mail :

Sequel to Group Theory

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We'll start with reminding the reader about the "Important Remark" near the end of Part III, a Remark concerning the direction of reading products of group elements, like, say, apq. We read such products (from that Remark onwards) from back to front. Thus (with respect to apq) first  q, then  p, and then  a.
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Cyclic Groups of composite order

When we come to groups of composite order -- and in the present document first of all to cyclic groups of composite order -- things are getting much more interesting. This is because groups of composite order -- in contrast to groups of prime order -- have SUBGROUPS. And many features of groups are directly associated with subgroups (The cyclic groups C4 and C6, both of composite order, have already been considered (previous Part) with respect to their crystallographic realizations).

Let us start with a rather large cyclic group of order twelve (C12), and discover its features. C12 can be realized in several ways. If we let it be realized as the numbers  0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11  under addition modulo 12, and if we arrange those numbers according to their natural sequence (as we had just presented them, namely 0, 1, 2, 3, etc.) then the following table will emerge (Recall that addition modulo 12 is just addition, but only retaining the remainders after division by 12) :

 + mod 12 0 1 2 3 4 5 6 7 8 9 10 11 period 0 0 1 2 3 4 5 6 7 8 9 10 11 1 1 1 2 3 4 5 6 7 8 9 10 11 0 12 2 2 3 4 5 6 7 8 9 10 11 0 1 6 3 3 4 5 6 7 8 9 10 11 0 1 2 4 4 4 5 6 7 8 9 10 11 0 1 2 3 3 5 5 6 7 8 9 10 11 0 1 2 3 4 12 6 6 7 8 9 10 11 0 1 2 3 4 5 2 7 7 8 9 10 11 0 1 2 3 4 5 6 12 8 8 9 10 11 0 1 2 3 4 5 6 7 3 9 9 10 11 0 1 2 3 4 5 6 7 8 4 10 10 11 0 1 2 3 4 5 6 7 8 9 6 11 11 0 1 2 3 4 5 6 7 8 9 10 12

Table 5.1

Table of C12   (version a).
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We can generate the whole group with the element  1, namely by repeated addition modulo 12, i.e. by repeated application of the element  1 :
12 = 1 + 1   (mod 12) gives the element 2.
13 = 1 + 1 + 1   (mod 12) gives the element  3.
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112 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12 mod 12 = 0.

This is because the element  1 has period  12.
Also each other element of period  12 will generate the whole group by repeated application.

But if we try to generate the whole group with the element  6, which has period 2, then we get by repeated application of  6 :
62 = 6 + 6 = 12 mod 12 = 0.
63 = 6 + 6 + 6 = 18 mod 12 = 6.
So only the elements  0  and  6  are generated. And this means that the set {0, 6} is a subgroup of our group {0, 1, 2, 2, 4, 5, 6, 7, 8, 9, 10, 11}, + mod 12  ( = C12). now, as we've said, with subgroups are connected some important features. In order to show these features we will rearrange the group elements in such a way that the subgroup {0, 6} stands out clearly, resulting in the next group table version of C12 :

 + mod 12 0 6 1 7 2 8 3 9 4 10 5 11 period 0 0 6 1 7 2 8 3 9 4 10 5 11 1 6 6 0 7 1 8 2 9 3 10 4 11 5 2 1 1 7 2 8 3 9 4 10 5 11 6 0 12 7 7 1 8 2 9 3 10 4 11 5 0 6 12 2 2 8 3 9 4 10 5 11 6 0 7 1 6 8 8 2 9 3 10 4 11 5 0 6 1 7 3 3 3 9 4 10 5 11 6 0 7 1 8 2 4 9 9 3 10 4 11 5 0 6 1 7 2 8 4 4 4 10 5 11 6 0 7 1 8 2 9 3 3 10 10 4 11 5 0 6 1 7 2 8 3 9 6 5 5 11 6 0 7 1 8 2 9 3 10 4 12 11 11 5 0 6 1 7 2 8 3 9 4 10 12

Table 5.2

Table of C12   (version b).
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In the above table (version b) we can see that the squares containing the elements  1  and  6  (i.e. the elements of the subgroup {0, 6}) are patterned such that they remind us of the pattern of identity elements in cyclic groups where, in the group table, the elements are ordered according to increasing numerical value, as we had it in version a of C12 above.

