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Sequel to Group Theory

Definition
n! ("n-faculty") is a number that is equal to  1  x  2  x  3  x  4  x ... x  n-1  x  n.

For example 4! = 1 x 2 x 3 x 4 = 24.

An important theorem is Cayley's Theorem, which we will give without proof :

Any finite group of order  n  (in which  n  is accordingly finite) is isomorphic to a subgroup of the group S_n of permutations of  n  objects.

This group S_n (or P_n) is the full group of all  n!  permutations of  n  objects (symbols, letters, numbers, etc.). The order of the group S_n is therefore  n!, and it is called the Symmetric Group, so that the full group of permutations of four letters, is denoted S₄ or P₄, and is of order 24.

In the foregoing we have already referred to cyclic groups, for example to C₁₂, inspired by a clock. The name "cyclic" derives from the fact that such groups appear whenever we have a set of cyclic permutations of a number of objects.
If we have for example the cycle p = (01234), meaning

0 becomes 1,
1 becomes 2,
2 becomes 3,
3 becomes 4, and
4 becomes 0 again,

we can derive three other cycles from it by following (01234) but taking only every second number :

0 becomes 2,
2 becomes 4,
4 becomes 1,
1 becomes 3, and
3 becomes 0 again.

So we have the cycle (02413), which is in fact an application of the cycle  two times, i.e.  p².
In the same way we can derive the cycles (03142) = p³ and (04321) = p⁴. If we, starting from the first cycle p = (01234), and pick out every fifth number, then we get

0 becomes 0,
1 becomes 1,
2 becomes 2,
3 becomes 3,
4 becomes 4.

And this is of course the identity.

These five cycles can be expressed as the following five cyclic permutations :

These five permutations can also be written down as follows :

For  1  (identity) we could write  p₀, and for  p  we could write  p₁.
In this way we can see that all the elements of this group (C₅) can be expressed in terms of a single element  p, namely p⁰, p¹, p², p³, p⁴, i.e. all the possible distinct 'powers' of p. This is true of all cyclic groups :

The cyclic group of order  n  (denoted C_n) has elements {1, x, x², x³, x⁴, ..., x^n-1} with  x satisfying the relation xⁿ = 1.
(x^m  is not  1  when  m  is equal to or greater than  1, but smaller than  n).

Cyclic groups may be divided into two kinds :  those for which the order is prime, and those for which it is composite.

A prime number is an integer that can only be divided by 1 and by itself (like 1, 2, 3, 5, 7, 11, etc). All other integers are called composite numbers (like 4, 6, 8, 9, 10, 12, etc.

Cyclic groups of which the order is prime, like the above group, C₅, are in most repects less interesting, chiefly because they have no subgroups.

While the group C₅ does not occur in crystals, other cyclic groups do, for instance the Cyclic Group C₆ (which is, consequently, a group of composite order), of which the abstract table (i.e. a table without its elements being interpreted as permutations, numbers, symmetry transformations or whatever) is :

This abstract table (abstract group) can show many realizations, in terms of permutations, numbers (for instance the complex sixth roots of unity), mechanical devices, or whatever.

The possible realizations of an abstract group should not be confused with the possible automorphisms of the group. The latter are just rearrangements of the symbols of the elements while retaining the same table structure.

We will concentrate on the interpretation (realization) of the abstract group C₆ in terms of symmetry transformations.
C₆ is, among others, realized in the promorph or stereometric basic form of crystals of the Class 6, i.e. crystals possessing a 6-fold rotation axis as their only symmetry element (concerning its point symmetry, i.e. its non-translational symmetry). This symmetry element contains six possible rotations. See Figure 1.

Figure 1.  The Regular Hexagonal Gyroid Pyramid as the Stereometric Basic Form (belonging to the Homogyrostaura hexamera, [Stauraxonia heteropola gyrostaura]) of crystals belonging to the Hexagonal-pyramidal Class (Class 6) of the Hexagonal Crystal System. Its symmetry content consists of a 6-fold rotation axis.

Realized as the symmetries of the mentioned gyroid pyramid, we can now interpret the individual elements of the group C₆ :

1  is the identity element of the group, i.e. a rotation about the main axis of the gyroid pyramid by 0⁰, or, equivalently, 360⁰ or a multiple thereof. Its period is 1, i.e. 1¹ = 1 = identity.

p  is an (by convention) anticlockwise rotation about the main axis by 60⁰. Its period is 6, i.e. p⁶ = 6 x 60 = 360⁰ = 1 = identity.

p²  is an anticlockwise rotation of 120⁰, i.e. it is two times the (transformation)  p. Its period is 3, i.e. (p²)³ = 3 x 120 = 360⁰ = 1 = identity.

p³  is an anticlockwise rotation of 180⁰, i.e. it is three times the (transformation)  p. Its period is 2, i.e. (p³)² = 2 x 180 = 360⁰ = 1 = identity.

p⁴  is an anticlockwise rotation of 240⁰ (or a clockwise rotation of 120⁰, i.e. the inverse of  p²), i.e. it is four times the (transformation)  p. Its period is 3, i.e. (p⁴)³ = 3 x 240 = 720 = 1 = identity.

p⁵  is an anticlockwise rotation by 300⁰ (or a clockwise rotation of 60⁰, i.e. the inverse of  p, i.e. five times the (transformation)  p. Its period is 6, i.e. (p⁵)⁶ = 6 x 300 = 1800 = 5 x 360⁰ = 1 = identity.

