Construction of Finite Groups (Zn )

of

Order 1 to Order 6

 I II We construct the groups based on 1 and its roots. 1= cos 2nπ + sin 2nπ =ei2mπ  . here m is a positive integer. is the group nomenclature & n is its order i.e. no. of elements in it. Binary operation is o and is ordinary multiplication. Based on figures of col II, we construct here Caley's Table. There is a pattern in the table. In any row or column, no element is repeated . In the column, the arrangements follow a cyclical order. For example, in Z5, the first column is (e,a,b,c,d). then the second column will be (a,b,c,d,e), third column will be (b,c,d,e,a) and so on. Similar patterns exist for rows.Construction of operators : This is the most interesting part. Each operator is represented by a square matrix of the order same as the order of the group. For example, for Z2, the matrix shall be 2x2 type. The matrices shall obey the binary operation rules as codified in the Caley Table. These groups are Abelian .The group is cyclic.*Any representation of a  finite group is equivalent to a Unitary Representation. *All representations of any finite group are completely reducible. * Finite groups can have elements which are discrete or continuous. Z1: no. of elements 1; element is  1=ei2mπ ; n=order of the group=1. since there is only 1 element, the same is the identity element & inverse of its own.       e e     e state is |e > [e]=[1] Z2: no. of elements 2; it involves square root of 1. elements are a=e(i2mπ/2)*1  ,  e=e(i2mπ/2) *2; n=order of the group=2. No. of arrangements=22 =4Caley's Table        e       a e     e       a a     a       e states are |a> , |e> . a o e =a=e o a  ;  e o e =ea o a =e ;|e > = 1  a> = 0           0           1 D[e]=1   0 0   1 D[a] = 0   1 1   0 We put |e > = 1            0          a> =   0            1 D(a)|e> =|a> D(a)|a> =|e> hence veracity of Caley Table [since a o a =e, a has to be involutary matrix. If the matrix is of type a   b c   d involutary condition demands Either b=c=0 and a=±1; d=±1 or b=c a=-d a=±√(1-bc) we put b=c=1 a=0; Exa- 1 0   0  1 ------ -1  0    0 -1 ----- 1  0 0 -1 ------ -1  0  0   1 ------------ √(1-bc)      b   c         √(1-bc) ------------------- -√(1-bc)      b   c         -√(1-bc) ---------------- √(1-bc)      b   c         -√(1-bc) ------------------ -√(1-bc)      b   c         √(1-bc) ---------------------- Z3: no. of elements 3; it involves cube  root of 1. elements are a=e(i2mπ/3)*1  ,  b=e(i2mπ/3) *2; e=e(i2mπ/3) *3;n=order of the group=3.No. of arrangements=32 =9Caley Table        e       a       b e     e       a       b a     a       b       e b     b       e       aAnother example of Z3 is (1,ω ,ω2 ). Another representation of Z3 is D(a), D(b) , D(e)There is another group A3  whose order is also 3 (n!/2 where n=3). It is known as Alternating group of order 3. It is a permutation group and consists of a set of even permutations. It is a normal sub group of S3 whose order is 3! . The group is a cyclic one and Caley table is same as that of Z3. Ratio of index of S3 to that of  A3 is 6/3=2 which is a prime number and so its quotient group is cyclic. The elements of A3  are:e=  1  2  3          a=  1   2    3       b=  1  3  2      1  2  3                2   3    1              3  2 1now a=  1  2  3 = 1   2  3    * 1  2   3  =(1,2)(2,3) =(1,2,3) -> no. of permutation=2              2  3  1    2   1  3       1  3   1now b=  1  3  2 = 1  2  3    * 1  2   3  = (1,3)(2,3) =(1,3,2)-> no. of permutation=2              3  2  1    3  2  1       1  3   2Now D(a) is same as that of Z3 . D(b) same as that of Z3. e is same as that of Z3.D(a) <-  (1,2,3) D(b) <-  (1,3,2)D(e) <-  (1)         (13) -> 0 0 1      (12) -> 0 1 0        (23) -> 1 0 0                     0 1 0                   1 0 0                    0 0 1                     1 0 0                   0 0 1                    0 1 0The Projection Operator P for Z3 is1  1  1 * 1/31  1  11  1  1such that D(g)P =P All representations in Z3 are fully reducible  and D(g)P =P  and PD(g)P=D(g)P is the test. If P projects onto an invariant sub-space but 1-P does not, then the representation is not fully reducible. If S= 1  1   1          1 ω2  ω                 1 ω   ω2 S-1 =    1/3        1/3               1/3             1/3   -(1+ω )/3ω           1/3ω              1/3        1/3ω              -(1+ω )/3ω states are |a> ,|b>, |e>a o e =a=e o a  ;b o e =b=e o b  ;     a o b =e=b o a  ; a o a =b b o b =ae o e =e ;|e> = 1  |a> =  0    |b> = 0         0             1             0         0             0             1taking from 1st,2nd , 3rd column of identity matrix. D(e)=1  0  0 0  1  0 0   0  1 D(a)=0  0  1 1  0  0 0  1  0 D(b)=0  1  0 0  0  1 1  0  0 the trick is in identity matrix, first column is (1,0,0) . so in a, make the first column, (0,1,0) by shifting 1 down. do the same in subsequent columns. The Similarity matrix for similarity transformation of D is S= 1   1   1 * 1/3 1   ω  ω2 1   ω2 ωand transformed D'(a)=1    0    00    