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  • Note 13
    Note 13 A primitive lattice has a unit cell with points only at its vertices back to main text

    Original URL path: http://www.metafysica.nl/crystals_rev_note13.html (2016-02-01)
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  • Note 14
    a Silicon atom surrounded by four Oxygen atoms in such a way that the Silicon atom resides in the center of an imaginary tetrahedron at the corners of which are

    Original URL path: http://www.metafysica.nl/crystals_rev_note14.html (2016-02-01)
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  • Note 15
    the structure coincide with itself In this case it is a reflection in a plane followed by a translation along a line in that plane A screw axis is a rotation followed by a translation along the axis of rotation

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  • Note 16
    in addition to its vertices two ore more faces of the chosen unit cell contain a lattice point in their center It can also mean that in addition to its vertices the center of the chosen unit cell contains a

    Original URL path: http://www.metafysica.nl/crystals_rev_note16.html (2016-02-01)
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  • Derivation of the Crystal Classes : Explanation
    the drawing Of course this is not evident just like that but the next Figures will show that indeed the motif pair in stereograms represented by the projections of face poles and then depicted as small solid circles standing for a projection from above or small open circles standing for a projection from below the projection plane is horizontally repeated about the vertical axis Figure 11aa Part of a stereogram see 3 in Figure 11a in the main document showing two pairs of projections of face poles sometimes called motifs related by a 2 fold rotation axis Also the members of each pair are related to each other by a 2 fold rotation axis The next Figure shows that these pairs are repetitions of each other i e they are congruent Figure 11aaa This Figure depicts the same situation as the above Figure but now seen in a direction perpendicular to that of the latter The projections of the face poles are now depicted by what they represent motifs drawn as comma s We see the three 2 fold rotation axes a b and c See Figure 11aa depicted as blue solid ellipses The axis b relating the two motif

    Original URL path: http://www.metafysica.nl/draaisymmetrie.html (2016-02-01)
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  • 3-D Crystals XIV
    identity element which we denote by the symbol 1 Finally we know that any subgroup H from which cosets are formed is itself also a coset namely 1H H1 H Well now the proofs Suppose H is a subgroup of the group G and let the elements of H be 1 h 1 h 2 h 3 h m 1 where H is of order m Take an element x of G which is not in H and form the left coset xH x xh 1 xh 2 xh 3 xh m 1 Now the elements of this coset must all be distinct for if xh r xh s then the cancellation law would lead to h r h s which is false since the elements of the subgroup must be distinct Moreover H and xH cannot possibly have any element in common For suppose that xh r h s Then xh r h r 1 h s h r 1 or x h s h r 1 But h r 1 is in H and so is the product h s h r 1 since H is a sub group But we deliberately chose x not to be in H so we have reached a contradiction Therefore H and xH are fully disjoint Now it is possible that the union of H and xH is equal to G i e to the whole group which would mean that all the elements of G have appeared either in H or in xH If this is the case then the order of G would be 2m since both H and xH each contain m elements and Lagrange s Theorem would then be proved i e the order of the subgroup H which stands for any subgroup of the group G is a factor of the order of the whole group G If however there are still elements in G which are not in H and not in xH choose any one of these y and form the left coset yH y yh 1 yh 2 yh 3 yh m 1 The cancellation law again convinces us that these are all distinct but we must still guard against appearances of elements already listed in the first two cosets H and xH An argument similar to the one which we used to show that there was no overlapping between H and xH will be used Suppose if possible that xh r yh s Then y xh r h s 1 and since h r h s 1 is an element of H we have y is an element of xH which is contrary to our certain knowledge that y is not in xH Thus xH and yH are fully disjoint Also H and yH are fully disjoint because when we suppose the contrary namely that h r yh s then again h r h s 1 y and since h r h s 1 is an element of H y would be in H which also is contrary to our knowledge that it is not so Thus there is no overlapping between these three cosets H xH and yH they are disjoint and contain 3m distinct elements of the group If by now the group is exhausted its order is 3m and Lagrange s Theorem is proved i e the order of the subgroup H is a factor of the order of the whole group G If not choose an element z not in H xH nor yH and form the left coset zH The argument proceeds as above at each stage it being necessary to prove a no repetitions in the newly formed coset and b no duplications in the new coset of any elements in the previously formed cosets Finally if the group is finite and after say r cosets H xH yH zH have been formed there remain no unused elements of G anymore from which to form new cosets we conclude that G contains exactly rm elements so that m must be a factor of the order n of the group and Lagrange s Theorem is proved The number of cosets r n m is called the index of the subgroup in the group G Of course when above we selected x y z to form the new cosets we had a free choice on each occasion from all the remaining elements of the group and had we made different choices we should have arrived at the same cosets but in a different order We might for example have selected y first then x In this case the coset yH would have been second H yH instead of third H xH yH on the list Or we might have selected the element zh 7 with which to form our second coset In that case we should have had zh 7 zh 7 h 1 zh 7 h 2 zh 7 h 3 zh 7 h m 1 i e z h 7 h 7 h 1 h 7 h 2 h 7 h 3 h 7 h m 1 But the set between now obtained is simply a rearrangement of the elements of the subgroup H so that the left coset of H by zh 7 is in fact zH As a final assurance not only that the cosets are distinct but also that every element in the group appears in some coset this is obvious since if g is any element whatever of G then it most certainly appears in the coset gH being the product of itself with the identity which is undoubtedly in H The above consideration points to the general result that if y is any element in xH then xH and yH are the same set Indeed y must be in yH where it is multiplied with the identity and because as we have seen left cosets cannot overlap so they must be identical The same goes for right cosets The main lines of the above argument are still sound when applied to infinite groups so far as the proof of the disjoint property of the cosets is concerned But of course either r or m or both must now be infinite and we do not have any result for infinite groups corresponding to Lagrange s Theorem BUDDEN p 347 Note that the non overlapping disjoint property of cosets i e that xH and yH do not have any element in common is entirely dependent upon H being a subgroup For taking the group table for D 6 See below suppose H 1 p which is not a subgroup because it is not closed pp is not in H Then Ha 1 p a a pa a b and Hc 1 p c c pc c a and we see that the element a occurs in both sets i e these sets are not disjoint 1 p p 2 a b c x y z u v w 1 1 p p 2 a b c x y z u v w p p p 2 1 b c a y z x v w u p 2 p 2 1 p c a b z x y w u v a a c b 1 p 2 p u w v x z y b b a c p 1 p 2 v u w y x z c c b a p 2 p 1 w v u z y x x x y z u v w 1 p p 2 a b c y y z x v w u p p 2 1 b c a z z x y w u v p 2 1 p c a b u u w v x z y a c b 1 p 2 p v v u w y x z b a c p 1 p 2 w w v u z y x c b a p 2 p 1 Table 14 2 Table of D 6 On the other hand 1 p p 2 is a sub group which can be seen by the fact of its