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The Total Symmetry of Three-dimensional Crystals

Part XVIII
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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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Sequel to Automorphisms

In the previous document we studied the automorphisms of the group D4 ,  in order to get to grips with the important concept of automorphisms of a group. We will continue this study with respect to other groups.

Earlier we remarked that the only inner automorphism of an Abelian group is the identity permutation of its elements, since every element is self-conjugate. All the automorphisms of Abelian groups are therefore outer. We will further see that the group D3  (which is isomorphic to S3 ,   the group of all permutations of three symbols) has six automorphisms whose group is D3 ,  and these are all inner ones. This group therefore has no outer automorphisms.
These examples show the two extreme cases.
That all the automorphisms -- regarded as permutations of the elements of the given group -- themselves form a group, is readily appreciated. For an automorphism is a permutation of the elements which preserves products. When one such permutation is followed by another, the result will be another permutation, with products preserved throughout. Associativity of permutations under successive composition is assured, and the identity permutation is present. The inverse of an automorphism also preserves structure (and thus as such is an automorphism, and thus every inverse belongs to the set of permutations preserving structure), so all the requirements of a group are fulfilled.
That the inner automorphisms of a group are a subgroup of the group of automorphisms, can be proved as follows (only establishment of closure is needed) :

We must in fact prove that the successive application of two inner automorphisms is again an inner automorphism. If we take a general element  g  of the group G, and transform this element by an element  y  of that same group, then the resulting transform,  ygy-1,  represents an inner automorphism (by the element  y ). When we now transform that result in turn by an element  z  of the group, we get  z(ygy-1)z-1,  which represents the successive application of two inner automorphisms, and which is equal to  (zy)g(y-1z-1) = (zy)g(zy)-1,  and this is clearly yet another inner automorphism, because  zy  is an element of G, so the set of inner automorphisms of the group G is closed, and that means that it is a subgroup of the group of automorphisms of the group G.

The stronger result, that the inner automorphisms are normal subgroup of the full group of automorphisms, will be proved later.

Intuitively we can define tha automorphisms of a group as follows:
The automorphisms of a group are those one-to-one correspondences between the elements of that group with those of a copy of it that preserve products.

We will now show that the group D3 has six automorphisms (and no more), and that these are all inner automorphisms.

The group D3 can be realized as the (group of) symmetries (direct and opposite) of an equilateral triangle or of a regular trigonal pyramid. Its defining relations are

3 = 1
a2 = 1
ar = r2a

(BUDDEN, p. 290)

The element  r  can be interpreted as a rotation of 1200 anticlockwise.
The identity element will be -- as usual -- called 1.
The element  2 is then a rotation of 2400 anticlockwise. But this is equal to a rotation of 1200 clockwise, which means that  r  and  2  are each other inverse.
So  r-1 = r 2. The period of  r  and also of  2  is 3.
The element  a  can be interpreted as a reflection in a line (in the equilateral triangle) or in a plane (in the regular trigonal pyramid). So its period is 2. And this means that aa = 1, which in turn means that  a-1 = a.
To understand the third defining relation  ar = r 2a,  we interpret it geometrically in the next Figure.

Figure 1.  Diagram, illustrating geometrically that  ar = r 2a.
a  = mirror line.

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We can generate the elements of the group D3 with the generators  r  and  a  :

• not doing anything yields the identity, 1
• r = r  (generator)
• r(r) = 2
• r(r 2) = r 3 = 1
• a = a  (generator)
• aa = a2 = 1
• r(a) = ra
• r(ra) = 2a
• r(r 2a) = r 3a = a.

So the sequence of elements so obtained is :

1,   r,   r 2,   a,   ra,   r 2a

Let's determine the periods of those elements :

• 1 has period 1.

• rr = r 2, and
2r = r 3 = 1.  So the period of the element  r  is 3.

• (r 2)(r 2) = r 4 = r 3r = 1r = r, and
r(r 2) ( = (r 2)3 ) = r 3 = 1.  So the period of  r 2  is 3.

• a2 = 1.  So the period of  a  is 2.

• (ra)2 = rara = r(ar)a = r(r 2a)a  (third defining relation) = rr 2aa = 1.1 = 1.  So the period of the element  ra  is 2.