Figure 2.  Regular Trigonal Gyroid Pyramid, as the Stereometric Basic Form of Crystals of the Trigonal-pyramidal Class of the Hexagonal Crystal System. This specific basic form belongs to the ( promorphological category of the ) Homogyrostaura trimera (Stauraxonia heteropola gyrostaura). Its symmetry content consists of one 3-fold rotation axis.

The elements in this table are interpreted as follows :
1  is the identity, namely a rotation of 0⁰ about the gyroid pyramid's main axis.
p  is an anticlockwise rotation of 120⁰ about the mentioned axis.
p²  is an anticlockwise rotation of 240⁰. It is equivalent to a clockwise rotation of 120⁰.

Although in handling Group Theory only insofar as we need it for the description of crystal symmetries, it is instructive to realize that groups can have many realizations. So our cyclic group C₃ can be realized in finite arithmetics as a table of multiplication modulo 14
(x mod 14) :

The multiplication modulo 14, is like ordinary multiplication but in the results only retaining the remainders after division by 14. It is commutative because ordinary multiplication is commutative. So 2 x 8 (mod 14) = 8 x 2 (mod 14) = 16 = 16 - 14 = 2. Further we see that 4 x 8 (mod 14) = 4, and 8 x 8 (mod 14) = 8, and this means that 8 is the identity element. We see that the table has the same structure as the previous table.

Figure 3.  The Regular Tetragonal Gyroid Pyramid as the general promorph of crystals of the Class 4, as such belonging to the (promorphological category of the) Homogyrostaura tetramera (Stauraxonia heteropola).

With respect to the symmetries of crystals we can interpret the elements of the table just given as follows :

1  is the identity element. It is a rotation of 0⁰, or, equivalently, of 360⁰ or a multiple thereof.
p  is an anticlockwise rotation of 90⁰ (about the main axis of the gyroid pyramid). Its period is 4, because p⁴ = 90 + 90 + 90 + 90 = 360⁰ = 0⁰ = identity = 1.
p²  is an anticlockwise rotation of 180⁰. Its period is 2, because (p²)² = 180 + 180 = 360⁰ = 0⁰ = identity = 1.
p³  is an anticlockwise rotation of 270⁰. It is equivalent to a clockwise rotation of 90⁰. Its period is 4, because (p³)⁴ = 270 + 270 + 270 + 270 = 1080 = 360 + 360 + 360 = 0 + 0 + 0 = 0 = identity = 1.

In the Crystallographic literature the asterisk is replaced by a horizontal score above the relevant numeral (indicating a rotation axis). So  1*  indicates a rotation of 0⁰ about the axis  1  immediately followed by an inversion across a point on that axis. Analogously  3*  indicates an anticlockwise rotation of 120⁰ about an axis labelled 3  immediately follwed by an inversion across a point on that axis.
See for "inversion across a point" the first Part of this website -- accessible by  back to homepage -- and there in the document "The Morphology of Crystals".

So we see the Cyclic Group C₂ crystallographically realized in the Classes  2, m, and  1*. And we see that the abstract group (C₂) crosses the boundary between two different Crystal Systems.

The group table of the group C₂ is as follows :

Here the element  1 is the identity element (Its period is, like all identity elements, 1).
The element p  has period  2, because  p² = 1. This element can be crystallographically realized either as a rotation of 180⁰, and then we have to do with the Sphenoidic Crystal Class of the Monoclinic System, or as a reflection across a mirror plane, and then we have to do with the Domatic Crystal Class of the Monoclinic System, or a point inversion, and then we have to do with the Pinacoidal Class of the Triclinic System.

The geometric solid generally representing geometrically the symmetry of crystals of the Sphenoidic Class (i.e. Class  2) of the Monoclinic Crystal System is the Amphitect Gyroid Pyramid. It is the promorph or stereometyric basic form of all the crystals of this Class and as such belongs to the (promorphological category of the) Heterogyrostaura tetramera (Stauraxonia heteropola gyrostaura). See next Figure.