closure the other group requirements are then automatically fulfilled because it is part of an existing overall group structure The cosets are indicated by colors and we see that they are disjoint Below we have isolated such a part of the group table representing the coset x y z x y z y z x z x y Note that the order of elements in a set and thus also in a coset does not make any difference so x y z y z x z x y This coset can be written as x 1 p p 2 y 1 p p 2 z 1 p p 2 Cosets and Equivalence Classes Given a subgroup H of a group G two elements of G not in H may or may not lie in the same coset of H For example taking H to be the subgroup 1 a b c d f of the group S 4 See for this group Table 13 2 of the previous document the elements p and s are in the same right coset whereas p and w are not Afgain in the same group we have the subgroup 1 i m with the right coset a k t and the left coset a j n So for example a and t do occur in the same right coset but are not in the same left coset The relationship between two elements of belonging to the same coset of a particular subgroup is an important one It is an equivalence relation An equivalece relation between two elements of a set means that the elements have something in common by reason of which they belong to the same class If we would classify all words according to the number of letters of which they are built then we can say that the word see belongs to the Class 3 of all words The word house then belongs to the Class 5 Also can we say that the words bad sad and red belong with the word see in same Class We say that the set of words has been partitioned into equivalence classes membership of these classes being described by the equivalence relation defined thus Two words belong to the same equivalence class if they have an equal number of constituent letters The relationship between the words is that of consisting of an equal number of letters and we shall denote this relationship by the symbol so that in the present case BAD SAD but SAD HOUSE in which is a negation An Equivalence Relation on a set must satisfy the following properties a a Reflexive a b implies b a Symmetric a b and b c implies a c Transitive Where a b and c are any members of the set These three requirements are also sufficient and constitute a definition for an equivalence relation The following Theorem can be easily proved An Equivalence Relation partitions a set into disjoint subsets It can be demonstrated that in a group membership of the same coset left or right relative to a given subgroup is an equivalence relation Further it is useful to show that we can express membership of the same coset in a slightly different manner Suppose a and b belong to the same left coset xH Note that a coset is not necessarily a group or subgroup then it is possible to find elements h r h s in H such that a xh r and b xh s Therefore a 1 b xh r 1 xh s h r 1 x 1 xh s generally pq 1 q 1 p 1 h r 1 h s becuse x 1 x 1 so that since h r and h s belong to the subgroup H so does h r 1 and so does the product h r 1 h s Thus a 1 b is an element of H So we have shown that if a and b belong to the same left coset then a 1 b is an element of H And now the other way around This condition is also sufficient to guarantee membership of the same left coset For a 1 b being an element of H implies that a 1 b h where h is an element of H Now we can say that the statement a 1 b h is the same as the statement aa 1 b ah and this is the same as the statement b ah so that b is in aH And because H is a sub group it contains the identity element which implies that aH contains a because it contains a1 which is equal to a and we just saw that b also is in aH thus both elements a and b are in the same left coset In the same way a necessary and sufficient condition for a and b to belong to the same right coset is ab 1 is an element of H Multiplication of subsets If A a 1 a 2 a 3 a r and B b 1 b 2 b 3 b s are two sub sets of a group G we shall use the notation AB to denote the set consisting of all possible products a 1 b 1 a 2 b 1 a r b 1 a 1 b 2 a 2 b 2 a r b 2 a r b s the number of which is clearly rs though some values may be repeated We can set out these products as an array in a multiplication table b 1 b 2 b 3 b s a 1 a 1 b 1 a 1 b 2 a 1 b 3 a 1 b s a 2 a 2 b 1 a 2 b 2 a 2 b 3 a 2 b s a r a r b 1 a r b 2 a r b 3 a r b s The object of the notation AB is for the purpose of abbreviation We shall call AB the product set of the subsets A and B Note that in general AB and BA are different unless the group is Abelian Note also that we have a binary operation on subsets which is non commutative We shall see and have already seen where we discuss homomorphism how cosets may themselves form a group under this operation Now consider a sub group H 1 h 1 h 2 h 3 h m 1 If we form the product set of H with itself which we may denote H 2 it is clear that since H is a sub group every element of the form h r h s is in H and so there are not m 2 different products but only m distinct products being the elements of H itself We may say in fact that H 2 H This condition does in fact express the closure of H under internal products and this means that H 2 H is also a sufficient condition for a sub set H to be a sub group of a finite group according to the theorem A subset H of a finite group G is a subgroup of G if and only if it is closed under the operation of the group Products of subsets from a finite group We now take some examples BUDDEN pp 357 from the group Q 6 whose table is shown below In order to clearly distinguish the numeral 1 from the letter l we indicate the latter as l 1 a b c d f g h j k l m period 1 1 a b c d f g h j k l m 1 a a l 1 d f g h j c m k b 6 b b 1 m j c d f g h l a k 6 c c j d k l a 1 b m g h f 4 d d c f m k l a 1 b h j g 4 f f d g b m k l a 1 j c h 4 g g f h 1 b m k l a c d j 4 h h g j a 1 b m k l d f c 4 j j h c l a 1 b m k f g d 4 k k m l g h j c d f 1 b a 2 l l k a f g h j c d b m 1 3 m m b k h j c d f g a 1 l 3 Table 14 3 Table of Q 6 Let A be the sub set a d m h then A 2 AA is given by the elements in the table below i e A 2 is the set l f b j c k g 1 and we may note that even if B were limited to the elements a d the product AB would still be the same set the first two rows of the table containing all eight elements of A 2 Also here In order to clearly distinguish the numeral 1 from the letter l we indicate the latter as l a d m h a l f b j d c k g 1 m b j l f h g 1 c k Again if we take C a b k and D d g j then the product of C and D contains the elements in the two tables below d g j a f h c b c f h k h c f CD a b k d c f h g f h c j h c f DC In this case it seems

    Original URL path: http://www.metafysica.nl/turing/d3_lattice_14.html (2016-02-01)
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  • 3-D Crystals XV
    off Conjugacy classes If for two elements y and z of a group it is possible to find an x in the group such that z xyx 1 or that y x 1 zx or that zx xy or that x 1 z yx 1 then as we have seen y and z are described as a pair of conjugate elements If y and z are both given there may not exist in the whole group a single element x such that zx xy and in this case y and z are not conjugate Thus conjugacy is a relation between elements of a group It is evidently reflexive since there is always an element in the group that transforms a given element into itself namely the identity y 1y1 1 and we have already shown that it is a symmetric relation which justifies the name conjugate To show that it is also transitive which then finally would establish the relation is conjugate to to be an equivalence relation suppose that the relation is conjugate to is denoted R Then if yRz and zRw we require to demonstrate that yRw Suppose we have a group G Now yRz implies yx 1 x 1 z for some x 1 in G And zRw implies zx 2 x 2 w for some x 2 in G From zx 2 x 2 w we can deduce zx 2 x 2 1 x 2 wx 2 1 an this implies z x 2 wx 2 1 Making use of this we can now say yx 1 x 1 x 2 wx 2 1 so that yx 1 x 2 x 1 x 2 wx 2 1 x 2 which implies yx 1 x 2 x 1 x 2 w Thus if x 1 x 2 x 3 the group G being closed we have yx 3 x 3 w and therefore yRw We have thus proved that conjugacy