• (r 2a)2 = (r 2a) (r 2a) = r 2ar 2a = r 2arra = r 2(ar)ra = r 2(r 2a)ra = r 22(ar)a =
4(ar)a = r 3r(ar)a = r(ar)a = r(r 2a)a = rr 2aa = r 3aa = 1.1 = 1.  So the period of the element  r 2a  is 2.
To summarize these periods (of the elements of the group D3 ) we get

 element period 1 1 r 3 r2 3 a 2 ra 2 r2a 2

In order to find automorphisms we must replace the generator pair  r  and  a  by another equivalent pair (also having respectively period 3 and 2), and then generate the remaining group elements. The resulting sequence of group elements is then an automorphism. Next we choose another pair of generators (of period 3 and 2 respectively), and generate again the remaining group elements, resulting in yet another automorphism.
Because there are two elements of period 3, and three of period 2, there are six possible choices of generator pairs, and among them there are none that effect duplication of elements (in the generated sequence of group elements) :

 period 4 period 2 r a r ra r r2a r2 a r2 r2a r2 ra

The sequence of group elements generated by  r  and  a  has been already performed.
Generating the sequence of group elements by  r  and  ra  goes as follows :

• not doing anything yields the identity 1.
• r = r  (generator).
• r(r) = 2.
• ra = ra  (generator).
• r(ra) = 2a.
• 2(ra) = r 3a = a.
• a(ra) = (ar)a = (r 2a)a = r 2aa = r 2.
• ra(a) = ra2 = r.
• (r 2a)a = r 2aa = r 2.
• a(r 2a) = (ar)ra = (r 2a)ra = rr(ar)a = rr(r 2a)a = r 4aa = r 4 = r.
So we have generated the sequence

1    r    r 2    ra     r 2a    a

And this constitutes a second automorphism (the first one is the original sequence of group elements, generated by  r  and  a,  which we set as identity automorphism).
In the same way we generate the other four sequences of group elements on the basis of the alternative generator pairs listed above. The total result is the following :

 1 r r2 a ra r2a 1 r r2 ra r2a a 1 r r2 r2a a ra 1 r2 r a r2a ra 1 r2 r r2a ra a 1 r2 r ra a r2a

These six sequences of group elements represent all automorphisms of the group D3.

To show that these are all inner automorphisms, we will derive all inner automorphisms of D3 ,  and see that there are six of them :

To do this we determine the transform of all six elements by all elements of the group.
Generally, if we take each element  x  of the group, and transform it with the transforming element  y,  i.e. we determine  yxy-1 ,  we will obtain an inner automorphism.
If we then do the same, but now with the transforming element  z,  i.e. we form  zxz-1 ,  we will obtain a second inner automorphism (which could of course be identical to the one obtained before -- or it is not).
If we let, in this way, figure each element of the group as transforming element (transforming each element of the group), we get all the possible inner automorphisms.

We start by remarking that the element 1 (the identity element) is always transformed into itself :  Let  x  be any element of the group. Then we have :  x1x-1 = xx-1 = 1.

Further, when 1 is the transforming element, it will transform any element  x  into itself :
1x1-1 = x1-1 = x1 = x  (because 1.1 = 1, implying that the inverse of 1 is 1,
that is 1-1 = 1).

In the following we will make use of the three defining relations (of the group D3 ) mentioned earlier :

3 = 1
a2 = 1
ar = r2a

Further we have  r-1 = r 2,  because  r-1  can be seen as a clockwise rotation of 1200, while r 2 can be seen as two times an anticlockwise rotation of 1200. And these rotations have exactly the same effect.

Let's now derive all inner automorphisms of the group D3.

As has been explained, the identity automorphism is obtained by transforming each element of the group by the transforming element 1, and we let this identity automorphism correspond with the first row of the above array (depicting the six automorphisms, derived by the interchange of generators).

Transformation of each non-identity element by the transforming element  r.

• r r-1 = r1 = r.

• 2 r-1 = r r r r-1 = 2.

• a r-1 = rar 2 = r(ar)r = r(r 2a)r = rr 2ar = r 3ar = ar = 2a.

• ra r-1 = r 2ar 2 = r 2(ar)r = r 2(r 2a)r = r 22ar = r 4ar = r 3rar = rar
= r(ar) = rr 2a = a.