Figure 4.  The Amphitect Gyroid Pyramid as the promorph of all the crystals of the Sphenoidic Class of the Monoclinic Crystal System. The tip of the pyramid is pointing to the beholder. Its only symmetry element (not to be confused with group element) is a 2-fold rotation axis going from the pyramid's tip to its base.

The geometric solid generally representing geometrically the symmetry of crystals of the Domatic Class (i.e. Class  m) of the Monoclinic Crystal System is Half a Rhombic Pyramid. It is the promorph or stereometyric basic form of all the crystals of this Class and as such belongs to the (promorphological category of the) Zygopleura eudipleura (Stauraxonia heteropola heterostaura allopola). See next Figure.

Figure 5.  Half a Rhombic Pyramid as the promorph of all the crystals of the Domatic Class of the Monoclinic Crystal System. The tip of the pyramid is pointing to the beholder. Its only symmetry element (not to be confused with group element) is a mirror plane as the median plane of the pyramid.

The geometric solid generally representing geometrically the symmetry of crystals of the Pinacoidal Class (i.e. the Class  1*) of the Triclinic Crystal System is a Triclinic Bipyramid or Oblique Rhombic Bipyramid. It is the promorph or stereometric basic form of all the crystals of this Class and as such belongs to the Anaxonia centrostigma (Anaxonia). See next Figure.

Figure 6   Triclinic Bipyramid, Stereometric Basic Form of the Pinacoidal Class of the Triclinic Crystal System.
The angle between the equatorial plane and the main axis is different from 90⁰.

Figure 7.  Triclinic Bipyramid, Stereometric Basic Form of the Pinacoidal Class of the Triclinic Crystal System.
The angle between the equatorial plane and the main axis is different from 90⁰. Triclinic axial system inserted. The three axes do not involve angles of 90⁰.

Figure 8.  Equatorial plane of a Triclinic Bipyramid, as seen from a direction perpendicular to that plane (and thus not seen along the direction of the bipyramid's main axis).
The angle between the equatorial plane and the main axis is different from 90⁰. Two triclinic axes inserted (red), they meet in the bipyramid's center of symmetry ( i ).

Figure 9.   Parallelopipedum, having the symmetry of the Pinacoidal Class of the Triclinic Crystal System. All angles are different from 90⁰. There are no symmetry axes, nor mirror planes.

Figure 10.  Parallelopipedum, having the symmetry of the Pinacoidal Class of the Triclinic Crystal System. All angles are different from 90⁰. There are no symmetry axes, nor mirror planes. When we nevertheless try out a 2-fold rotation axis, we see that such an axis will map the face (of the parallelopipedum) on which it is perpendicular, indeed onto itself. But such an axis will not map the whole parallelopipedum onto itself, because that axis is not perpendicular to the opposite face (it is not parallel to AB). This opposite face will not be mapped onto itself by an oblique 2-fold rotation axis.

Figure 11.  Half a Triclinic Bipyramid, the Stereometric Basic Form of the crystals of the Asymmetric Class of the Triclinic Crystal System. The depicted geometrical solid is (in fact) a single pyramid :   a totally irregular pyramid.

Figure 12.  Equatorial plane of Half a Triclinic Bipyramid.

The Cyclic Group C₁ has only one element, the identity. This element can be interpreted as a rotation of 0⁰ about any axis of the given solid.

So the (abstract) cyclic groups realized in (single non-twinned) crystals are accordingly :  C₁, C₂, C₃, C₄ and C₆.

Cyclic Groups having a number of elements which is prime, we already met : C₁ and C₃.
We will now discuss a cyclic group of order 7 (which is also prime), C₇, realized in finite arithmetics as addition modulo 7 :

The period of the identity element is (as it is always)  1.
The period of all other elements of the group (C₇) is  7.
Another way of saying this is that any element of the group may be used to generate the whole group.
In the above table we used the element  1  to generate the group. We did this by successively applying the element  1  :
1 + 1 = 1² = 2  gave the element 2.
1 + 1 + 1 = 1³ = 3  gave the element 3.
1 + 1 + 1 + 1 = 1⁴ = 4  gave the element 4.
1 + 1 + 1 + 1 + 1 = 1⁵ = 5  gave the element 5.
1 + 1 + 1 + 1 + 1 + 1 = 1⁶ = 6  gave the element 6.
finally, 1 + 1 + 1 + 1 + 1 + 1 + 1 = 1⁷ = 7 = 0 (mod 7), which is the identity element.
(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 1⁸ = 8 = 1 (mod 7), etc.).
While the generator element was  1, we've now generated the elements 2, 3, 4, 5, 6 and 0, i.e. we now have all the elements of the group.
But we can, alternatively, use the element  3  to generate the same group :

So, given the generator element 3, the rest of the elements can be generated as follows :
3 + 3 = 3² = 6.
3 + 3 + 3 = 3³ = 9 = 2 (mod 7).
3 + 3 + 3 + 3 = 3⁴ = 12 = 5 (mod 7).
3 + 3 + 3 + 3 + 3 = 3⁵ = 1 (mod 7).
3 + 3 + 3 + 3 + 3 + 3 = 3⁶ = 4 (mod 7).
3 + 3 + 3 + 3 + 3 + 3 + 3 = 3⁷ = 21 = 0 (mod 7), which is the identity.
(3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 3⁸ = 24 = 3 (mod 7)).
(3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 3⁹ = 27 = 6 (mod 7), etc.).