is an equivalence relation We may now proceed to set up equivalence classes For a fixed element y all those elements z 1 z 2 z 3 which are conjugate to y will be in the same equivalence class with y We can state the concept of a conjugacy class in a different but equivalent way Those elements which belong to the same conjugacy class may be transformed into each other Note that in an Abelian group in which all elements commute the question of conjugacy is trivial The only element which is conjugate to y is y itself so each element is the sole member of its particular conjugacy class In all groups including non Abelian groups it is obvious that the identity is going to be in a class on its own it cannot be transformed into another element of the group because it has a different nature from all the others Indeed it is the only element of period 1 and since we have seen that the period of an element is unchanged by conjugation by any other element it is clear that no other element of the group can possibly be conjugate to the identity An element is said to be self conjugate when it commutes with every element of the group i e belongs to the center See later for the concept of center For if xy yx for all y in the group then x yxy 1 for all y i e x is transformed into itself by all the elements of the group no other element of the group is conjugate to it its conjugacy class contains it alone Finding conjugacy classes Let us show how we can find conjugacy classes in a given group Now it is evidently going to be a very long and tedious process to work out xyx 1 for every possible pair of values of x and y for a group of order twelve up to 144 such computations would be necessary But the work can be greatly reduced To begin with we may easily find pairs of conjugate elements as follows find any element p in row x say p xy Next find p in column x so that p zx Then p xy zx so that yRz This is a suitable procedure only when the transforming element x is given It becomes a different matter when we want to know whether two given elements are conjugates Suppose we are trying to discover whether or not two given elements y and z are conjugate See next Figure Figure 6 Finding out whether the two elements z and y are conjugates This means that we are seeking an x in the group such that zx xy u say The element u will occur in row z column x and also in column y row x To find it we look along row z and down column y at each element in turn thus in such a way that we compare elements which find themselves every time at the same location in row z and in column y We first compare the first element of row z with the first element in column y and if they are not identical we compare the second element of row z with the second element in column y etc See diagram above till we find a pair the same i e two identical elements In the diagram it is supposed that the first five elements in row z and column y disgree till we reach the sixth which are both u s The corresponding elements that are being compared are linked by red lines in the diagram This common element u has the property that it lies in row z column x say and also in column y row x so that zx xy or z xyx 1 and zRy It would also be possible to have used column z in conjunction with row y since zRy is equivalent to yRz Having found a single example of zx xy we need go no further because we now know that y and z are conjugate elements We will however do so in the example that follows because it is instructive The example will be the group A 4 its table is 1 a b c p p 2 q q 2 r r 2 s s 2 period 1 1 a b c p p 2 q q 2 r r 2 s s 2 1 a a 1 c b s r 2 r s 2 q p 2 p q 2 2 b b c 1 a q s 2 p r 2 s q 2 r p 2 2 c c b a 1 r q 2 s p 2 p s 2 q r 2 2 p p r s q p 2 1 r 2 b s 2 c q 2 a 3 p 2 p 2 s 2 q 2 r 2 1 p c s a q b r 3 q q s r p s 2 b q 2 1 p 2 a r 2 c 3 q 2 q 2 r 2 p 2 s 2 c r 1 q b s a p 3 r r p q s q 2 c s 2 a r 2 1 p 2 b 3 r 2 r 2 q 2 s 2 p 2 a s b p 1 r c q 3 s s q p r r 2 a p 2 c q 2 b s 2 1 3 s 2 s 2 p 2 r 2 q 2 b q a r c p 1 s 3 Table 15 2 Table of A 4 Now one way in which the work can be reduced is that we know that yRz implies that y and z have the same period as we found out ABOVE Therefore it is a waste of time even to make the above check for elements of different periods In the group A 4 we begin by checking whether or not aRb i e whether the element a is conjugate to the element b note both a and b are of period 2 The best way to do this is to take row a and column b and write the latter horizontally below the former Above these we will place the group elements as they are arranged along the top of the group table as well as along its left side As such this top row does not belong to the group table proper 1 a b c p p 2 q q 2 r r 2 s s 2 Row a a 1 c b s r 2 r s 2 q p 2 p q 2 Column b b c 1 a s q 2 r p 2 q s 2 p r 2 How must we read and interpret this diagram We know that a product xy must be read off from a group table as follows Look in row x and then in column y and see what element we find at the intersection On the other hand if we want to know the factors of a product z given somewhere in the group table we determine first the row in which that product is say that it is row x and then we determine the column in which that product is say it is y And now we know that z xy When we want to know whether a is conjugate to b we look into row a and column b which we align horizontally as done above and see whether there are two identical elements products which are situated at corresponding locations of the row and column for example both at the third entry i e whether we can find in the above diagram identical products one precisely on top of the other one upper product and one lower product If we have found two such products then we know that a and b are conjugated So when we want to identify in our present example the factors of the upper product s which is in row a we see that it is at the same time in column p so we have s ap If on the other hand we want to determine the factors of the lower product s present in column b we must now interpret the arrangement of group elements written down on top of the above diagram as the headers of the twelve rows In this way we find that the lower product s is in row p while it is at the same time in column b and so we find s pb So from the above diagram we see that ap pb s so app 1 pbp 1 and this is equivalent to a pbp 1 which means that a and b are conjugates that is aRb As has been said this is sufficient for establishing aRb but it is instructive to analyse the above diagram further In it we see that there are some more identical products elements at corresponding locations of the row and column On the basis of this we can establish that aq qb r also implying aRb ar rb q also implying aRb Although we already know that aRb bRa we can check it out by using instead of row a and column b row b and column a 1 a b c p p 2 q q 2 r r 2 s s 2 Row b b c 1 a q s 2 p r 2 s q 2 r p 2 Column a a 1 c b r s 2 s r 2 p q 2 q p 2 The identical elements at corresponding positions are indicated red They are s 2 r 2 q 2 p 2 From this we can see that bp 2 p 2 a s 2 implying bp 2 p 2 1 p 2 a p 2 1 which is equivalent to b p 2 a p 2 1 meaning bRa Further we see that bq 2 q 2 a r 2 also implying bRa br 2 r 2 a q 2 also implying bRa bs 2 s 2 a p 2 also implying bRa In the same way we can verify aRc In order to do so we take row a and column c of the group table of the group A 4 1 a b c p p 2 q q 2 r r 2 s s 2 Row a a 1 c b s r 2 r s 2 q p 2 p q 2 Column c c b a 1 q r 2 p s 2 s p 2 r q 2 From this we can see that ap 2 p 2 c r 2 implying aRc aq 2 q 2 c s 2 implying aRc ar 2 r 2 c p 2 implying aRc as 2 s 2 c q 2 implying aRc And we know that cRa In the same way we can find that bRc and of course cRb And now we know that a b and c are each others conjugate and we will see that they are not conjugates of any other element of the group A 4 so we have the conjugacy class a b c we will continue to find more conjugacy classes of the group A 4 in due course Figure 7 The conjugate class a b c of the group A 4 In addition to being conjugate to each other these elements are also conjugate to themselves not expressed