• 2a r-1 = ar-1 = ar 2 = (ar)r = r 2ar = r 22a = ra.
In this way we have found a second inner automorphism :

1 ==> 1
r ==> r
2 ==> r 2
a ==> r 2a
ra ==> a
2a ==> ra

It corresponds with the third row of the above list of automorphisms found by replacing generators.

From the consideration of  r  as a clockwise rotation of 1200, we can deduce
that  (r 2)-1 = r. We will use this in the derivation of the next inner automorphism :

Transformation of each non-identity element by the transforming element  r 2.

• 2 r (r 2)-1 = r 2rr = r.

• 2 2 (r 2)-1 = r 22r = 2.

• 2 a (r 2)-1 = r 2ar = r 22a = ra.

• 2 ra (r 2)-1 = (r 2r)a(r 2)-1 = a(r 2)-1 = ar = 2a.

• 2 2a (r 2)-1 = rar = r(ar) = rr 2a = a.
With this we have found a third inner automorphism :

1 ==> 1
r ==> r
2 ==> r 2
a ==> ra
ra ==> r 2a
2a ==> a

And this corresponds with the second row of the above list of automorphisms found by replacing generators.

Transformation of each non-identity element by the transforming element  a.

• r a-1 = r 2aa-1 = 2.

• 2 a-1 = arra-1 = (r 2a)ra-1 = r 2(ar)a-1 = r 22aa-1 = r 4 = r.

• a a-1 = a.

• ra a-1 = r 2aaa-1 = 2a.

• 2a a-1 = arraa-1 = arr = r 2ar = r 22a = ra.
This gives us the fourth inner automorphism :

1 ==> 1
r ==> r 2
2 ==> r
a ==> a
ra ==> r 2a
2a ==> ra

And this corresponds to the fourth automorphism in the above list of automorphisms found by replacing generators.

Transformation of each non-identity element by the transforming element  ra.

To begin with, we see that  (ra)-1 = a-1r-1.
Further we know that  a2 = aa = 1, which implies  a-1 = a.  Also we saw that  r-1 = r 2.

• ra r (ra)-1 = r(ar)a-1r-1 = rr 2aa-1r-1 = rr 2r-1 = r-1 = 2.

• ra 2 (ra)-1 = rar 2a-1r-1 = r(ar)ra-1r-1 = rr 2ara-1r-1 = arar 2 = r 2aar 2 =
22 = r 4 = r.

• ra a (ra)-1 = ra-1r-1 = rar-1 = rar 2 = r(ar)r = rr 2ar = ar = 2a.

• ra ra (ra)-1 = rr 2aaa-1r-1 = a-1r-1 = ar 2 = arr = r 2ar = r 22a = r 4a = ra.

• ra 2a(ra)-1 = rar 2aa-1r-1 = rar 2r-1 = rar 22 = rar = rr 2a = a.
With this we have the fifth automorphism :

1 ==> 1
r ==> r 2
2 ==> r
a ==> r 2a
ra ==> ra
2a ==> a

And this corresponds to the fifth automorphism in the above list of automorphisms found by replacing generators.

Finally, transformation of each non-identity element by the transforming element  r 2a.

Earlier we saw that  (r 2)-1 = r, and that  a-1 = a.
Before we proceed, we see that  (r 2a)-1 = a-1(r 2)-1 = a(r 2)-1 = ar = 2a.

• 2r (r 2a)-1 = r 2ara-1(r 2)-1 = r 2(ar)a-1r = r 22aa-1r = r 22r = 2.

• 22 (r 2a)-1 = r 2ar 2a-1(r 2)-1 = r 2ar 2a-1r = r 2ar 2(ar) =
2ar 22a = r 2ara = r 22aa = r 22 = r.

• 2a (r 2a)-1 = r 2(r 2a)-1 = r 2a-1(r 2)-1 = r 2a-1r = r 2ar = r 22a = ra.

• 2ra (r 2a)-1 = r 2(ar)aa-1(r 2)-1 = r 2(ar)(r 2)-1 = r 22a(r 2)-1 =
ra(r 2)-1 = rar = rr 2a = a.

• 22a (r 2a)-1 = r 2ar 2aa-1(r 2)-1 = r 2ar 2r = 2a.
This gives the sixth inner automorphism :

1 ==> 1
r ==> r 2
2 ==> r
a ==> ra
ra ==> a
2a ==>r 2a

And this corresponds to the sixth automorphism in the above list of automorphisms found by replacing generators.