Starting with  0, then  3, and then the generated elements in the above order gives us one of the automorphisms of C₇, i.e. instead of the original order of elements

0 1 2 3 4 5 6

We have now :

0 3 6 2 5 1 4

Indeed when we construct the table for this latter sequence we see that the structure is preserved :

We can interpret this table also geometrically, namely as a tour of the vertices of the star heptagon :

The vertices are visited in the order of the cycle (following the arrowed lines), 0, 3, 6, 2, 5, 1, 4, and back to 0 again. The fact that every vertex is visited in turn corresponds to the fact that, if you change your socks every (say) three days, then whichever day you start on, you must make a subsequent change on every possible day of the week, i.e. there will be some monday on which you change your socks, but also some tuesday on which you do that, some wednesday, some thursday, some fryday, some saturday and some sunday. Here we are dealing with the modulo 7 arithmetic, and this statement about changing socks would remain true if  3  were replaced by 1, 2, 4, 5, 6, or for that matter by any number not divisible by 7. That's why cyclic groups of prime order do not have subgroups (i.e. there is no subcycle that restricts its visits to just a part of the vertices).
If we were dealing with a composite modulus, e.g. modulo 12, it would be a different story. If receive (and pay) your telephone bill once a quarter, and the first account is received in February, then subsequent payments will be due in May, August, November, February, ... and you will never receive the bill in any other month.

The fact that every element in C₇, except the identity element, is primitive, i.e. each of them can generate the whole group, because they each have a period of 7, and may be used as a generator, means that here we have a system in which all the elements (apart from the identity element) are of equal status - every member of this 'democracy' is just as good as every other, except the identity, who is a sort of president. This means that the group C₇ has six automorphisms : 
If we take the element 1 as generator, a certain sequence of the group elements results, let us call it the original sequence.
If we take instead of 1, the element 2 as generator, another sequence of the group elements will result. This other sequence will, however, preserve structure, i.e. with respect to the original sequence :  products are preserved, which in turn means that a product in a group table based on the original sequence of group elements, corresponds exactly to a product in a group table based on the second sequence of group elements. Such an alternative sequence of group elements, in which structure is preserved, is called an automorphism of the group.
If we take the element 3 as generator, yet another sequence will result, which constitutes yet another automorphism, and so on, with respect to the remaining elements of the group. The original sequence of group elements we call the identity automorphism. So the following replacements of the original generator by another generator

1 -- 1
1 -- 2
1 -- 3
1 -- 4
1 -- 5
1 -- 6

result in six different but structure preserving generated sequences of group elements, i.e. six reshufflings of group elements, and that means : six automorphisms. These specific reshufflings are based on the fact that we can generate the whole group either with element 1, or with 2, or with 3, or 4, or 5, or 6. We already have done so with element 1 and with element 3, resulting in two different, but structure preserving, element sequences -- autmorphisms.

Each of such an automorphism thus is a permutation (reshuffling) of the group elements (These are in our case the elements 0, 1, 2, 3, 4, 5 and 6), for example the automorphism resulting from the replacement of the generator  1  by the generator  3  ( The automorphism thus being an element sequence generated by the element 3 ) can be expressed by the following permutation of the group elements :

The elements in the sequential order  0 1 2 3 4 5 6  means that the group elements are successively generated by the element  1  :
1 + 1 (mod 7) gives element 2.
1 + 1 + 1 (mod 7) gives element 3.
1 + 1 + 1 + 1 (mod 7) gives element 4.
1 + 1 + 1 + 1 + 1 (mod 7) gives the element 5.
1 + 1 + 1 + 1 + 1 + 1 (mod 7) gives the element 6.
1 + 1 + 1 + 1 + 1 + 1 + 1 (mod 7) gives the element 0.
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 (mod 7) gives the element 1 again.
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 (mod 7) gives the element 2 again.
etc.

When we place the element 0 in front  [We are allowed to do so, because the element 0 does not partake in the permutations generating automorphisms -- it cannot do so because its period is different from that of the other ones],  the resulting sequence of the (generated) elements is :
0 1 2 3 4 5 6. This corresponds to the permutation (of the group elements) :

The corresponding group table is (as given earlier) :

2 + 2 (mod 7) gives the element 4.
2 + 2 + 2 (mod 7) gives the element 6.
2 + 2 + 2 + 2 (mod 7) gives the element 1.
2 + 2 + 2 + 2 + 2 (mod 7) gives the element 3.
2 + 2 + 2 + 2 + 2 + 2 (mod 7) gives the element 5.
2 + 2 + 2 + 2 + 2 + 2 + 2 (mod 7) gives the element 0 .