in the Figure Now it so happens that this class contains all the elements of period 2 But do not imagine that just because two elements have the same period they are bound to be conjugate For instance consider p and q 2 both of period 3 In order to find out whether they are conjugates we investigate row p and column q 2 1 a b c p p 2 q q 2 r r 2 s s 2 Row p p r s q p 2 1 r 2 b s 2 c q 2 a Column q 2 q 2 s 2 r 2 p 2 b s 1 q a p c r Here we do not see the same products at corresponding locations in the row and the column so we conclude that p and q 2 are not conjugate elements despite the fact that they have the same period On the other hand p and q are conjugate there being three occasions for detecting it as we go along row p and column q 1 a b c p p 2 q q 2 r r 2 s s 2 Row p p r s q p 2 1 r 2 b s 2 c q 2 a Column q q r p s r 2 c q 2 1 s 2 b p 2 a From this we can see that pa aq r implying pRq pr rq s 2 also implying pRq ps 2 s 2 q a also implying pRq It turns out that pRqRrRs giving a conjugacy class p q r s and also which can be verified in the same way that p 2 Rq 2 Rr 2 Rs 2 giving the class p 2 q 2 r 2 s 2 The group A 4 is therefore partitioned into the following conjugacy classes 1 a b c p q r s p 2 q 2 r 2 s 2 The identity element is always transformed into itself for any element y of whatever group we have y1y 1 yy 1 1 The results obtained above i e the discovering of all the conjugacy classes of the group A 4 should be interpreted in the context of the direct symmetries of the Regular Tetrahedron We have seen that p q r and s are in one class and p 2 q 2 r 2 and s 2 in another Interpreted as symmetries of the Regular Tetrahedron these conjugate elements represent rotations through 120 0 which implies a period of 3 one class being clockwise and the other class anticlockwise Those which belong to the same conjugacy class e g p q r s may be transformed into each other each being a rotation about an altitude of the tetrahedron in the same sense Note however that while in A 4 the elements of period 3 fall into two classes in S 4 the group consisting not only of all even permutations of four symbols as is the case in A 4 but also of all odd permutations of those four symbols these eight elements are all in the same class More generally it is true that rotations of the same magnitude and sense about different axes are conjugate if an operation exists in the group which would map one axis into the other Thus in A 4 for example p and q are conjugate with xpx 1 q and the transforming element x is one which maps the directed altitude p into the directed altitude q of the tetrahedron But there is no element in the group which transforms p into q 2 in other words there is no operation in the group which transforms the p altitude into the q altitude and reverses its sense rotating backwards about the q altitude is the operation q 2 Before we continue to discuss the conjugacy of group elements and especially say some more about the transforming elements we will explain two important group theoretical concepts namely center and centralizer Center of a non Abelian group Sometimes in a non Abelian group we find an element say k which commutes with every element of the group Elements of the group which have this property form a special subset known as the center of the group It is clear that the identity always belongs to the center because it commutes with every element of the group 1y y1 where y is any element of the group In an Abelian group all elements commute with all the others so there the center is the whole group itself We shall now demonstrate that this set of elements which commute with every element of the group do in fact form a sub group To establish this we only need to demonstrate closure Now suppose that H is the center of the group G i e it is the set of all those elements of G which commute with every element of G We note first that the center of every group is non empty since we saw that the identity element commutes with every element of a group A group whose center consists of the identity only is described as having a trivial center Suppose that h 1 and h 2 are any two elements of H Now we must prove that h 1 h 2 also belongs to H i e that H is closed under the group operation and consequently is a subgroup of G Let x be any element of G Then h 1 being an element of H implies h 1 x xh 1 h 2 being an element of H implies h 2 x xh 2 Then h 1 h 2 x h 1 h 2 x h 1 xh 2 h 1 x h 2 xh 1 h 2 x h 1 h 2 Thus h 1 h 2 commutes with x any element of G and so h 1 h 2 is an element of H by definition Therefore the center H is a subgroup of G and is of course an Abelian subgroup Although a group such as Q 6 is a group of order 12 it has a very small center consisting of the identity and one more element i e it is of order 2 There are some groups whose center contains only the identity It can be proved for example that the center of S n group of all permutations of n symbols is merely the identity for all n greater than 2 and this may be surprising Another way of saying this is that there is no permutation of n symbols other than the identity which commutes with every other permutation or that for any given permutation one can always find a permutation which does not commute with it BUDDEN p 218 Normalizers and centralizers If we select one particular element a of a group G and write down the set of all elements which commute with the given element i e the set x ax xa where x is element of G meaning the set of all elements x such that ax xa this set is called the centralizer or normalizer of a and we can prove that it is bound to be a subgroup For if x 1 and x 2 are in the centralizer of a then ax 1 x 1 a and also ax 2 x 2 a Hence x 1 x 2 a x 1 x 2 a x 1 ax 2 x 1 a x 2 ax 1 x 2 a x 1 x 2 Thus x 1 x 2 also commutes with the given element a and so is in the normalizer of a So closure is established and with it that the centralizer is a subgroup Similarly of course x 2 x 1 is also in the centraliser because now we know that the centralizer is a group but note that x 1 x 2 is not necessarily equal to x 2 x 1 so that one should not expect the centralizer to be Abelian Now we shall contrinue our discussion concerning conjugates The transforming elements Now when we were looking for conjugacy classes in the group A 4 we did not bother to check say row p against column p because we know that pRp i e conjugacy is reflexive which means that there is always some element namely the identity that transforms p into itself 1p1 1 1p1 p1 p It would be interesting to note which elements of a group can transform an element into itself In the case of p in the group A 4 we have 1 a b c p p 2 q q 2 r r 2 s s 2 Row p p r s q p 2 1 r 2 b s 2 c q 2 a Column p p s q r p 2 1 s 2 c q 2 a r 2 b We can see that p1 1p p implying p11 1 1p1 1 and this is equivalent to p 1p1 1 and this means pRp pp pp p 2 implying ppp 1 ppp 1 and this is equivalent to p ppp 1 and this means pRp pp 2 p 2 p 1 implying pp 2 p 2 1 p 2 p p 2 1 which is equivalent to p p 2 p p 2 1 and this means pRp Hence p is transformed into itself by the elements 1 p and p 2 Or xpx 1 p when x 1 x p or x p 2 Again consider the values of x for which c xcx 1 1 a b c p p 2 q q 2 r r 2 s s 2 Row c c b a 1 r q 2 s p 2 p s 2 q r 2 Column c c b a 1 q r 2 p s 2 s p 2 r q 2 From this we see that c1 1c c implying c 1c1 1 which means cRc ca ac b implying c aca 1 which means cRc cb bc a implying c bcb 1 which means cRc cc cc 1 implying c ccc 1 which means cRc So c is self conjugate under transformation by the elements 1 a b and c But cp r and pc q so cp is not equal to pc which implies that c is not equal to pcp 1 which means that c is not transformed into itself by the element p So c is self conjugate only with respect to certain transforming elements Only if it would be transformed into itself by all the elements of the group it would be truly self conjugate see BUDDEN p 374 5 You will realize that for a given element y of any group if y is self conjugate under transformation by an element x so that y xyx 1 then this means that yx xy in other words that x will commute with y All those elements which do commute with the given