So we have demonstrated that all automorphisms of the group D3 are inner automorphisms.

We can now write down the above determined permutations (representing automorphisms) as cycles, in order to determine their periods :

The permutation

1 ==> 1
r ==> r
2 ==> r 2
a ==> r 2a
ra ==> a
2a ==> ra

can be expressed as the cycle (a    r 2a     ra), so the period is 3.

The permutation

1 ==> 1
r ==> r
2 ==> r 2
a ==> ra
ra ==> r 2a
2a ==> a

can be expressed by the cycle (a    ra    r 2a), so the period is 3.

The permutation

1 ==> 1
r ==> r 2
2 ==> r
a ==> a
ra ==> r 2a
2a ==> ra

can be expressed by the cycle (r    r 2) (ra    r 2a), so the period is 2.

The permutation

1 ==> 1
r ==> r 2
2 ==> r
a ==> r 2a
ra ==> ra
2a ==> a

can be expressed by the cycle (r    r 2) (a    r 2a), so the period is 2.

Finally, the permutation

1 ==> 1
r ==> r 2
2 ==> r
a ==> ra
ra ==> a
2a ==>r 2a

can be expressed by the cycle (r    r 2) (a    ra), so the period is 2.

The group of automorphisms of D3 therefore contains :

One element of period 1  (identity automorphism).
Two elements of period 3.
Three elements of period 2.

This distribution of periods is the fingerprint of the group D3, so the automorphism group of the group D3 is isomorphic to D3 :  Aut(D3) = D3 .

Now, having seen several examples of automorphisms, we characterize automorphisms once again, in order to get more and more to grip with this rather difficult concept :

Two groups, G and H, are isomorphic to each other, if there can be found a one-to-one correspondence between the elements of G and those of H, such that products are preserved (implying that the two groups have identical structure).
Of course a group G is always isomorphic to itself, because one can always find a one-to-one correspondence between the elements of G and those of (the same group) G, such that products are preserved, namely the identity permutation of the group elements.
If, moreover, there exists, in addition to this permutation, yet another one-to-one correspondence between the elements of G and those of (the same group) G, i.e. yet another permutation of G's elements, such that products are preserved, then we have an automorphism (other than the identity automorphism) of the group. The set of all possible automorphisms of the group G itself forms a group under successive application of those permutations, and is called the automorphism group of G, i.e. Aut(G). Sometimes this automorphism group is isomorphic with the group itself, Aut(G) = G, but very often it is a different group.

It can happen that the automorphism group of an Abelian group itself is non-Abelian. However, the automorphisms of a cyclic group (which is always Abelian) do form an Abelian Group which, however, need not be cyclic. For example, Aut(C12) = D2 . (See later on.)

If we have a cyclic group of prime order, all the elements, except the identity, can serve as generator of the group, and they all generate a different sequence of elements. So the number of possible automorphisms is one less than the order of the group, and, moreover, the automorphism group is in this case also cyclic :  Aut(Cp) = Cp-1 , when  p  is prime. In a relevant section of Part IV we have given an example of such a cyclic group of prime order, namely C7 , considered its automorphisms, and spoke about automorphism generally.

We will continue to study the automorphisms of the Abelian groups C12 and C6 x C2 .
The cyclic group of order 12 ( = composite order), i.e. the group C12 ,  can be represented by the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} under addition modulo 12. Because the order of this cyclic group is not prime, but composite, not all non-identity elements can serve as generators. Only those of period 12 can do so. These are the elements 1, 5, 7 and 11. When we now first generate the group elements by the generator 1, then generate them with the generator 5, then do the same with respect to the generator 7, and finally do it with the generator 11, we get four different sequences of group elements, which are thus permutations of the twelve group elements, and because they generated sequences they are automorphisms of our group C12 .
The first sequence we get by setting the element 0 (identity element), then setting the given element 1 (generator), and then repeat the generator till all the group elements are generated, which here means that we keep adding (modulo 12) 1 till all elements are generated.
The second sequence is obtained when setting 0, then 5 (generator), and then adding 5 (modulo 12), till all elements are generated.
In the same way the other two sequences are obtained. All in all we get four sequences, each consisting of the twelve group elements :

 0 1 2 3 4 5 6 7 8 9 10 11 0 5 10 3 8 1 6 11 4 9 2 7 0 7 2 9 4 11 6 1 8 3 10 5 0 11 10 9 8 7 6 5 4 3 2 1

These four sequences are thus permutations of the group elements, and they correspond to the following cycles  :

(1 5) (2 10) (4 8) (7  11).
(1 7) (3 9) (5  11).
(1  11) (2  10) (3 9) (5 7).

