1. NCM LECTURE NOTES ON LATIN SQUARES.
NOW WE SHALL SEE THE MOST BEAUTIFUL RELATION S BETWEEN , ALGEBRA , NUMBER THEORY AND
COMBINATORICS , WHEN WE SOLVE PROBLEMS BASED ON “ LATIN SQUARES “. THEY TOO HAVE VERY
IMPORTANT CONNECTION IN “ CRYPTOGRAPHY “ ALSO.
NOW , WE WILL MOVE TO THE FORMAL DEFINITION OF “ LATIN SQUARES “.
LET “ n “ BE A GIVEN POSITIVE INTEGER AND LET “ S “ BE A GIVEN SET OF “ n “ DISTINCT
ELEMENTS .
S = { s1 , s2 , s3 , . . . , sn – 1 , sn } . A LATIN SQUARE OF ORDER “ n “ BASED ON “ S “ IS AN
“ n - by – n array “ , each of whose entries is an element of “ S “ SUCH THAT EACH OF THE “ n “
elements of “ S “ OCCURS EXACTLY ONCE IN EACH ROW AND EXACTLY ONCE IN EACH COLUMN .
THUS EACH OF THE ROWS AND EACH OF THE COLUMNS OF A LATIN SQURE IS A PERMUTATION OF
THE ELEMENTS OF “ S “. LET “ G “ BE A GIVEN FINITE GROUP OF “ n “ ELEMENTS.
SUPPOSE “ f “ IS A FUNCTION : S x S …… > S
SUCH THAT f ( s i , s j ) = si . sj LIES IN “ S “ WITH THE FOLLOWING PROPERTY
FOR A FIXED “ i “ , si . sj IS A PERMUTATION ON “ S “ FOR ALL “ j “ VARIES FROM 1 TO “ n “.
Similarly for a fixed “ j “ , si . sj IS A PERMUTATION ON “ S “ FOR ALL “ i “ varies from 1 TO “ n “.
SUCH A FUNCTION “ f “ IS CALLED A LATIN FUNCTION DEFINED ON S x S TO S .
G = { g1 , g2 , g3 , … , gi , … , gj , … , gn – 1 , gn } . WITHOUT LOSS OF GENERALITY , ASSUME g1 = e , the
identity element of “ G “. NOW WE SHALL DEFINE AN “ n – by – n “ MATRIX “ A “ AS FOLLOWS :
A = [ ai j ] = [ gi . gj ]. Now it is easy to verify “ A “ IS A LATIN SQUARE OF ORDER “ n “ BASED ON
“ G “
THE i th ROW OF “ A “ = { gi . g1 , gi . g2 , gi . g3 , … , gi . gi , … , gi . gj , … ,gi . gn }.
THE j th COLUMN OF “ A “ = { g1 . gj , g2 . gj , g3 . gj , … , gi . gj , … , gj . gj , … , gn . gj } .
THEOREM – 1 . LET “ G “ BE A GIVEN ( ANY ) FINITE GROUP OF ORDER | G | .
G = { g1 = e , g2 , g3 , … , gi , … , gj , … , g| G | } . LET “ m “ , “ n “ BE GIVEN POSITIVE INTEGERS SUCH
THAT ( m , | G | ) = ( n , | G | ) = 1 .
DEFINE | G | x | G | matrix “ A “ AS A = [ gi
m
. gj
n
] ; i , j belongs to { 1 , 2 , 3 , … , | G | } .

2. “ A “ IS A LATIN SQUARE OF ORDER | G | , BASED ON “ G “.
THEOREM – 2 : LET “ a “ , “ b “ BE ANY TWO FIXED ELEMENTS OF “ G “ , GIVEN FINITE GROUP.
DEFINE A = [ ( a . gi ) . ( gj . b ) ] , B = [ ( a . gi ) . ( b . gj ) ] , C = [ ( gi . a ) . ( gj . b ) ] ,
D = [ ( gi . a ) . ( b. gj ) ] . A , B , C , D ARE ALL “ LATIN SQUARES “ OF ORDER | G | , BASED ON “ G “ .
THEOREM - 3
LET T1 , T2 BE ANY TWO AUTOMORPHISMS OF A FINITE GROUP “ G “ .
LET “ X “ BE A MATRIX OF ORDER | G | x | G | , DEFINED BY
X = [ T1 ( g i ) . T 2 ( g j ) ] . THEN “ X “ IS A LATIN SQUARE OF ORDER | G | , BASED ON
“ G “.