element y constitute what we have described as the centralizer or normalizer of y and we saw that the centralizer is a subgroup Confirmation of this is seen in the case of A 4 by noting that the centralizer of p is the subgroup 1 p p 2 See row p and column p above where we see the elements x of the group A 4 namely 1 p and p 2 such that xpx 1 p which is equivalent to xpx 1 x px which in turn is equivalent to xp px The centralizer of c is the subgroup 1 a b c See row c and column c above Indeed we can see in the group table of A 4 ABOVE that pp p 2 pp 2 1 p1 1p p p 2 1 1p 2 p 2 p 2 p 2 p 11 1 So all possible products of the subset 1 p p 2 are again elements of this subset so this subset is a subgroup For the set 1 a b c it is directly evident from the group table that it is a subgroup Conjugacy classes and cosets Now cosets of a group only exist in relation to some pre selected subgroup Nevertheless cosets seem somehow to be emerging in the considerations concerning conjugacy even though no subgroups are identified apart from what we did in the last paragraph With respect to the group A 4 we can see that those elements which transform r into p by the relations pb br s pq 2 q 2 r b ps sr q 2 are b s and q 2 Now b s q 2 is in fact one of the cosets in A 4 being a left coset of 1 r r 2 b s q 2 b 1 r r 2 s 1 r r 2 q 2 1 r r 2 or a right coset of 1 p p 2 b s q 2 1 p p 2 b 1 p p 2 s 1 p p 2 q 2 All this can be verified by consulting the group table of A 4 Now just as subgroups arise from a consideration of centralizers so their cosets arise in an analogous way from a consideration of conjugacy classes We now proceed to investigate this more fully BUDDEN pp 380 Now we will expound the relationship between coset centralizer and conjugation First we shall just state what this relationship is and then prove it Recall that when we have the relation pbp 1 c we say that the element b is transformed by the transforming element p and when this results in the element c then we say that b is conjugate to c which can be denoted by bRc When we have a centralizer H of an element x in a group G then each element h of the centralizer transforms x into itself because xh hx which implies xhh 1 hxh 1 which is equivalent to x hxh 1 And when we take as transforming element an element z not in H then x will not necessarily be transformed into itself i e zxz 1 is not necessarily equal to x Let us say it is y And when we take as transforming elements the elements of the left coset of H by the element z i e zH then all these elements will transform x into y So given the coset zH where H is the centralizer of x all the elements of this coset will transform the element x into a same element namely y when indeed zxz 1 y And conversely if we have kxk 1 y zxz 1 then k is an element of zH i e of the left coset of H by the element z Thus we reason from coset to conjugation and then from conjugation to coset Well let us now prove all this Let a be a fixed element of a group G and let H be the centralizer of a i e the set H 1 h 1 h 2 h 3 such that ah ha for all h element of H which implies a hah 1 meaning that all elements h of H transform the element a into itself Now let b be a second fixed element of G which is not in H and let bab 1 c Consider the transform of a by any element bh of the left coset bH bh a bh 1 bhah 1 b 1 because bh 1 h 1 b 1 b hah 1 b 1 bab 1 because since h is an element of H hah 1 a c Hence all elements of the left coset bH transform a into c For example in A 4 See the group table of A 4 ABOVE the centralizer of r is the subgroup 1 r r 2 and its left coset by b is b 1 r r 2 b s q 2 which can be read off from the group table and each of these three elements transforms r into p brb 1 brb p because b 2 1 which is equivalent to bb 1 implying b b 1 Further rb q bq p srs 1 srs 2 p q 2 r q 2 1 q 2 rq p because q 2 1 q so q 2 r q 2 1 q 2 rq q 2 s 2 p We have shown above the example just given that in the general case all the elements of the left coset bH transform a into bab 1 c We shall also prove the converse that if xax 1 c then x is an element of bH where H is the centralizer of a For c bab 1 xax 1 given so that ab 1 b 1 xax 1 which implies a b 1 xax 1 b b 1 x a b 1 x 1 because genrally ab 1 b 1 a 1 This means by definition that b 1 x lies in the centralizer of a i e b 1 x is an element of H so finally x is an element of bH because b 1 x being an element of H is equivalent to bb 1 x being an element of bH which in turn is equivalent to x being an element of bH which was what we wished to prove We may add that since all the cosets of a particular subgroup contain the same number of elements it follows that the number of elements which will transform a given element a alike i e every time into the same element is equal to the order of the centralizer of a The transform of a given subgroup conjugate subgroups When H is a given subgroup of a group G we now consider the transform of H by a fixed element a of G not in H This will be denoted aHa 1 and refers to the set a1a 1 ah 1 a 1 ah 2 a 1 ah 3 a 1 where 1 h 1 h 2 h 3 are the elements of H We shall demonstrate that this new set is also a subgroup of G and to do this in the case of finite groups it will only be necessary to establish closure Well if ah r a 1 and ah s a 1 are any two elements of aHa 1 then ah r a 1 ah s a 1 ah r a 1 a h s a 1 ah r h s a 1 because a 1 a 1 But h r and h s are in the subgroup H and thus also their product h r h s and we can call this product h t And now we know that ah t a 1 which is equal to the product ah r a 1 ah s a 1 is undoubtedly one of the elements of aHa 1 Thus aHa 1 is a subgroup of G It will be of the same order as H provided there are no repetitions of elements This possibility is ruled out by the fact that ah r a 1 ah s a 1 would imply that h r h s because applying the cancellation law two times ah r a 1 ah s a 1 is equivalent to a 1 ah r a 1 a 1 ah s a 1 which implies h r a 1 h s a 1 which is equivalent to h r a 1 a h s a 1 a implying h r h s Therefore H and aHa 1 are subgroups of the same order and the group aHa 1 is called the conjugate subgroup of the group H both subgroups of G by the element a The remaining question is Are they the same subgroup Well they re generally not For instance in the group A 4 See Table 15 2 we have the subgroup 1 p p 2 Calling this H then pHp 1 p1p 1 ppp 1 pp 2 p 1 1 p pppp 1 1 p pp 1 p p 2 H However aHa 1 1 q q 2 and this is not the subgroup H That aHa 1 is indeed equal to 1 q q 2 can be seen as follows aHa 1 a1a 1 apa 1 ap 2 a 1 and because a 2 1 See table we can say a 1 a Therefore we have aHa 1 a1a apa ap 2 a which is See table equal to 1 q q 2 We now prove that conjugate subgroups are isomorphic Suppose we have a group G with H being one of its subgroups Suppose further that k is an element of G Then the conjugate of the subgroup H by the element k is the subgroup kHk 1 Above we have established that the order of H and its conjugate is the same So we can set up a 1 1 one to one correspondence between their respective elements Let this 1 1 correspondence be the mapping PHI from H to kHk 1 where PHI means that we assign to each element of H the conjugate by the element k of that element and that conjugate is an element of aHa 1 Let p and q be any two elements of H Then PHI p kpk 1 and PHI q kqk 1 Now PHI p PHI q kpk 1 kqk 1 kp k 1 k qk 1 kpqk 1 PHI pq which means that products are preserved when we go from H to kHk 1 and establishes the isomorphism between them Clearly in the general case a subgroup H is transformed into itself by any of its own elements i e hHh 1 H when h is an element of H For if h r h s are any two elements of H then because H is a group h s 1 the inverse of h s must also be an element of H and so is the product h

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  • 3-D Crystals XVI