As can be seen, all these three cycles are of period 2.
Together with the identity permutation they form the group D2, so Aut(C12 = D2 .

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Next we take the direct product group C6 x C2 ,  defined by

6 = 1
a2 = 1
ar = ra

(BUDDEN, p. 423)

where  r  and  a  are independent generators (both -- or alternative pairs -- are needed to generate all group elements) of periods 6 and 2 respectively. In terms of these generators, the group (Also discussed in HERE in Part IX ) contains :

Six elements of period 6 (arranged here in inverse pairs) :

r,  r 5
ar,  ar 5
ar 2,  ar 4

Two elements of period 3 :

2,  r 4

Three elements of period 2 :

3,  a,  ar 3

To seek automorphisms, we replace the generators  r  and  a  by other pairs of independent generators of period 6 and 2.
Because there are six elements of period 6 and three of period 2, there must be 18 :

 r a r r3 r ar3 r5 a r5 r3 r5 ar3 ar a ar r3 ar ar3 ar5 a ar5 r3 ar5 ar3 ar2 a ar2 r3 ar2 ar3 ar4 a ar4 r3 ar4 ar3

Among these possible pairs there are, however, six pairs of dependent generators, i.e. two generators which can generate each other, but cannot generate all elements of the group. In the above listing these pairs are indicated by red coloring. Let us consider these six pairs, show that the members of those pairs are dependent on each other, i.e. one can generate the other, while -- in the calculations -- realizing that the group is Abelian, that is all products commute :

5,  r 3.
55 = r 10 = r 4,  and  r 45 = r 9 = r 3.

ar 5,  ar 3.
ar 5ar 5 = aar 10 = r 10 = r 4,  and  r 4ar 5 = ar 9 = ar 3.

ar 2,  a.
ar 2ar 2 = r 4,  and  r 4ar 2 = a.

ar 4,  a.
ar 4ar 4 = r 8 = r 2,  and  r 2ar 4 = a.

ar,  ar 3.
arar = r 2,  and  r 2ar = ar 3.

r,  r 3.
rr = r 2,  and  r 2r = r 3.

The twelve remaining pairs are pairs of independent generators. They generate twelve different sequences of group elements, which are -- because they are generated, and thus preserve products -- at the same time the twelve automorphisms of our group, i.e. of the group C6 x C2 .

Before we tabulate the twelve element sequences (automorphisms) generated by the twelve pairs of alternative independent generators, we show how -- as an example -- the pair (of independent generators)  (ar 2,  ar 3)  generates all elements of the group, and -- as an automorphism -- results in a certain sequence of those elements. But before that, we generate the original sequence of group elements by the original pair of generators  r  and  a  :

The original sequence of group elements is generated by the generators  r  and  a  as follows :

First, we have -- as being given -- the identity element, 1, which does not partake in any swapping of group elements in order to generate an automorphism, because its period, which is 1, is (always) unique.
Secondly we have
r,  which is given. Then we have
rr = 2,
2r = 3,
3r = 4,
4r = 5,
5r = r 6 = 1. Further we have
a,  which is given, and then we obtain
ar,  from multiplication with  r,
arr = ar 2,
ar 2r = ar 3,
ar 3r = ar 4,
ar 4r = ar 5,
ar 5r = ar 6 = a.

So the initial sequence of group elements, representing the identity permutation, and thus the identity automorphism, is :

1,  r,  r 2,  r 3,  r 4,  r 5,  a,  ar,  ar 2,  ar 3,  ar 4,  ar 5.

An alternative sequence of group elements can be generated by the independent pair of generators (ar 2,  ar 3) as follows :

First, of course, we have 1. And then
ar 2,  which is (now) given.
ar 2ar 2 = 4,
4ar 2 = a,
aar 2 = 2,
2ar 2 = ar 4,
ar 4ar 2 = 1.
ar 3  (given),
ar 3ar 2 = 5,
5ar 2 = ar 7 = ar,
arar 2 = 3,
3ar 2 = ar 5,
ar 5ar 2 = r 7 = r.