NOW LET US DO SOME GOOD “ NUMBER THEORY “ .
LET “ N “ BE A GIVEN POSITIVE INTEGER.
Z N = { 0 , 1 , 2 , 3 , … , N – 2 , N – 1 } BE THE RING OF INTEGERS MOD ( N ) .
+ N , x N IS THE ADDITION MOD N AND MULTIPLICATION MOD N DEFINED ON “ Z N “ .
DEFINE Z N
* = Z N - { 0 } WITH “ a “ LIES IN Z N
* IF AND ONLY IF ( a , N ) = 1 .
SELECT FIXED a , b LIES IN Z N
* [ ( a , N ) = ( b , N ) = 1 ].
SELECT “ r “ , “ s “ [ FIXED , LIES IN Z N ] .
NOW DEFINE A BIJECTIVE MAP f a , r : Z N ……..> Z N
By f a ,r ( x ) = a . x + r ( MOD N ) . SIMILARLY
We can define g b , s ( x ) = b . x + s ( MOD N ) .
NOW DEFINE A = [ f a , r ( i ) + g b , s ( j ) ( MOD N ) ] .
A IS A LATIN SQUARE OF ORDER “ N “ BASED ON Z N .
AS A SIMPLE CORLLARY A = [ ( i + j ) ( mod N ) ] , ALWAYS DEFINES A LATIN SQUARE
OF ORDER “ N “ BASED ON Z N

3. ALSO DEFINE A MATRIX B = [ ( a . i + j ) ( MOD N ) ] , ALWAYS DEFINE A LATIN
SQUARE OF ORDER “ N “ BASED ON Z N .
DEFINE “ M = 2 . N + 1 “ . CONSTRUCT AN M by M MATRIX “ C “ AS FOLLOWS :
C = [ ( N + 1 ) . ( i + j ) ( mod M ) ] IS A LATIN SQUARE OF ORDER “ M “ BASED ON ZM
THIS IS THE BEAUTIFUL EXAMPLE OF IDEMPOTENT SYMMECTRIC LATIN SQUARE OF ODD
ORDER.
DEFINE FOR ANY FIXED NON NEGATIVE INTEGER “ k “ , M = 2 k + 1 + 5
CONSTRUCT AN “ M by M “ matrix D = [ ai j ] = [ ( 2k + 3 ) ( i + j ) ( MOD M ) ]
FOR i , j belongs TO { 0 , 1 , 2 , 3 , 4 , … , 2k + 1 + 4 }.
THIS “ D “ ALSO DEFINES AN IDEMPOTENT SYMMECTRIC LATIN SQUARE OF ORDER “ M “
BASED ON “ ZM “
NOW LET US CONSTRUCT MORE GENERALLY IDEMPOTENT SYMMECTRIC LATIN SQUARE
AS FOLLOWS : DEFINE M = 2 k + 1 + p , where “ p “ is ODD PRIME.
DEFINE A MATRIX OF ORDER “ M “ AS X = [ ai j ] =
[ [ 2k + ( p + 1 / 2 ) ] ( i + j ) ( MOD M ) ]
Here i , j BELONGS TO { 0 , 1 , 2 , 3 , . . . , 2k + 1 + p - 1 }. X IS IDEMPOTENT AND SYMMECTRIC
LATIN SQUARE OF ORDER “ M “ BASED ON ZM .
LET “ p “ BE ANY GIVEN ODD PRIME. LET US SELECT A POSITIVE INTEGER “ e “ SUCH THAT
G.C.D ( e , p - 1 ) = 1. NOW DEFINE A MATRIX “ X “ OF ORDER “ p x p “ as follows.
X = [ ai j ] = [ ( p + 1 / 2 ) . ( ie + je ) ( MOD p ) ] , i , j LIES IN { 0 , 1 , 2 , 3 , . .. , p – 1 } is
a symmectric LATIN SQUARE.

4. NOW LET US CONSTRUCT R.S.A ( PUBLIC KEY CRYPTOGRAPHY ) LATIN SQUARE.
SELECT TWO DISTINCT VERY LARGE ODD PRIMES “ p , q “ .
Select two fixed positive integers e , d such that ( e , ( p - 1 ) . ( q - 1 ) ) = 1
And ( d , ( p - 1 ) . ( q - 1 ) ) = 1
Now define a matrix A = [ ie + jd ( MOD N ) ] , N = p . q .