ω   0 0    0   ω2 D'(b)=1    0    00    ω2   0 0    0   ω D'(e)=1  0  0 0  1  0 0   0  1 D & D' are equivalent representations in Z3. Z4: no. of elements 4; it involves 4th  root of 1. elements are a=e(i2mπ/4)*1  ,  b=e(i2mπ/4) *2; c=e(i2mπ/4) *3;e=e(i2mπ/4) *4;n=order of the group=4. No. of arrangements=42 =16       e       a       b       c e     e       a       b       c a     a       b       c       e b     b       c       e       a c     c       e        a      bThere is another finite group of order 4 known as Klein Group V4 with binary operation of addition of integers mod 2. The Caley table is       e       a       b       c               (0,0)    (1,0)    (0,1)    (1,1) e     e       a       b       c    (0,0)   (0,0)     (1,0)    (0,1)    (1,1) a     a       e       c       b    (1,0)   (1,0)     (0,0)    (1,1)    (0,1) b     b       c       e       a    (0,1)   (0,1)     (1,1)    (0,0)    (1,0) c     c       b        a      e    (1,1)   (1,1)      (0,1)   (1,0)    (0,0)Here e=(0,0)  a=(1,0)  b=(0,1)   c=(1,1). Suppose we want to find out b o c=(0,1) o (1,1) =(1,0) as 0+1=1 and mod 2 =1. also 1+1=2 and mod 2 is 0. Klein group is a non-cyclic group.Another example of Z4 is (1,-1,i,-i) under binary operation of multiplication. states are |a> ,|b>,|c>, |e>a o e =a=e o a  ;b o e =b=e o b  ;     c o e =c=e o c  ;     a o b =c=b o a  ; b o c =a=c o b  ; a o c =e=c o a  ; a o a =b  ; b o b =e  ; c o c =b  ;e o e =e ;|e> = 1  |a> =  0    |b> = 0  |c>= 0         0             1             0          0         0             0             1          0         0             0             0          1taking from 1st,2nd , 3rd column,4th column of identity matrix. D(e)=1  0  0   0 0  1  0   0 0  0  1   0 0  0  0   1 follow the same method as above. D(a)=0  0  0   1 1  0  0   0 0  1  0   0 0  0  1   0 D(b)= 0  0  1   0 0  0  0   1 1  0  0   0 0  1  0   0 D(c)=0  1  0   0 0  0  1   0 0  0  0   1 1  0  0   0 Z5: no. of elements 5; it involves 5th  root of 1. elements are a=e(i2mπ/5)*1  ,  b=e(i2mπ/5) *2; c=e(i2mπ/5) *3;d=e(i2mπ/5) *4;e=e(i2mπ/5) *5;n=order of the group=5.No. of arrangements=52 =25       e       a       b       c      d e     e       a       b       c      d a     a       b       c       d      e b     b       c       d       e      a c     c       d        e      a      bd    d        e       a       b     c Example: D(c)|d> = |b>0 0 1 0 0   *   0  = 0 = |b> 0 0 0 1 0       0      0 0 0 0 0 1       0      1 1 0 0 0 0       0      0 0 1 0 0 0      1       0 states are |a> ,|b>,|c>,|d>, |e>a o e =a=e o a  ;  b o e =b=e o b  ;     c o e =c=e o c  ;  d o e =c=e o d  ;   a o b =c=b o a  ; b o c =e=c o b  ; c o d =b=d o c  ; d o a =e=a o d  ; a o c =d=c o a  ; b o d =a=d o b  ; a o a =b  ; b o b =d  ; c o c =a  ; d o d =c  ; e o e =e ; |e>=1 |a>=0  |b>=0 |c>=0 d>=0       0         1         0        0        0       0         0         1        0        0       0         0         0        1        0       0         0         0        0        1 D(e)= 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 D(a)= 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 D(b)= 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 D(c)=0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 D(d)= 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 Z6: no. of elements 6; it involves 6th  root of 1. elements are a=e(i2mπ/6)*1  ,  b=e(i2mπ/6) *2; c=e(i2mπ/6) *3;d=e(i2mπ/6) *4;e'=e(i2mπ/6) *5;e=e(i2mπ/6) *6;n=order of the group=6.No. of arrangements=62 =36       e       a       b       c      d       e' e     e       a       b       c      d        e' a     a       b       c       d      e'       e b     b       c       d       e'     e        a c     c       d        e'      e      a        bd    d        e'       e       a     b         c e'   e'        e       a       b     c         d states are |a> ,|b>,|c>,|d>,|e'>, |e>. |e'> is made bold to distinguish it from identity element.a o e =a=e o a  ;  b o e =b=e o b  ;     c o e =c=e o c  ;  d o e =c=e o d  ;   e' o e =e'= e o e';   a o b =c=b o a  ; b o c =e'=c o b  ; c o d =a=d o c  ; d o e' =c=e' o d  ; e' o a =e=a o e'  ;   a o c =d=c o a  ; b o d =e=d o b  ; c o e' =b=e' o c  ; e' o b =a=b o e'  ;   a o d =e'=d o a  ; a o a =b  ; b o b =d  ; c o c =e'  ; d o d =b  ; e' o e' =d  ;e o e =e  ;|e>=1 |a>=0  |b>=0 |c>=0 d>=0       0         1         0        0        0       0         0         1        0        0       0         0         0        1        0       0         0         0        0        1       0         0         0        0        0|e'>= 0         0         0         0         0         1 D(e)=1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 D(a)=0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 D(b)=0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 D(c)=0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 D(d)=0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 D(e')=0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0