    I and PHI kk 1 I PHI k 1 hence I PHI k 1 I so PHI k 1 I which means that k 1 belongs to K and the inverse of each element k of K is contained in K Therefore K is a subgroup Associativity is already guaranteed because we have to do with the same group operation To show that K is moreover a normal subgroup we must demonstrate that the left and right cosets of K coincide i e that if x is any element whatsoever in G then xK Kx or equivalently xKx 1 K Well xKx 1 is the set xk 1 x 1 xk 2 x 1 xk 3 x 1 where k 1 k 2 k 3 are elements of K and we must now show that this set is equal to K in which it is not necessarily so that k 1 xk 1 x 1 k 2 xk 2 x 1 etc we must show that any of these elements are contained in K i e that xk r x 1 is element of K And we do this by showing that xh r x 1 is mapped by PHI in the element I of the image group In other words we must show that for any element k r in K the transform of k r by the element x that is xk r x 1 is mapped by PHI in I and so must also be in K Now PHI xk r x 1 PHI x PHI k r PHI x 1 properties of homomorphism But PHI k r I since k r is element of K Hence PHI xh r x 1 PHI x I PHI x 1 PHI x PHI x 1 PHI xx 1 property of homomorphism PHI 1 I as proved above So PHI xh r x 1 I which indeed means that xh r x 1 belongs to K which establishes that xKx 1 K and thus that xK Kx for every element x of G and consequently we have shown that the kernel K is a normal subgroup of the parent group Normal subgroups groups of cosets quotient groups In previous documents we have seen that when it is possible to partition the group table of a group G in a certain way a homomorphism is possible from G to the sets that are represented by those partitions We will show under what conditions such partitions are possible and thus under what conditions a homomorphism is possible Let us consider a group of even order 2n which has a subgroup of order n for example the dihedral group D 4 which has 8 elements and which has a subgroup of order 4 namely C 4 1 r r 2 r 3 a b c d 1 1 r r 2 r 3 a b c d r r r 2 r 3 1 b c d a r 2 r 2 r 3 1 r c d a b r 3 r 3 1 r r 2 d a b c a a d c b 1 r 3 r 2 r b b a d c r 1 r 3 r 2 c c b a d r 2 r 1 r 3 d d c b a r 3 r 2 r 1 Table 16 1 Table of D 4 The subgroup C 4 is 1 r r 2 r 3 We see this subgroup represented in the top left hand corner red block of the group table The elements of its coset a b c d are adjacent to it blue block The table is divided into four blocks and it will be seen that the bottom right hand block also consists entirely of the elements of the subgroup 1 r r 2 r 3 The remaining two squares are entirely filled with the elements from the coset a b c d so that the table possesses the broad overall structure of the C 2 group K A K K A A A K where K denotes the subset 1 r r 2 r 3 and A denotes the set a b c d The reason why the elements of D 4 can be so arranged such that this overall structure becomes directly apparent is because of the Latin Square Property of all finite groups It can be proved that The structure table for any finite group is always a Latin Square that is to say each row and each column contains all the elements of the set with no repetitions This does not mean however that a table in the form of a Latin Square is necessarily a group table Associativity will usually fail somewhere along the line But when a table is a group table of a finite group then it possesses the Latin Square Property Since the top left hand square of the table of D 4 contains the subgroup 1 r r 2 r 3 each of its four rows must be a permutation of these four elements because multiplications of the elements with each other always result in elements of that same subgroup And if an element x of any subgroup is multiplied with two different elements of that subgroup the result is always two different elements For if xy u and xz u then we have xy xz which is equivalent to x 1 xy x 1 xz which is equivalent to y z and this is contrary to the supposition namely that they are different So xy and xz always result in the production of different elements So no row connected with the subgroup 1 r r 2 r 3 contains repetitions and we see the Latin Square Property at work Therefore for the whole group since each of the first four rows must be a permutation of all eight elements Latin Square Property it follows that the remaining four elements of each row must be drawn from a b c d and thus the whole of the top right hand block must consist of a s b s c s and d s A similar consideration of the Latin Square Property applied to the first four columns convinces us that the bottom left hand block must also consist entirely of a s b s c s and d s Finally the remaining four elements of the last four rows or columns are bound to be 1 r r 2 or r 3 again by the Latin Square Property In previous documents we already showed several examples of group tables revealing their overall C 2 structure for example with respect to the group D 6 in Part VII Table 7 10 We may say that the group has been mapped on to the group C 2 by a homomorphism which takes the elements of the subgroup K into the identity element of C 2 while the elements of its coset are carried into the other element of C 2 In all this we can interpret the whole subgroup K which itself is its right or left coset by the identity element as one element of the image group and also the whole coset of K as the other element It is obvious that this arrangement can always succeed when the order of the subgroup K is one half of the order of the group G i e when the index of the subgroup is 2 Normal subgroups of index 3 We next ask the question if the index of the subgroup is 3 will it always be possible to map the group onto the group C 3 in a similar way as is done in the next table where the group is A 4 and the subgroup D 2 Here A 4 is mapped homomorphically by the function let us again call it PHI onto the group C 3 by the correspondence 1 a b c K p q r s P p 2 q 2 r 2 s 2 Q 1 a b c p q r s p 2 q 2 r 2 s 2 1 1 a b c p q r s p 2 q 2 r 2 s 2 a a 1 c b s r q p r 2 s 2 p 2 q 2 b b c 1 a q p s r s 2 r 2 q 2 p 2 c c b a 1 r s p q q 2 p 2 s 2 r 2 p p r s q p 2 r 2 s 2 q 2 1 b c a q q s r p s 2 q 2 p 2 r 2 b 1 a c r r p q s q 2 s 2 r 2 p 2 c a 1 b s s q p r r 2 p 2 q 2 s 2 a c b 1 p 2 p 2 s 2 q 2 r 2 1 c a b p s q r q 2 q 2 r 2 p 2 s 2 c 1 b a r q s p r 2 r 2 q 2 s 2 p 2 a b 1 c s p r q s 2 s 2 p 2 r 2 q 2 b a c 1 q r p s Table 16 2 Table of A 4 K P Q K K P Q P P Q K Q Q K P Table of C 3 We now attempt to do the same sort of thing in the case of the subgroup 1 b c g C 4 of the group Q 6 The table of this group is shown below 1 a b c d f g h j k l m 1 1 a b c d f g h j k l m a a l 1 d f g h j c m k b b b 1 m j c d f g h l a k c c j d k l a 1 b m g h f d d c f m k l a 1 b h j g f f d g b m k l a 1 j c h g g f h 1 b m k l a c d j h h g j a 1 b m k l d f c j j h c l a 1 b m k f g d k k m l g h j c d f 1 b a l l k a f g h j c d b m 1 m m b k h j c d f g a 1 l Table 16 3 Table of Q 6 In trying to find a partition of the group table of Q 6 on the basis of the subgroup H 1 b c g in the same way as was accomplished above with the group A 4 we rearrange the elements of Q 6 as given in the above group table such that we get the subgroup 1 b c g in the top left hand corner of the new group table All this should lead to the establishing of a homomorphism of the group Q 6 onto a smaller group isomorphic to C 3 Well let s try 1 b c g a l j f d h k m 1 1 b c g a l j f d h k m b b 1 g c l a f j h d m k c c g b 1 j f l a k m h d g g c 1 b f j a l m k d h a a l d h l l a h d j j f k m f f j m k d d h h k k m m Part of Table of Q 6 Now the right coset of H 1 b c g by the element a i e Ha is 1 b c g a a l j f where l means the letter l to distinguish it from the identity element symbolized by the numeral 1 so we place these elements next in order across the top margin of the new table and complete the row with d h k and m Everything goes smoothly for the first four rows Recall that all the products are determined according to the structure of the group Q 6 displayed in Table 16 3 given above but when we come to rows a l j f we run up against trouble for the left hand block of the second super row contains not only the elements a l j and f but also four different alien elements indicated by red coloring in the above table d h k and m It is clear that the reason the above case does not work is precisely because 1 b c g is not a normal subgroup of Q 6 the left coset a 1 b c g a l d h is different from the right coset 1 b c g a a l j f and it is from this