The next group table of again C12 has the same arrangement of elements as in version b, but now we highlight the different SUBSETS as we see them appear in the table, including the subset that is the subgroup {0, 6}. Every subset consists of two different group elements (two 'individuals' of each sort), for example

2 8
8 2

which should be written as the set {2, 8}.

 + mod 12 0 6 1 7 2 8 3 9 4 10 5 11 period 0 0 6 1 7 2 8 3 9 4 10 5 11 1 6 6 0 7 1 8 2 9 3 10 4 11 5 2 1 1 7 2 8 3 9 4 10 5 11 6 0 12 7 7 1 8 2 9 3 10 4 11 5 0 6 12 2 2 8 3 9 4 10 5 11 6 0 7 1 6 8 8 2 9 3 10 4 11 5 0 6 1 7 3 3 3 9 4 10 5 11 6 0 7 1 8 2 4 9 9 3 10 4 11 5 0 6 1 7 2 8 4 4 4 10 5 11 6 0 7 1 8 2 9 3 3 10 10 4 11 5 0 6 1 7 2 8 3 9 6 5 5 11 6 0 7 1 8 2 9 3 10 4 12 11 11 5 0 6 1 7 2 8 3 9 4 10 12

Table 5.3

Table of C12   (version b).
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Each colored square (containing four symbols) of the above group table represents a subset (not necessarily a subgroup!), each containing two elements :
{0, 6},   {1, 7},   {2, 8},   {3, 9},   {4, 10},   {5, 11}.
These sets are called COSETS. Especially they are cosets of the subgroup {0, 6}.
Let me explain.
In doing so I reproduce again the table of version  b with adapted coloring. The subgroup {0, 6} is highligted :

 + mod 12 0 6 1 7 2 8 3 9 4 10 5 11 period 0 0 6 1 7 2 8 3 9 4 10 5 11 1 6 6 0 7 1 8 2 9 3 10 4 11 5 2 1 1 7 2 8 3 9 4 10 5 11 6 0 12 7 7 1 8 2 9 3 10 4 11 5 0 6 12 2 2 8 3 9 4 10 5 11 6 0 7 1 6 8 8 2 9 3 10 4 11 5 0 6 1 7 3 3 3 9 4 10 5 11 6 0 7 1 8 2 4 9 9 3 10 4 11 5 0 6 1 7 2 8 4 4 4 10 5 11 6 0 7 1 8 2 9 3 3 10 10 4 11 5 0 6 1 7 2 8 3 9 6 5 5 11 6 0 7 1 8 2 9 3 10 4 12 11 11 5 0 6 1 7 2 8 3 9 4 10 12

Table 5.4

Table of C12   (version b).
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Cosets of a subgroup
If we take the SUBGROUP {0, 6}, and multiply (which here, in our present group means addition modulo 12) this whole group -- i.e. its set of elements -- with some element  x  of the group (in our case some element  x  of our group {0, 1, 2, ..., 11}, + mod 12  ( = C12)), then we can do this in two ways :
x{0, 6}  or  {0, 6}x. We then get respectively {x + 0 (mod 12), x + 6 (mod 12)}, which is a set consisting of two elements, and which we call the left coset of the subgroup {0, 6} by the element  x,  and (we get) {0 + x (mod 12), 6 + x (mod 12)}, which is also a set consisting of two elements, and which we call the right coset of the subgroup {0, 6} by the element  x. If we indicate the subgroup {0, 6} by the symbol K, then its left coset by the element  x  will be written down as  xH, while its right coset by the element  x  will be written down as  Hx. And when Hx = xH for all elements of the group, let us call it G (thus not only for the element  x), then the subgroup H is called a normal subgroup in G. The latter is an extremely important concept in Group Theory. We will return to it further down.

So when, in our present example, we 'multiply' the subgroup {0, 6} with the element  1 of our group, then we either get 1{0, 6} (left coset by the element 1), or {0, 6}1  (right coset by the element 1). Let us work out these multiplications (which are here, remember, addition modulo 12).
1{0, 6} = {1 + 0 (mod 12) = 1, 1 + 6 (mod 12) = 7}, which is the set {1, 7} and which is the left coset of {0, 6} by the element  1. This set is indicated in light blue in the above group table.
{0, 6}1 = (0 + 1 (mod 12) = 1, 6 + 1 (mod 12) = 7, which is the set {1, 7} and which is the right coset of {0, 6} by the element  1. This set is indicated green in the above group table.
So the left and right cosets of the subgroup {0, 6} by the element  1 are equal : 1{0, 6} = {0, 6}1, or, when we indicate the subgroup {0, 6} by H, we get 1H = H1.