So with the pair (ar 2, ar 3) we have generated the alternative sequence of group elements :

1,  ar 2,  r 4,  a,  r 2,  ar 4,  ar 3,  r 5,  ar,  r 3,  ar 5,  r.

Seen as a permutation of group elements (with respect to the original sequence of those same elements) we can determine what  cycle  it represents :

1 ==> 1
r ==> ar 2
2 ==> r 4
3 ==> a
4 ==> r 2
5 ==> ar 4
a ==> ar 3
ar ==> r 5
ar 2 ==> ar
ar 3 ==> r 3
ar 4 ==> ar 5
ar 5 ==> r

This implies the cycle ( r   ar 2   ar   r 5   ar 4   ar 5 ) ( r 2   r 4 ) ( r 3   a   ar 3 ), and from this we can see that the automorphism, defined by this cycle, is of period  6.

In the same way we can determine the remaining ten alternative sequences (permutations) of the group elements, by generating them by the remaining alternative pairs of independent generators, and we can also determine their periods. The result will be the twelve automorphisms of the group C6 x C2 .  Each row -- representing a permutation (automorphism) concludes with the period of that permutation :

 1 r r2 r3 r4 r5 a ar ar2 ar3 ar4 ar5 1 1 r r2 r3 r4 r5 ar3 ar4 ar5 a ar ar2 2 1 r5 r4 r3 r2 r a ar5 ar4 ar3 ar2 ar 2 1 r5 r4 r3 r2 r a3 ar2 ar a ar5 ar4 2 1 ar r2 ar3 r4 ar5 a r ar2 r3 ar4 r5 2 1 ar r2 ar3 r4 ar5 r3 ar4 r5 a r ar2 3 1 ar5 r4 ar3 r2 ar a r5 ar4 r3 ar2 r 2 1 ar5 r4 ar3 r2 ar r3 ar2 r a r5 ar4 6 1 ar2 r4 a r2 ar4 ar3 r5 ar r3 ar5 r 6 1 ar2 r4 a r2 ar4 r3 ar5 r ar3 r5 ar 2 1 ar4 r2 a r4 ar2 ar3 r ar5 r3 ar r5 3 1 ar4 r2 a r4 ar2 r3 ar r5 ar3 r ar5 2

These twelve automorphisms form a group. From the distribution of the periods over these elements we can identify it as D6 ,  so Aut(C6 x C2) = D6 .

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We have noted that the group C2 x C2 x C2 has 168 automorphisms. The reason for this large number is that its seven elements og period 2 are, so to speak, indistinguishable. Now C3 x C3 has eight elements of period 3, these being four pairs of inverses :  pp-1 --  qq-1 --  rr-1 --  ss-1.  Evidently this group will also have a large number of automorphisms, since any pair of the eight elements may be taken as generators provided they are not inverses.
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Automorphisms of non-Abelian groups

In the case of non-Abelian groups, we may obtain all the inner automorphisms systematically by transforming all the elements of the group by each element in turn. An interesting example is provided by Q4 ,  and we use the notation  1, r, a, s, t, u, v, w,  for its elements. Its group table is :

We will now show that, and how, the group (elements) can be generated by the element pair  r  and  t,  both of period 4 :

 1 r a s t u v w period 1 1 r a s t u v w 1 r r a s 1 v t w u 4 a a s 1 r w v u t 2 s s 1 r a u w t v 4 t t u w v a s r 1 4 u u w v t r a 1 s 4 v v t u w s 1 a r 4 w w v t u 1 r s a 4

Table 18.1

Table of Q4
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A group table of a certain group G, can be seen as the complete definition of the group, and so is the above group table with respect to the group Q4.

We will now show, that, and how, all the group (elements) can be generated by the element pair  r  and  t,  both of period 4.
The identity element 1 is given, as also is the element  r  (and also the element  t ). When we now consider the product  r r,  we see, by checking the group table, that we get the element  a, and we then add this new element to what we already have, namely the elements  1  and  r  :

 1 r a 1 1 r a r r a a a

If we determine  ra,  i.e. now let the new found element  a  to be involved, we find, by consulting the group table, the new element  s,  and we must add it to what we already have :

 1 r a s 1 1 r a s r r a s a a s s

From the group table, Table 18.1 above, we see that  aa = 1,  and that  ar = s,  sr = 1,  rs = 1,  as = r,  sa = r,  and  ss = a.  Filling all this in will result in :