left coset that two of these new alien elements d and h have come The other two k and m appearing in the left cosets j H and f H Homomorphic images of Abelian groups Now every subgroup of an Abelian group is obviously normal Every element x of an Abelian group commutes with every other so we have ab ba for every two elements of such a group This implies abb 1 bab 1 which gives a bab 1 for any two elements of the group And this implies bHb 1 H for any subgroup So for any subgroup of an Abelian group we have bHb 1 b Hb which gives bH Hb for any element b of the group which in turn means that for any subgroup of an Abelian group its left and right cosets are the same meaning that those subgroups are normal in the group And so it is always possible to produce homomorphic images of an Abelian group of composite order groups of prime order do not have subgroups at all whereby any subgroup may be mapped into the identity of the homomorphic image group and thus can serve as kernel for a particular homomorphism We have shown this in the case of the cyclic group C 12 where all its subgroups induce subdivisions of the group table and so allow for homomorphisms See Parts V and VI Another case of failure when subgroup is not normal Before drawing together the threads of the foregoing discussions let us give one more example to demonstrate the failure of a group to yield a homomorphic image by a non normal subgroup by considering 1 p p 2 in the group A 4 See Table 16 2 above The left cosets are 1 p p 2 a s r 2 b q s 2 and c r q 2 The right cosets are 1 p p 2 a r s 2 b s q 2 and c q r 2 Let us derive two of these cosets namely the left and right cosets of our subgroup by the element a a 1 p p 2 a ap ap 2 reading off from the group table a s r 2 1 p p 2 a a pa p 2 a a r s 2 So we have one example of the non equality of a right and left coset of 1 p p 2 by a same element a and this is sufficient for this subgroup not to be a normal subgroup Now we can show that the system of cosets with respect to our subgroup is not closed under set multiplication which implies that that system does not form a group that is to say it cannot figure as a homomorphic image of our group A 4 a quotient group cannot be formed on the basis of this particular subgroup we spoke about quotient groups in earlier documents and will return to them in due course a s r 2 c b q s 2 r p p 2 1 q 2 r 2 a s Because the subgroup 1 p p 2 has three different elements every coset of this subgroup therefore has three different elements The system of these cosets to be closed means that any product of any two of these cosets according to set multiplication must yield another coset also belonging to set of cosets of 1 p p 2 Well when we multiply the cosets c r q 2 and a s r 2 both cosets of our subgroup the result is see table above the set b q s 2 p p 2 1 r 2 a s and this is a set consisting of nine different elements so it cannot be any coset of 1 p p 2 and thus is the set of cosets of 1 p p 2 not closed and is therefore not a group and consequently cannot figure as a homomorphic image of our group A 4 In contrast we see in the Table of A 4 that the cosets of the subgroup 1 a b c are perfectly orderly which we can see when we look to the colored squares in the table representing these cosets each square consists of four different elements only the same number as the number of elements of the subgroup 1 a b c For example if we multiply the coset p q r s which is the left coset of 1 a b c by the element p as well as by the elements q r s with the coset p 2 q 2 r 2 s 2 which is the right coset of 1 a b c by the element p 2 as well as by the elements q 2 r 2 s 2 we get the square containing only the four elements 1 a b c Products of cosets of normal subgroups general theory We will now prove that the system of cosets of a normal subgroup forms a group For this we must show That the set of such cosets is closed under set multiplication That one of these cosets figures as identity element That for every coset in this set there exists an inverse which is also present in this set Associativity is evident because set multiplication in a group involves multiplication of group elements and this multiplication is associative because we have to do with a group namely our original group So if K is a normal subgroup of the group G and aK and bK are any two of its left cosets then we have bK Kb Hence the product set determined by the multiplication of sets on the basis of how the individual elements combine according to the group table of G aKbK aKKb But KK K as K is a subgroup the mutual multiplication of its elements always yields elements of that same subgroup Therefore aKbK aKb abK ab K and this is one of the left cosets of K Hence the system of cosets is closed under the operation of the formation of product sets Further we see that K which itself is also a coset namely 1K is an identity element within the system of cosets because for every element x of G we have xK K xKK xK because KK K and K xK KxK KKx because K is a normal subgroup Kx xK Finally because we have already supposed a to be an element of G and figuring as such in the coset aK a 1 must also be an element of G And so there exists also a coset a 1 K of K And because a 1 K aK a 1 KKa a 1 Ka a 1 aK K the identity element of the closed set of cosets we can say that a 1 K is the inverse of aK under set multiplication and we already know that this inverse is a genuine coset of K because a 1 is an element of G So every coset of the set of cosets of the normal subgroup K has an inverse which is one of the elements of that set of cosets So now we have proved that the set of cosets of a normal subgroup is a group in which the group operation is the set multiplication of those cosets And this group of cosets we call the quotient group or factor group G K G slash K of the group G with respect to the normal subgroup K And this quotient group is a homomorphic image of the group G The above important proof can also be given in an expanded form We shall give it with respect to closure First a preliminary remark We make use of the property of K being a normal subgroup of G as follows When k x is an element of K and p is an element of G then this means that pk x is an element of the left coset pK of the subgroup K by the element p But K is a normal subgroup which means that this left coset of K by the element p pK is equal to the right coset Kp of K by that same element p So the element pk x of pK is equal to some element k y p of Kp We write pk x k y p The product set aKbK means the set of all elements such as ak r bk s in which k r and k s are elements of K Now we know that bk s is in bK but then since bK Kb it is also in Kb so bk s k t b for some element k t of K Hence a typical element of aKbK is ak r bk s ak r k t b ak u b because K is a sub group the product of two of its elements k r and k t is another element of it and we call this element k u But again we require the normal property of K to be able to say that k u b bk v for some element k v in K Hence a typical element of aKbK is abk v and this is an element of the left coset of K by the element ab i e of ab K So any product of two left cosets of K results in some other left coset of K which means that the set of cosets of the normal subgroup K is closed under set multiplication In Part XIV where we discussed the products of sets we stated without proof that only when two cosets of a normal subgroup of order m are used to form products do we get a product set containing m elements only and thus can be a coset of that normal subgroup The illustrations there were all drawn from the group Q 6 We have now generalized the result examined the theory behind it and provided the proof Quotient groups or factor groups When H is a normal subgroup of a group G we shall write H G In the literature instead of a small triangle is used We now return to the question of the mapping of a group G which contains a normal subgroup K of index m meaning order G order K onto a smaller group of order m as we already have seen on many occasions As has been said the group of cosets is called the Quotient Group or the Factor Group and is