A is a LATIN SQUARE OF ORDER “ N “ BASED ON Z N .
Y = [ ( a . ie + j ) ( MOD N ) ] , ( a , p . q ) = 1 IS A LATIN SQUARE OF ORDER
“ N = p. q “ BASED ON ZN ; i , j BELONGS TO { 0 , 1 , 2 , 3 , … ,N – 1 }. HERE a , e ARE FIXED.
NOW WE ARE GOING TO USE TWO BEAUTIFUL RESULTS , NAMELY PERMUTATION
POLYNOMIALS OVER FINITE FIELDS.
LET US OBSERVE THE FOLLOWING FACTS :
FACT – 1 : LET “ K “ BE A FINITE FIELD WITH “ q = pn “ elements . THEN xq = x FOR ALL
“ x “ LIES IN “ K “ .
FACT – 2 : IF “ f [ t ] , g [ t ] “ are two polynomials over “ K “ OF DEGREES < q , and
If f ( a ) = g ( a ) , for all “ a “ LIES IN “ K “, THEN f = g .
FACT – 3 : LET “ f “ be a given polynomial over “ K “ . f [ t ] LIES IN K [ t ]. THEN THERE
EXISTS A POLYNOMIAL f* LIES IN K [ t ] OF DEGREE < q , SUCH THAT f * ( a ) = f ( a ) , for all
“ a “ LIES IN “ K “ .
FACT - 4 : LET “ T “ BE ANY FUNCTION FROM “ K “ TO ITSELF . T : K ……> K. THEN THERE
EXISTS A POLYNOMIAL f [ t ] LIES IN K [ t ] SUCH THAT T ( a ) = f ( a ) for all “ a “ LIES
IN “ K “ .

5. DEFINITION : A POLYNOMIAL “ f “ OVER A FINITE FIELD “ K “ IS CALLED A PERMUTATION
POLYNOMIAL IF THE MAPPING f : K ……. > K , DEFINED BY “ a …….> f ( a ), for “ a “ LIES
IN “ K “ IS ONE - TO – ONE.
NOW OUR AIM IS TO CONSTRUCT A LATIN SQUARE , USING PERMUTATION POLYNOMIALS
OVER A FINITE FIELD “ K “ with “ q “ elements.
THEOREM – 1 ( R.A . MOLLIN AND C. SMALL ) . LET K = GF ( q ) , GIVEN FINITE FIELD
WITH “ q “ elements. ASSUME THE CHARACTERISTIC OF K IS DIFFERENT FROM “ 3 “.
THAT IS CHAR ( K ) NOT EQUAL TO 3. THEN f ( x ) = a x3 + b x2 + c x + d ( a LIES IN K* )
THAT IS “ a “ is non –zero element of “ K “ is a PERMUTATION POLYNOMIAL ON “ K “
I F AND ONLY IF “ b 2 = 3 . a .c “ and q IS CONGRUENT TO 2 ( MOD 3 ) [ q = 2 ( MOD 3 ) ]
THEOREM – 2 ( RAJESH PRATAP SINGH AND SOUMEN MAITY ) . LET “ p “ = ODD PRIME.
LET “ K = Zp = FIELD OF INTEGERS MOD “ p “ .
DEFINE f [ t ] LIES IN Zp [ t ] such that f [ t ] = ta ( t ( p – 1 / 2 ) + b ) with ( a , p – 1 ) = 1
And “ b LIES IN Zp* “ ( non – zero element in Zp ). Then f [ t ] is a permutation polynomial
ON Zp if and only if ( b2 - 1 )( p – 1 / 2 ) = 1 ( MOD p ) .
IF WE WANT TO USE THESE TWO THEOREMS EFFECTIVELY FOR THE CONSTRUCTION OF LATIN
SQUARE MOD “ p = ODD PRIME “ , WE NEED TO UNDERSTAND THE BASIC CONCEPTS
ABOUT “ QUADRATIC RESIDUES “
THEREFORE , NOW WE ARE CONCERNED WITH QUADRATIC CONGRUENCES OF THE FORM
“ x 2 = n ( MOD p ) “ , where p = ODD PRIME AND “ n “ is NOT CONGRUENT TO
0 ( MOD p ) . IF “ x “ is a solution so is “ – x “ . HENCE THE NUMBER OF SOLUTIONS IS
EITHER “ 0 “ OR “ 2 “.