So placing again the element 0 in front we get the sequence
0 2 4 6 1 3 5. This corresponds to the permutation :

The corresponding group table is :

We can clearly see that the table structure of this group table is the same as the previous one (corresponding to the identity permutation).

With the element 3 we can generate the group elements as follows :

3 + 3 (mod 7) gives the element 6.
3 + 3 + 3 (mod 7) gives the element 2.
3 + 3 + 3 + 3 (mod 7) gives the element 5.
3 + 3 + 3 + 3 + 3 (mod 7) gives the element 1.
3 + 3 + 3 + 3 + 3 + 3 (mod 7) gives the element 4.
3 + 3 + 3 + 3 + 3 + 3 + 3 (mod 7) gives the element 0.

So placing the element 0 in front, we get :
0 3 6 2 5 1 4.

This corresponds to the permutation :

The corresponding table (as given earlier) is :

Also here the structure has remained the same.

With the element 4 we can generate the group elements as follows :

4 + 4 (mod 7) gives the element 1.
4 + 4 + 4 (mod 7) gives the element 5.
4 + 4 + 4 + 4 (mod 7) gives the element 2.
4 + 4 + 4 + 4 + 4 (mod 7) gives the element 6.
4 + 4 + 4 + 4 + 4 + 4 (mod 7) gives the element 3.
4 + 4 + 4 + 4 + 4 + 4 + 4 (mod 7) gives the element 0.

So placing the element 0 in front, we get :

0 4 1 5 2 6 3.

This corresponds to the permutation :

The corresponding table is :

Also here the structure is unchanged.

With the element 5 we can generate the group elements as follows :

5 + 5 (mod 7) gives the element 3.
5 + 5 + 5 (mod 7) gives the element 1.
5 + 5 + 5 + 5 (mod 7) gives the element 6.
5 + 5 + 5 + 5 + 5 (mod 7) gives the element 4.
5 + 5 + 5 + 5 + 5 + 5 (mod 7) gives the element 2.
5 + 5 + 5 + 5 + 5 + 5 + 5 (mod 7) gives the element 0.

So when placing the element 0 in front we get the sequence :
0 5 3 1 6 4 2.

The corresponding permutation is :

And the table is :

Also here the structure is fully preserved.

Finally, with the element 6 we can generate the group elements as follows :

6 + 6 (mod 7) gives the element 5.
6 + 6 + 6 (mod 7) gives the element 4.
6 + 6 + 6 + 6 (mod 7) gives the element 3.
6 + 6 + 6 + 6 + 6 (mod 7) gives the element 2.
6 + 6 + 6 + 6 + 6 + 6 (mod 7) gives the element 1.
6 + 6 + 6 + 6 + 6 + 6 + 6 (mod 7) gives the element 0.

So when placing element 0 in front we get the sequence :
0 6 5 4 3 2 1

This corresponds to the permutation :

And the table is :

Also here the structure is preserved.

Writing down all six automorphisms as permutations of the six mentioned elements, while writing down only the lower part of the permutation symbol, we get :

0 1 2 3 4 5 6  (identity)
0 2 4 6 1 3 5
0 3 6 2 5 1 4
0 4 1 5 2 6 3
0 5 3 1 6 4 2
0 6 5 4 3 2 1

It is perhaps instructive to show a case -- i.e. a permutation of the group elements -- that does not yield an automorphism. We try the permutation:

The table would be accordingly :

We see clearly that the structure is not preserved (not only with respect to the pattern of identity elements, but also with respect to the NE--SW diagonals). So if we depart from the sequence 0 1 2 3 4 5 6 of the group elements, then this last (departure) that we've just tried out, is not a automorphism of the group C₇.

Let's try yet another case, namely the permutation :

The table is then :

The structure is again totally different.

Above we found out that the following permutations yielded automorphisms of the group C₇. They should themselves form a group, the automorphism group of the group C₇, notated Aut(C₇) :

0 1 2 3 4 5 6  (identity)
0 2 4 6 1 3 5
0 3 6 2 5 1 4
0 4 1 5 2 6 3
0 5 3 1 6 4 2
0 6 5 4 3 2 1

The  0  element does not partake in the permutations, so in fact we have the permutations :

1 2 3 4 5 6  (identity)
2 4 6 1 3 5
3 6 2 5 1 4
4 1 5 2 6 3
5 3 1 6 4 2
6 5 4 3 2 1

i.e. six permutations of 6 symbols.
Let's look to the periods of those six permutations.
The period of the identity permutation is of course 1.
We now will apply the second permutation repeatedly till we arrive at the identity :

1 2 3 4 5 6
2 4 6 1 3 5  (1)
4 1 5 2 6 3  (2)
1 2 3 4 5 6  (3)

So the period of the second permutation is 3.