indicated G K Thus for example when G is A 4 and K is D 2 the quotient group G K is C 3 See Table 16 2 Again when G is C 4 xC 2 and K is C 4 the quotient group G K is C 2 See Table 10 4 in Part X and it is this latter sort of example which probably accounts for the name quotient But the name and the notation are somewhat unfortunate for if the quotient group of G by the normal subgroup K is a group H i e G K isomorphic H the notation strongly suggests that G KxH i e K and H multiplied but this is not always so Thus A 4 D 2 C 3 is not equivalent to A 4 D 2 xC 3 i e A 4 is not the direct product of the groups D 2 and C 3 since in any case the latter i e D 2 xC 3 is Abelian whereas A 4 is not In Table 5 3 of Part V and Table 6 5 of Part VI we can see that C 12 C 2 C 6 and C 12 C 6 C 2 yet C 12 is different from C 6 xC 2 See HERE in Part IX Again S 4 A 4 C 2 but S 4 is not the direct product of A 4 and C 2 More remarkable is that both C 6 C 3 and D 3 C 3 give the same quotient group C 2 However it may be proved BUDDEN p 407 that if A G B G and the sets A and B have only the element 1 in common and oder A order B order G then G isomorphic AxB Where means is a normal subgroup of Which means that if A and B are two normal subgroups of a group G which have only the identity in common and the order of A multiplied with the order of B is equal to the order of G then the group G is the direct product of A and B Quotient groups of direct product groups We have noted that C A B does not imply that C isomorphic AxB One may wander however whether the converse is true If C AxB is it true that C A B Before we can answer this question it is first necessary to discover whether A is a normal subgroup of C otherwise C A is meaningless Suppose A is the group 1 a 1 a 2 a 3 and B is the group 1 b 1 b 2 b 3 The elements of the direct product group AxB are 1 1 a 1 1 a 2 1 a 3 1 subgroup A 1 b 1 a 1 b 1 a 2 b 1 a 3 b 1 1 b 2 a 1 b 2 a 2 b 2 a 3 b 2 1 b 3 subgroup B Note that we here have to do with the direct product of two g r o u p s We should not confuse this with the product of two subsets of one and the same group See Part XIV Multiplication of subsets and for the direct product of groups see Part IX Direct product groups Because C is the direct product of the groups A and B A and B are subgroups of the group C Now the first row of this array constitutes an isomorphic image of the group A i e it can represent this group and also its coset by the element 1 as a subgroup of C while the successive rows are the cosets by the elements 1 b 1 1 b 2 etc For instance the left coset of A by the element 1 b 1 is 1 b 1 1 1 a 1 1 a 2 1 a 3 1 which is the set 1 b 1 a 1 b 1 a 2 b 1 a 3 b 1 and this is the second row of the array So every row of the above array is a coset of A and they are all the cosets of A The corresponding left and right cosets are here always the same which we shall demonstrate below We shall explain why they are all the cosets of A where A is represented in the product group by its isomorphic image 1 1 a 1 1 a 2 1 a 3 1 At first sight we could note that the rows of the array only represent cosets of A by the elements 1 1 1 b 1 1 b 2 1 b 3 etc But of course every element of our group C of which the elements are enumerated in the above array can form a coset of A So let us form a left coset of A by the element a 2 b 1 a 2 b 1 1 1 a 1 1 a 2 1 a 3 1 a 2 1 b 1 a 2 a 1 b 1 a 2 a 2 b 1 a 2 a 3 b 1 We see a set of ordered pairs and we focus on the first member of each pair Such a first member is a product of a 2 with an element of A i e with the elements 1 a 1 a 2 a 3 etc These latter elements are all the elements of the group A But a 2 the element with which each element of A the first member of the original pairs is multiplied is also an element of A And because A is a group the resulting products the first members of the resulting set of pairs must all be elements of A again And because all the enumerated elements of A are different elements multiplying each of them with a same element namely a 2 the result will be as many elements as we had before and again different elements which means that we get all the elements of A back again So all the first members of the above resulted pairs are just elements of A again while the second member being b 1 This means that the resulting set of pairs must be equal to the second row of the above array so no new coset is produced by the multiplication of the element a 2 b 1 with the set 1 1 a 1 1 a 2 1 a 3 1 The same reasoning can be applied to all the other elements of the row from which we took the pair a 2 b 1 by taking those other elements and multiplying each of them in turn with the set 1 1 a 1 1 a 2 1 a 3 1 In this way we can handle all rows of the above array and discover that no new cosets are being produced From all this we can see that the rows of the above array represent all left cosets of A where A is represented by 1 1 a 1 1 a 2 1 a 3 1 We will now show that the left cosets are identical to the corresponding right cosets since in any case we have a r 1 1 b s 1 b s a r 1 a r b s for all r and s which indeed means that the left coset is identical to the corresponding right coset We are therefore certain that A and especially its isomorphic image 1 1 a 1 1 a 2 1 a 3 1 is a normal subgroup of C and so C A is meaningful and we now need to know whether C A has the structure of B Recall that we are investigating the question that if C AxB is it true that C A B Suppose the rows of the above array are denoted A B 1 B 2 B 3 i e they are all the cosets of A and of its isomorphic image so that B r denotes the set of ordered pairs 1 b r a 1 b r a 2 b r a 3 b r Consider the product of the cosets B r B s letting a i b r be any element of B r and a j b s be any element of B s we have fully in the context of multiplying sets which here means multiplication of their respective elements these being ordered pairs a i b r a j b s a i a j b r b s a k b t Because a i and a j are elements of the group A 1 a 1 a 2 a 3 their product a i a j is also belonging to A so we can call this product a k Similarly b r and b s are elements of the group B 1 b 1 b 2 b 3 so their product b r b s is also an element of B so we can call it b t Therefore the product of any two elements of B r and B s is an element of B t which we can see regarding the structure of the last term a k b t of the above relation and we may express this B r B s B t The behavior of the cosets of A or of its isomorphic image in the direct product group therefore exactly mimics the behavior of the elements of the group B itself For the group B 1 b 1 b 2 b 3 we can say b r b s b t which indeed mimics B r B s B t This means that the group of cosets C A is isomorphic to the group B Hence it is true that C A B which is equivalent to AxB A B And because the group B is not different in status from the group A we can also state AxB B A Chains of normal subgroups We have seen that D 2 A 4 meaning that D 2 is a normal subgroup of A 4 See Table 16 2 the subgroup consisting of 1 a b c in the case of the group table just referred to But D 2 is Abelian so any of its subgroups is normal to it e g 1 a 1 a b c See next group table 1 a b c 1 1 a b c a a 1 c b b b c 1 a c c b a 1 Table 16 4 Table of D 2 So we have C 2 D 2 A 4 a chain of normal subgroups We may extend the chain one stage further to the left by including the identity 1 as a trivial normal subgroup of C 2 See next tables 1 a 1 1 a a a 1 Table 16 5 Table of C 2 1 1 1 Table 16 6 Table of C 1 Table 16 6 depicts the group consisting of the identity element alone An extension to the right is also possible because A 4 S 4 Hence we obtain the chain C 1 C 2 D 2 A 4 S 4 Note the curious fact that though C 2 D 2 S 4 yet C 2 is not normal in S 4 Referring to the group table for S 4 See Table 13 2 Part XIII we have the normal subgroup 1 h r y Though each of the subgroups 1 h 1 r 1 y is normal in D 2 neither of them is normal in S 4 This must be so because if 1 h is to be normal in a group h must commute with every element of the group only true for subgroups of order 2 This we can show as follows If

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