From that table we can verify that the left and right cosets of our subgroup H ( = {0, 6}) by the element  7  are both {7, 1}, i.e. 7H = H7 = {7, 1}. And these cosets are equal to 1H and to H1 [Recall that for sets to be equal, the arrangements of their elements does not matter, thus the set {1, 7} is equal to the set {7, 1}].
So 1H = H1 = 7H = H7.

The left coset of H by the element  2, i.e. the set 2H, is 2{0, 6}, which is 2 + 0 (mod 12) = 2, 2 + 6 (mod 12) = 8} which is the set {2, 8}, brown in the above table.
The right coset of H by the element  2, i.e. the set H2, is {0, 6}2, which is {0 + 2 (mod 12) = 2, 6 + 2 (mod 12) = 8} which is the set {2, 8}, dark green in the above table.
So 2H = H2 = {2, 8}.

The left and right cosets of H by the element  8  both are also {8, 2} ( = {2, 8}), so 2H = H2 = 8H = H8 = {2, 8}.

In the same way we can see that
3H = H3 = 9H = H9 = {3, 9}.
4H = H4 = 10H = H10 = {4, 10}.
5H = H5 = 11H = H11 = {5, 11}.
0H = H0 = 6H = H6 = {0, 6}.

We see that for every element  x  of our group G  xH = Hx, which means that H is a normal subgroup in  G.

Homomorphic image of a group
In Table 5.3 we see that these cosets are repeated throughout the group table.
In fact we see that these cosets are multiplied with each other (according to the group operation which is addition modulo 12), as is illustrated in the next Table (still of our group G = {0, 1, 2, 3, ..., 11}, + mod 12,   C12).

 + mod 12 0 6 1 7 2 8 3 9 4 10 5 11 period 0 0 6 1 7 2 8 3 9 4 10 5 11 1 6 6 0 7 1 8 2 9 3 10 4 11 5 2 1 1 7 2 8 3 9 4 10 5 11 6 0 12 7 7 1 8 2 9 3 10 4 11 5 0 6 12 2 2 8 3 9 4 10 5 11 6 0 7 1 6 8 8 2 9 3 10 4 11 5 0 6 1 7 3 3 3 9 4 10 5 11 6 0 7 1 8 2 4 9 9 3 10 4 11 5 0 6 1 7 2 8 4 4 4 10 5 11 6 0 7 1 8 2 9 3 3 10 10 4 11 5 0 6 1 7 2 8 3 9 6 5 5 11 6 0 7 1 8 2 9 3 10 4 12 11 11 5 0 6 1 7 2 8 3 9 4 10 12

Table 5.5

Table of C12   (version b).
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In the above Table we see that multiplication of the sets {2, 8} and {3, 9} results in the set {5, 11}. Let us check this :
{2, 8}*{3, 9} = {2 + 3 (mod 12), 2 + 9 (mod 12), 8 + 3 (mod 12), 8 + 9 (mod 12)} = {5, 11, 11, 5} = {5, 11}.

In Table 5.3 we see all the possible products of the cosets, and also see that multiplication of these cosets does not bring us outside the set of cosets (of the subgroup H = {0, 6}):

{0, 6} = 0H = H0 = 6H = H6
{1, 7} = 1H = H1 = 7H = H7
{2, 8} = 2H = H2 = 8H = H8
{3, 9} = 3H = H3 = 9H = H9
{4, 10} = 4H = H4 = 10H = H10
{5, 11} = 5H = H5 = 11H = H11

So multiplication of the cosets of the subgroup H of our group G is closed.
There is also a coset that functions as an identity element, namely the coset 0H = H0 = H = {0, 6}, i.e. the left (or right) coset of our subgroup by the element  0, which is the subgroup {0, 6} itself. See Table 5.2 where we can see it be distributed in the same way as we see it in cyclic groups. Indeed we see that multiplication of {0, 6} with any coset X results in coset X again. And in that same Table 5.2 we can see that for every coset X there is a coset X-1, such that XX-1 = H. So in our set of cosets for every coset there is an inverse coset. The multiplication of these cosets is associative, which means that (XY)Z = X(YZ) for every three cosets in our set of cosets.
All this implies that the set of cosets of the subgroup H of our group G = {0, 1, 2, 3, ..., 11}, + mod 12,  is itself a group.