Now the third permutation :

1 2 3 4 5 6
3 6 2 5 1 4  (1)
2 4 6 1 3 5  (2)
6 5 4 3 2 1  (3)
4 1 5 2 6 3  (4)
5 3 1 6 4 2  (5)
1 2 3 4 5 6  (6)

So the period of the third permutation is 6.

Now the fourth permutation :

1 2 3 4 5 6
4 1 5 2 6 3  (1)
2 4 6 1 3 5  (2)
1 2 3 4 5 6  (3)

So the period of the fourth permutation is 3.

Now the fifth permutation :

1 2 3 4 5 6
5 3 1 6 4 2  (1)
4 1 5 2 6 3  (2)
6 5 4 3 2 1  (3)
2 4 6 1 3 5  (4)
3 6 2 5 1 4  (5)
1 2 3 4 5 6  (6)

So the period of the fifth permutation is 6.

Finally, the sixth permutation :

1 2 3 4 5 6
6 5 4 3 2 1  (1)
1 2 3 4 5 6  (2)

So the period of the sixth permutation is 2.

Summarizing those periods gives :

1, 3, 6, 3, 6, 2

and we will show that this is the 'signature' of the group C₆, i.e. the cyclic group of order 6. Its elements are here represented by the above permutations (of the elements of the group C₇). We will give these permutations 'names', i.e. indicate them by letters :

1 2 3 4 5 6  =  i
2 4 6 1 3 5  =  a
3 6 2 5 1 4  =  b
4 1 5 2 6 3  =  c
5 3 1 6 4 2  =  d
6 5 4 3 2 1  =  f

These elements are indeed generated by an element of period 6, for example the element
5 3 1 6 4 2 which we can now call  d (The other element is b).
The repeated application of d yielded -- as we saw above -- the following permutations :

1 2 3 4 5 6  =  i
5 3 1 6 4 2  =  d
4 1 5 2 6 3  =  c
6 5 4 3 2 1  =  f
2 4 6 1 3 5  =  a
3 6 2 5 1 4  =  b

So we have an element -- the permutation  d -- which generates all the other elements, so these permutations must form a cyclic group of order six (C₆). So we can say that the automorphism group of the group C₇ is the group C₆, i.e. Aut(C₇) = C₆.
Let's analyze this automorphism group further. Its elements are the permutations  i, a, b, c, d and f. These six permutations can be seen as the six permutations of the vertices of a regular hexagon under rotation of the latter, i.e. they can be seen as the direct symmetries of the Regular Hexagon, i.e. its rotational symmetries only. The next Figure depicts this hexagon and the appropriate labelling of its vertices in order for to express the permutations.

Figure 13.  A regular hexagon with marked vertices. As such it represents the identity permutation   1 2 3 4 5 6  =  i ( i.e. 1 remains 1, 2 remains 2, etc.) of the six (non-identity) elements of the group C₇ , forming the group C₆ ( = Aut(C₇ ). This identity permutation is equivalent to a rotation of the hexagon (and thus its vertices) of  0⁰ about the hexagon's mid-point.

which we had called  b  above.

Figure 14.  Anticlockwise rotation of the regular hexagon of Figure 13 by 60⁰, according to permutation  b. We must read this rotation as follows :
1 goes to position 3.
2 goes to position 6.
3 goes to position 2.
4 goes to position 5.
5 goes to position 1.
6 goes to position 4.
The result of this rotation is depicted in the right image.

Figure 15.  Anticlockwise rotation of the hexagon of Figure 13 by 120⁰, according to the permutation  a. We must read this as follows :
1 goes to position 2
2 goes to position 4
3 goes to position 6
4 goes to position 1
5 goes to position 3
6 goes to position 5

Figure 16.  Anticlockwise rotation of the hexagon of Figure 13 by 180⁰, according to the permutation  f. We must read this as follows :
1 goes to position 6
2 goes to position 5
3 goes to position 4
4 goes to position 3
5 goes to position 2
6 goes to position 1

Figure 17.  Anticlockwise rotation of the hexagon of Figure 13 by 240⁰, according to the permutation  c. We must read this as follows :
1 goes to position 4
2 goes to position 1
3 goes to position 5
4 goes to position 2
5 goes to position 6
6 goes to position 3

Figure 18.  Anticlockwise rotation of the hexagon of Figure 13 by 300⁰, according to the permutation  d. We must read this as follows :
1 goes to position 5
2 goes to position 3
3 goes to position 1
4 goes to position 6
5 goes to position 4
6 goes to position 2

i  period 1, rotation of  0⁰.
b  period 6, rotation of  60⁰.
a  period 3, rotation of  120⁰.
f  period 2, rotation of  180⁰.
c  period 3, rotation of  240⁰.
d  period 6, rotation of  300⁰.