6. DEFINITION : SUPPOSE “ p “ is an ODD PRIME and “ a “ is an integer. “ a “ is defined to
be a quadratic residue MOD p if
1. a is NOT congruent to 0 ( MOD p )
2. THE CONGRUENCE x 2 = a ( MOD p ) has a solution “ x “ LIES IN Zp*
“ a “ is said to be a quadratic non – residue MOD p if “ a “ IS NOT CONGRUENT
TO 0 ( MOD p ) and “ a “ is NOT a quadratic residue MOD p .
FACT - 1 : LET “ p “ be an ODD PRIME. THEN “ EVERY REDUCED SYSTEM MOD p “
Contains exactly ( p - 1 / 2 ) quadratic residues and exactly ( p - 1 / 2 )
quadratic non residues MOD p.
THE QUADRATIC RESIDUES BELONG TO THE RESIDUE CLASSES CONTAING THE NUMBERS
12 ( MOD p ) , 22 ( MOD p ) , 32 ( MOD p ) , . . . , ( ( p - 1 ) / 2 ) 2 ( MOD p )
Let us have the following example. TAKE p = 1 1 . THEN QUADRATIC RESIDUES MOD 1 1
ARE { 12 = 1 , 22 = 4 , 32 =9 , 42 = 5 , 52 = 3 } AND QUADRATIC NON – RESIDUES ARE
{ 2 , 6 , 7 , 8 , 10 }. NOW LET US PROVE THIS FACT – 1 BY GROUP THEORY.
PROOF : Z p
* = { 1 , 2 , 3 , . . . , p - 2 , p – 1 } forms an ABELIAN GROUP ( IN FACT CYCLIC
GROUP ) WITH RESPECT TO THE BINARY OPERATION “ MULTIPLICATION MOD p “ ( = x p )
| Zp
* | = p - 1 . CONSIDER A MAP T : Z p
* ………..> Zp
* , defined by
T ( x ) = x 2. THIS MAP “ T “ IS A GROUP HOMORPHISM FROM Z p
* to ITSELF.
KERNEL ( T ) = { x LIES IN Z p
* | x2 = 1 }. SINCE | KERNEL ( T ) | = 2 , “ T “ IS NOT
ONE – TO – ONE ( THEREFORE “ T “ IS NOT “ ONTO “ )
LET “ H = T ( Z p
* ) “ , proper subgroup of Z p
*
H = { x 2 | x LIES IN Zp
* } = set of all quadratic residues MOD p AND
| H | = ( p – 1 ) / 2 . MORE OVER H IS THE NORMAL SUBGROUP OF INDEX “ TWO “.
THEREFORE Zp
* = H U H .y ( y is a quadratic non – residue ( MOD p ) ). THIS IS THE
IMPORTANT OBSERVATION , HERE.

7. IN THIS CONTEXT , WE WOULD LIKE TO REGISTER TWO CELEBRATED RESULTS, NAMELY
EULER’S CRITERIAN AND THE FAMOUS “ THE QUADRATIC RECIPROCITY LAW
NOW WE SHALL DEFINE Legendre ‘ s symbol . LET “ a “ be any given integer
( a / p ) = 0 if n = 0 ( MOD p )
( a / p ) = 1 if n is a quadratic residue MOD p
( a / p ) = - 1 if n is a quadratic non residue MOD p
EULER ‘ S CRITERIAN : LET “ p “ be given ODD PRIME. THEN FOR ANY INTEGER “ a “,
We have ( a / p ) = a ( p – 1 ) / 2 ( MOD p )
THE QUADRATIC RECIPROCITY LAW
IF “ p , q “ are distinct odd primes , then ( p / q ) . ( q / p ) = ( - 1 )( p – 1 ) . ( q – 1 ) / 4
THE Legendre ‘ s symbol ( a / p ) is a completely multiplicative function of “ a “.
LET “ p “ be a given fixed ODD PRIME. LET a 1 , a2 be any given positive integers .
THEN ( a1 . a2 / p ) = ( a 1 / p ) . ( a 2 / p ) .
NOW ,LET US BEGIN OUR MAIN TASK OF CONSTRUCTING A LATIN SQUARE BASED ON Zp , p is
ODD PRIME. AGAIN , LET US RECALL ,THE BEAUTIFUL RESULT OF R.A. MOLLIN AND C.SMALL.