If we look to the cosets of H -- which include H itself, which is the left or right coset by the element  0, or by the element  6 -- we see that in all cases the difference between the two elements is 6, so we could represent these cosets by, say, the smallest element, because then the other element is already determined. So we can rename these cosets as follows :

{0, 6} --   0
{1, 7} --   1
{2, 8} --   2
{3, 9} --   3
{4, 10} -- 4
{5, 11} -- 5

In fact this is a mapping of the cosets onto the set {0, 1, 2, 3, 4, 5}, which we will call B. Said differently :
The elements 0 and 6 of group G are mapped onto element 0 of B.
The elements 1 and 7 of group G are mapped onto element 1 of B.
The elements 2 and 8 of group G are mapped onto element 2 of B.
The elements 3 and 9 of group G are mapped onto element 3 of B.
The elements 4 and 10 of group G are mapped onto element 4 of B.
The elements 5 and 11 of group G are mapped onto element 5 of B.

And if we now replace all the cosets in our above tables by these numbers, i.e. if we actually carry out the above mapping of group G onto the set B, we get the following table, which is -- as we found out above -- a group table :

 + mod 6 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4

Table 5.6

Table of C6
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As the cosets formed a group, so do their images by the above mapping. The new group is -- as it turns out -- a cyclic group of order 6, i.e. C6, which is realized in the above table as the set {0, 1, 2, 3, 4, 5} under addition modulo 6, so it is the group B = {0, 1, 2, 3, 4, 5, }, + mod 6.
This group B we call the homomorphic image of group G by the subgroup H. This subgroup itself is mapped onto the identity of the image group, namely 0 (as element of H = {0, 6}, and because of this we call the subgroup H = {0, 6} the kernel of the present homomorphism.
This homomorphism we can call  h, and it was defined by :

{0, 6} --   0
{1, 7} --   1
{2, 8} --   2
{3, 9} --   3
{4, 10} -- 4
{5, 11} -- 5

Next we summarize this homomorphic mapping of G onto B :

 + mod 12 0 6 1 7 2 8 3 9 4 10 5 11 0 0 6 1 7 2 8 3 9 4 10 5 11 6 6 0 7 1 8 2 9 3 10 4 11 5 1 1 7 2 8 3 9 4 10 5 11 6 0 7 7 1 8 2 9 3 10 4 11 5 0 6 2 2 8 3 9 4 10 5 11 6 0 7 1 8 8 2 9 3 10 4 11 5 0 6 1 7 3 3 9 4 10 5 11 6 0 7 1 8 2 9 9 3 10 4 11 5 0 6 1 7 2 8 4 4 10 5 11 6 0 7 1 8 2 9 3 10 10 4 11 5 0 6 1 7 2 8 3 9 5 5 11 6 0 7 1 8 2 9 3 10 4 11 11 5 0 6 1 7 2 8 3 9 4 10
 + mod 6 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4
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While an isomorphism between two groups G1 and G2 means that the elements of the one group stand in a 1,1 correspondence to each other, and products are preserved,  a homomorphism between two groups implies a many-to-one correspondence between the elements of the original group and those of the image group and products are preserved.
Our above homomorphism consisted of correspondences like {0, 6} -- 0,  {1, 7} -- 1, etc., i.e. two-to-one correspondences. Products being preserved means that the image (under either a homomorphism or an isomorphism) of the product xy is equal to the product of the images of x and y (images, also under that same homomorphism or isomorphism).
So if the two groups G1 and G2 are connected to each other by the homomorphism  h, then for any two elements  x  and  y  of G1 the following relation (expressing the preservation of products) holds :

h(xy) = h(x) h(y)

This in turn means that to the product  xy  in G1 corresponds the product  h(x)h(y) in G2.

In the same way, when two groups G1 and G2 are connected by the isomorphism  i, then for any two elements  x  and  y  of G1 the following relation (expressing the preservation of products) holds :

i(xy) = i(x) i(y)

Also here the product  xy  in G1 corresponds to the product i(x)i(y) in G2.

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In the next document we will continue our study of cyclic groups of composite order, for the time being elaborating still further the group C12, especially the other possible subgroups, their cosets, and the corresponding homomorphic mappings.
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