In order to make a group table it is convenient to express the just mentioned elements  i, b, a, f, c  and  d  in terms of a generator. In our case element b, which is of period 6, can function as generator.
We will call element  i  element  1. And if we call the element  b (60⁰ rotation)  r, then the element  a  (120⁰ rotation) will be  r², element  f (180⁰) will be  r³, etc. In this way we get the following renaming :

i -- 1
b -- r
a -- r²
f -- r³
c -- r⁴
d -- r⁵

We will now give the table for this automorphism group (C₆), i.e. the table for Aut(C₇) :

So this table depicts the automorphism group (C₆) of the group C₇, in terms of the generator  r, which (generator) can be interpreted as a rotation of 60⁰ of a regular hexagon about its center.

Figure 19.  Diagram illustrating preservation of products.
The image of the product is identical to the product of the images of the factors, implying that to the original factors and product there not only correspond images, but also that those images themselves relate to each other as factors and product.

So according to this permutation each element of the group corresponds with a certain (other) element of that same group :

0 ==> 0
1 ==> 2
2 ==> 4
3 ==> 6
4 ==> 1
5 ==> 3
6 ==> 5

This is a function from the group to itself. Let us call this function PHI.
Then the image of (say) 2, under the function PHI, is 4. In short :  PHI(2) = 4.
The group operation is addition modulo 7. And  "2 + 4 = 6"  we call a product of the elements  2  and  4. These latter two elements are then the factors. We could also write :  2.4 = 6,  (2)(4) = 6,  or 24 = 6, for that matter.

Let us now consider the product 3 + 5 = 1, in which  3,  5  and  1 are elements from the original sequence  0 1 2 3 4 5 6  of group elements. So here  1  is the product. Then we can say :

PHI(1) = 2  (image of the product)
PHI(3) = 6  (image of factor)
PHI(5) = 3  (image of factor), then
PHI(3) PHI(5) = 6 + 3 = 9 = 2  (because 9 mod 7 = 2) (product of the images of factors)

So the image of the product is equal to the product of the images, which means that that particular product is exactly reflected in the alternative sequence of elements according to the function PHI. And this in turn means that that particular product is preserved. When this holds for all possible products, we have an automorphism of the group.

Let us now prove that in any sequence of group elements, generated by one of the possible generators of the group C₇, namely the generators 1, 2, 3, 4, 5 or 6,  resulting in the above listed six permutations of group elements,  products are preserved , and thus these sequences representing automorphisms of the group C₇  :

The sequence of group elements generated by the element  1  can be written as

0, 1, 1², 1³, 1⁴, 1⁵, 1⁶

in which, generally, 1^y,  in the present case, means 1 + 1 + 1 + . . . , where this sequence contains  y  1's.  For example  1⁴ is equal to 1 + 1 + 1 + 1 = 4  (Recall that the group operation is addition modulo 7 ).
Let's now choose a different generator out of the remaining possible generators 2, 3, 4, 5, 6 ( all of the same period, namely 7 ), namely the generator  x.  It will generate the sequence

0, x, x², x³, x⁴, x⁵, x⁶

in which, for instance,  x³ = x + x + x = 3x, and which is a new sequence of group elements of the group C₇ , i.e. a permutation of those group elements. The correspondence of the elements of the original sequence with those of the new sequence can be indicated by the function THETA, and is the following one-to-one correspondence between the elements of our group :

0 ==> 0
1 ==> x
1² ==> x²
1³ ==> x³
1⁴ ==> x⁴
1⁵ ==> x⁵
1⁶ ==> x⁶

A product of elements of the original sequence could be, for instance, 1²1⁴ = 1⁶ (which means 2 + 4 = 6 ), or, generally, 1^a1^b = a1 + b1 = a + b.  And  a + b  is certainly an element of the group, let us call it c,  which is equal to  1^c.  So the product of  1^a  and  1^b  is equal to 1^c, or, equivalently,  a + b = c. Now we can say :

1^a1^b = 1^c  (product)
THETA(1^c ) = x^c = cx  ( image of the product )
THETA(1^a ) = x^a  ( image of factor )
THETA(1^b ) = x^b  ( image of factor ), then
THETA(1^a ) THETA(1^b ) = x^ax^b = ax + bx = (a + b)x = cx  ( = x^c ) (product of images of factors).

So, now we have proved, generally for the group C₇ ,  that the image, under the general function THETA, of any product is equal to the product of the images of the factors, which means that for all six permutations of the group elements (of the group C₇ ), resulting from an alternative choice of generator, products are preserved, so that these six permutations (including the identity permutation) are indeed as many automorphisms of the group C₇ .