TAKE GF ( q ) = Z p , where “ p “ is ODD PRIME.
HAVE A CLOSE LOOK AT THE CUBIC POLYNOMIAL f ( x ) = a x3 + bx2 + cx + d LIES IN Z p [ x ].
Here 1. We observe b 2 = 3 . a . c and p = 2 ( MOD 3 )
THEREFORE THE Legendre ‘s symbol ( 3.a.c / p ) = 1
This ,further reduces to ( 3 / p ) . ( a / p ) . ( c / p ) = 1 AND p = 2 ( MOD 3 )
CASE - 1 : ( 3 / p ) = 1 . THIS IMPLIES p = 1 ( MOD 12 ) OR p = 11 ( MOD 12 ).
SINCE p = 2 ( MOD 3 ) , HERE p = 1 1 ( MOD 12 ).

8. THEREFORE IN THIS CASE OUR ODD PRIME p = 1 2 .k + 1 1, for some suitable integer “ k “.
Now ( a / p ) . ( c / p ) = 1 implies ( a / p ) = ( c / p ) = 1 ( OR ) ( a / p ) = ( c / p ) = - 1.
Again let us recall the beautiful decomposition “ Zp = H U H. y “.
THIS FORCES US TO SELECT a , c LIES IN H ( OR ) SELECT a , c LIES IN H.y.
THEN , b 2 = 3a c IN Z p , where p = 11 ( MOD 12 ).
AT THIS TIME , LET US CONSTRUCT TWO POLYNOMIALS f( x ) , g ( x ) LIES IN Z p [ x ] .
That is f ( x ) = a x3 + bx2 + cx + d with a , c non – zero and a , c LIES IN “ H “.
THEN WE CAN SOLVE b , such that b 2 = 3 ac.
Define g ( x ) = a’ x3 + b’ x2 + c’ x + d’, here a’ , c’ are non – zero LIES IN H. y.
Then b’2 = 3 a’ c’ is SOLVABLE IN Z p.
Now define a matrix “ M “ OF ORDER “ p x p “ , by
M = [ f (i ) + g ( j ) ], i , j LIES IN Z p , p = 12 k + 1 1 .
Then “ M “ IS A LATIN SQUARE OF ORDER “ p “ .
NOW CONSIDER THE CASE , WHERE ( 3 / p ) = - 1. USING
“ QUADRATIC RECIPROCITY LAW “ , WE MUST CONCLUDE p = 5 ( MOD 12 ) ( OR )
p = 7 ( MOD 12 ) . SINCE p = 2 ( MOD 3 ) , WE MUST HAVE p = 5 ( MOD 12 )
CASE – 2 : ( 3 / p ) = - 1 AND p = 5 ( MOD 1 2 ). p = 12 k + 5.
Then ( a . c / p ) = - 1. THIS IMPLIES ( a / p ) . ( c / p ) = - 1 .
Here also , we have Zp
* = H U H . y.
If “ a “ LIES IN H , THEN “ c “ lies in H .y
If “ a “ LIES IN H .y , THEN “ c “ LIES IN H.
THEN ONLY b 2 = 3.ac , is solvable , for “ b “ LIES IN Z p
* , here p = 12 k + 5 .
IF WE CONSTRUCT f [ x ] , g [ x ] with the above conditions,

9. .
THEN M = [ f [i ] + g [ j ] ] defines a LATIN SQUARE OF ORDER “ p = 12 k + 5 “.
Now , again , let us have a close look at theorem – 2 of R. P.Singh and Soumen Maity
HERE “ p “ is any ODD PRIME . Select “ a “ such that ( a , p – 1 ) = 1 . b LIES IN Zp
*
With ( ( b 2 - 1 ) / p ) = 1 . here f [ t ] = ta ( t( p – 1 ) / 2 + b ) .
Now ( ( b 2 - 1 ) / p ) = 1 IMPLIES ( ( b – 1 ) / p ) . ( ( b + 1 ) / p ) = 1.
There SELECT “ b “ such that , b – 1 AND b + 1 LIES IN “ H “ ( OR ) b – 1 and b + 1 LIES IN
H .y. IN THIS CASE ALSO , WE CAN CONSTRUST M = [ f[ i ] + g [ j ] ] , LATIN SQUARE OF ORDER
“ p “
NOW LET US LEARN TWO IMPORTANT CONCEPTS FROM FINITE GROUP THEORY.