 1 r a s 1 1 r a s r r a s 1 a a s 1 r s s 1 r a

We will now let the second generator -- the element  t  -- to be involved. In the group table (of our group Q4 ) we see that  st = u,  so in addition to the element  t  (given), the new element  u  must be added :

 1 r a s t u 1 1 r a s t u r r a s 1 a a s 1 r s s 1 r a u t t u u

Further we see in the group table that  rt = v,  and  at = w,  and these two new elements can be added :

 1 r a s t u v w 1 1 r a s t u v w r r a s 1 v a a s 1 r w s s 1 r a u t t u u v v w w

Now all group elements of Q4 are generated, which means that we have shown that the elements  r  and  t  form a possible pair of generators of the group Q4 . To find all automorphisms we can consider all alternative pairs of elements, both (elements of such a pair) of period 4, to be an alternative generator pair, provided such a pair not being a pair of inverses (because in that case we do not have to do with two independent generators -- and it takes two independent generators to create the whole group). Before we do this, we first determine the special automorphisms that are present among the total of automorphisms :  Because the group is non-Abelian there are distinct inner automorphisms, and we can find these by transforming each element of the group by each element (as transforming element) into its conjugate. In left column of the next array we let  x  successively stand for the elements  1, a, r, s, t, w, u, v.  The rows that come after that first column are the transforms of the respective group elements.

 1x1-1 1 a r s t w u v axa-1 1 a r s t w u v rxr-1 1 a r s w t v u sxs-1 1 a r s w t v u txt-1 1 a s r t w v u wxw-1 1 a s r t w v u uxu-1 1 a s r w t u v vxv-1 1 a s r w t u v

In the above array we find three different group element sequences, representing three inner automorphisms. They are permutations of group elements and can be expressed as cycles, from which we can determine the periods of the permutations (In the permutations only the six elements of period 4 partake, because the elements  1  and  a  are not replaced by other elements, as can be seen in the above tabulation) :

So, in addition to the identity, we have three permutations of period 2, so the group of inner automorphisms of Q4 can be identified as D2 .

But the group has six elements all of period 4, so one expects a large number of automorphisms in addition to the inner automorphisms just found. All these new automorphisms are outer automorphisms. We will now set out to find them, by means of replacing generators by alternative generators. As we have seen, this procedure produces all the automorphisms of the group, inner ones and outer ones.
The elements  1  and  a  cannot serve as generators because they do not, by them being combined, create other elements beyond themselves. They form the subgroup {1, a} of the group Q4 . Further, the element  a  needs not to be explicitly introduced, it can be produced by the other elements of the group. Indeed, in Table 18.1 we can see that  ss = a,  tt = a,  uu = a,  vv = a,  and  ww = a.  Further we see that {1, r, a, s} is a subgroup. The combining of its elements does not bring us outside this subgroup, which means that no new elements are created. So if we want to generate all elements we need more than one generator. And in fact, as we showed above, two generators are sufficient. Such a pair can be chosen from all elements of the group, excluding the elements  1  and  a,  i.e. such a generator pair can be chosen from the set {r, s, t, w, u, v}, provided that such a pair is not a pair of inverses.
Above we have seen that one such pair, namely  r  and  t,  generates all the group elements. The resulting sequence of group elements can be considered as one of the group's possible automorphisms. If we relate all other automorphisms of the present group to this particular one, we can call the latter the identity automorphism. The other automorphisms (inner and outer) can be found by replacing this generator pair by another non-inverse pair from the set {r, s, t, w, u, v}. Let us illustrate this by replacing the pair (r  t) by the pair (w  u), and see what automorphism we obtain.
Because according to te group table of Q4  rs = 1,  we can say that  s = r-1.  In the same way  tw = 1  implies  w = t-1.  Further we can see that  u = tr,  and  v = rt.  So we can express the elements  r,  s,  t,  w,  u  and  v  all in terms of  r  and  t,  namely, respectively as  r,  r-1,  t,  t-1,  tr  and  rt.
In the table we can also see that  wt = 1  implies  w-1 = t,  and  uv = 1  implies  u-1 = v.  Further we see that  uw = s,  and  wu = r.
Let us now replace the generators  r  and  t  as follows :

r ==> w
t ==> u

Given the above relations between elements, taken from the group table, we can now determine the effect of the just mentioned replacement of the generators  r  and  t  by the generators  w  and  u  (reading from the top down in following the generation of the effect) :

 r s t w u v r r-1 t t-1 tr rt w w-1 u u-1 uw wu w t u v s r

The first row of the above array represents the initial sequence of the six group elements of period 4.
The second row depicts these same elements in terms of  r  and  t.
The third row executes the replacement of  r  by  w  and of  t  by  u.
The fourth row results from the above given relations between elements, and is the final outcome (effect) of the mentioned replacement of generators.