Because there are six equally valid and independent generators of the group C₇ ,   we get six automorphisms.
Generally we can say that a cyclic group C_p ,  where  p  is prime, has  p-1  equally valid generators, and thus it has  p-1  automorphisms, so the order of the automorphism group (i.e. the group of all automorphisms) of C_p is  p-1.

The set of automorphisms is a set of permutations of group elements, and these particular permutations preserve products. That this set is itself a group is readily appreciated :
In every one such permutation a generator is replaced by another (one generator by one other, in cyclic groups), and the remaining elements are then generated by this new generator. So a general element  x^a  becomes  y^a,  where the generator  x  is substituted by the generator  y.  When we now apply another permutation (again one that substitutes a generator) on the result, say a replacement of the generator  y  by the generator  z,  then the general element becomes  z^a.  So the successive application of two permutations (each replacing a generator) effects  x^a ==> z^a,  and this involves yet another replacement of generators, namely  x  being replaced by  z,  resulting in yet another automorphism of the group C_p . So the set of automorphisms is closed under the operation of successive application.
The set also contains an identity element, namely the identity permutation, i.e. the identity automorphism.
Further, the inverse of every automorphism is itself an automorphism :  Suppose we have the automorphism in which the generator  x  is replaced by  y.  Then the inverse consists of a replacement of the generator  y  by the generator  x.
Finally, associativity is guaranteed by the fact that the successive application of permutations is always associative :
For three permutations named S, T, U, we have (ST)U = S(TU), because for a general element  s  of S,  a certain element  t  of T, and a certain element  u  of U, we have :

(s ==> t) ==> u   =   s ==> (t == > u)

because when we first replace  s  by  t,  and then  t  by  u,  we in fact have replaced  s  by  u.  And when, on the other hand, we replace  s  by the (result of the) replacement of  t  by  u,  we also have in fact replaced  s  by  u.

So the set of automorphisms of a group is indeed itself a group.

We now will demonstrate that the automorphism group of a cyclic group of prime order is also a cyclic group.
In order to do that, let's first again consider the group C₇ .  Because the group is of prime order, every non-identity element is a potential generator, i.e. every such element can generate all other elements of the group. So the generator  x,  where  x⁷ = the identity element, generates the following sequence of non-identity elements :  x, x², x³, x⁴, x⁵, x⁶. The generator  x²  generates the sequence  x², x⁴, x⁶, x, x³, x⁵.  And so on.
In this way six sequences of all the non-identity elements of the group are generated :

These six sequences are six permutations (of six objects), and it is obvious that we can represent them just as well by the indices ('powers') :

So here we have again six permutations, their inherent cycles and periods. Together they (as do the equivalent permutations above) form the group C₆ ,  isomorphic to the automorphism group of the group C₇. That this automorphism group is indeed  cyclic,  is appreciated by realizing that -- as has been said -- in a cyclic group of prime order every non-identity element can be a generator of the whole group. And because of this, looking to the list of permutations directly above -- permutations that came to be created by the generation of the group elements by the successive generators -- we see that the first row (being the identity permutation) of that list is just the natural number sequence  1 2 3 4 5 6.  In the second row all numbers have been replaced by another number (but always taken from the set {1, 2, 3, 4, 5, 6}). And in all the other rows this is also the case. This is because of the fact that each row is generated by another generator. Every permutation from the above list replaces every symbol of the initial sequence by another symbol, and this means that all the listed permutations can be interpreted as rotations of some two-dimensional figure, and this in turn implies that the group is cyclic.
All of this can easily be generalized to all cyclic groups of prime order :
Each and every non-identity element of such a group can generate all elements of such a group. The resulting sequences of group elements are each produced by a different generator. The first is generated by  x,  the second by  x²,  the third is generated by  x³,  and so on. Sequences produced by such generators will not show overlaps, and thus are permutations that can potentially represent rotations of some figure. And because all sequences must represent rotations, the group they form must be cyclic.

In the case of the automorphism group of the group C₇ ,  we can show -- with the aid of the representation of them by cycles (as indicated in the above list) -- that, and in what way, the five non-identity permutations (being elements of Aut(C₇)) can represent rotations. See next Figure.

Figure 20.  The five non-identity autmorphisms of the group C₇ interpreted geometrically as (anticlockwise) rotations.

In a cyclic group of prime order  p, every element (except the identity) is of period  p. (BUDDEN, p. 170)

There exists only one type of group of prime order, namely the cyclic group. (BUDDEN, p. 170)

Aut(C_p) is isomorphic to C_p-1 when  p  is prime, i.e. the automorphism group of a cyclic group of prime order has an order that is one lower than that of the original group. (BUDDEN, p. 423)

In the next document we will study cyclic groups of composite order.

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