DEFINITION – 1 : COMPLETE MAPPING OF “ G “
LET “ G “ BE A GIVEN FINITE GROUP. LET f : G …….> G BE A PERMUTATION.
WE CALL “ f “ A COMPLETE MAPPING OF “ G “ IF THE MAPPING “ T “ : g …… > g. f ( g ) is
also a permutation on “ G “.
DEFINITION - 2 : WE CALL THE MAPPING “ f “ ( which is a permutation on “ G “ ) , AN
ORTHOMORPHISM OF “ G “ IF THE MAPPING “ U “ : g ……. > g – 1 . f ( g ) is a permutation
on “ G “
WE CALL A MAPPING “ f “ OF “ G “ , A STRONG COMPLETE MAPPING OF “ G “ IF IT IS
BOTH A COMPLETE MAPPING AND AN ORTHOMORPHISM OF “ G “
THEOREM – 1 : “ f “ IS A COMPLETE MAPPING OF “ G “ I f and only if the mapping “ T “
DEFINIED BY T ( g ) = g. f ( g ) is an orthomorphism of “ G “.
A mapping “ f “ is an orthomorphism of “ G “ I f and only if THE MAPPING “ U “ ,
DEFINED BY U ( g ) = g – 1 . f ( g ) IS A COMPLETE MAPPING OF “ G “ .
NOTE : LET G = { g 1 , g2 , g3 , …. gn – 1 , gn } be a given group. LET “ f “ be a given
permutation on “ G “ .

10. DEFINE A MATRIX A = [ ai , j ] = [ f ( gi ) . gj ] IS A LATIN SQUARE OF ORDER | G | , BASED
ON “ G “ .
LEMMA - 1 : SUPPOSE “ f “ IS A COMPLETE MAPPING OF “ G “.
DEFINE A = [ f ( gi ) . gj ] and B = [ gi . f ( gi ) . gj ] . THEN “ A “ IS ORTHOGONAL
TO “ B “ .
LEMMA – 2 : LET “ G “ BE A GIVEN GROUP OF ORDER “ n “ .
Define “ M “ BE THE “ n x n “ CAYLEY TABLE WITH “ i j “ th ENTRY = gi . gj
M = [ gi . gj ] . LET “ f “ BE A GIVEN PERMUTATION ON “ G “.
Mf = [ gi . f ( gj ) ]. IT IS EASY TO SEE THAT M IS A LATIN SQUARE , AND THAT Mf IS OBTAINED
FROM “ M “ BY PERMUTING ITS COLUMNS.
“Mf “ IS ORTHOGONAL TO “ M “ I f and only if “ f “ IS AN “ORTHOMORPHISM “ OF “ G .
LEMMA - 3 : LET “ f , g “ BE ANY TWO PERMUTATIONS OF “ G “.
NOW CONSTRUCT “ Mf AND Mg “ . THEN “ Mf IS ORTHOGONAL TO Mg “
I F AND ONLY I F THE MAPPING “ T “ DEFINED BY T ( x ) = ( g ( x ) )– 1 . f ( x ) IS A
PERMUTATION OF “ G “ . [ x is an element of “ G “ ].
AT THIS POINT OF TIME I WOULD LIKE TO INTRODUCE A VERY IMPORTANT DEFINITION
DEFINITION : LET “ f , g “ BE ANY TWO MAPPINGS OF A GROUP “ G “ .
WE SAY “ f AND g “ ARE ORTHOGONAL
IF THE MAPPING “ T “ , DEFINED BY T ( x ) = ( g ( x ) ) – 1 . f ( x ) IS A PERMUTATION
ON “ G “ .
FACT – 1 : A MAPPING f : G ---- G IS A “ COMPLETE MAPPING “ OF “ G “ IF IT IS
ORTHOGONAL TO THE MAPPINGS “ k* “ AND “ l* “
Where k* ( x ) = 1 ( the identity element of “ G “ ) FOR ALL “ x “ LIES IN “ G “
AND l* ( x ) = x – 1 FOR ALL “ x “ LIES IN “ G “ .
FACT – 2 : A MAPPING “ f “ IS AN ORTHOMORPHISM IF IT IS ORTHOGONAL TO “ k* “ AND
IDENTITY MAPPING “ IG “ ( IG ( x ) = x FOR ALL “ x “ LIES IN “ G “ ) .