So the result is the following permutation of the six elements :

And this is equivalent to the cycle (r w v) (s t u), which implies that the period is 3.
So with  w  t  u  v  s  r  we have found (by the method of replacing generators, yielding all automorphisms) a first automorphism (apart from the trivial one, the identity automorphism). When we include the elements  1  and  a,  this first, and fully written out, automorphism is

1  a  w  t  u  v  s  r.

Indeed, because (See group table)  a = rr = ss = tt = uu = vv = ww,  the element  a  will not be affected by any swapping of the elements from the set {r, s, t, w, u, v}, so  a  remains  a,  and stays at the same place. And the element  1  cannot participate in any swapping-that-preserves-products and consequently yields an automorphism, because its period is unique (in all groups). So also the element  1  remains 1, and thus stays at the same place. So all automorphisms (of our group Q4 ) are sequences of group elements having the elements  1  and  a  at the same place within the sequence.

In the same way we will -- by the method of replacing generators -- determine all automorphisms (inner and outer) of our group Q4 . We start with determining all possible generator pairs, and then see what permutations (of group elements) they imply.

Because, as has been said, the elements  1  and  a  cannot be generators, we recrute the generator pairs from the set of the remaining group elements, i.e. the set {r, s, t, w, u, v}. In order to determine all the appropriate pairs, we first tabulate all possible pairs whatsoever, that can be formed by the elements of that set (pairs, consisting of two identical members are marked by red background coloring) :

 r s t w u v r rr rs rt rw ru rv s sr ss st sw su sv t tr ts tt tw tu tv w wr ws wt ww wu wv u ur us ut uw uu uv v vr vs vt vw vu vv

Now, in order to isolate those pairs that consist of two different and independent generators, we will first eliminate all pairs consisting of two equal members (already marked with red coloring in the above array) :

 r s t w u v r rs rt rw ru rv s sr st sw su sv t tr ts tw tu tv w wr ws wt wu wv u ur us ut uw uv v vr vs vt vw vu

Now some pairs consist of inverses (i.e. the members of such a pair are each other's inverse), and so do not represent two independent generators. These pairs must be eliminated. Let's determine them (with the help of the group table) :

rs = 1,  so  rs is a pair of inverses.
tw = 1,  so  tw is a pair of inverses.
uv = 1,  so  uv is a pair of inverses.

When,  generally,  xy  is a pair of inverses, then, of course,  yx  also is. So we have six pairs to be removed from our above array.
In the group table we see eight ones. One such one is the product of 1 and 1,  and another is the result of the product  aa. The remaining six are already accounted four, so we have found all the inverse pairs ocurring among the elements r, s, t, w, u  and  v.  Let's indicate the pairs of inverses in our array :

 r s t w u v r rs rt rw ru rv s sr st sw su sv t tr ts tw tu tv w wr ws wt wu wv u ur us ut uw uv v vr vs vt vw vu

We're now going to remove these inverse pairs :

 r s t w u v r rt rw ru rv s st sw su sv t tr ts tu tv w wr ws wu wv u ur us ut uw v vr vs vt vw

So we are left with twenty-four genuine generator pairs :

 rt rw ru rv st sw su sv tu tv wu wv tr ts wr ws ur us ut uw vr vs vt vw

Our initial generator pair was  (r  t),  and we replaced that pair by  (w  u),  and obtained a new sequence of group elements, namely  w, t, u, v, s, r,  representing one of the group's automorphisms (Recall that we're still considering the automorphisms of the group Q4 ).

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In the next document we will determine the remaining automorphisms (that automatically include inner as well as outer ones), by successively replacing the initial generator pair  (r  t) by the remaining pairs from the last obtained array of generator pairs.
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To continue click HERE for further group theoretic preparation to the study of the structure of three-dimensional crystals
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