IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 6 Issue 12, December 2019 ISSN (Online) 2348 – 7968 | Impact Factor (2019) – 6.248 www.ijiset.com FOUNDATIONS ON SEMI-PRIMITIVE ROOTS ALIREZA ALIZADEH MOGHADDAM ∗ Abstract. Z∗ Let n denoted as a multiplicative group of inte- 2 Z∗ gers modulo n. Any element a n is said to be a semi- primitive root if the set of H = f1; a; a2; :::; a('(n)=2)−1g is Z∗ the cyclic subgroup of n. On this article, we have dis- Z∗ cussed about some cyclic subgroup of the set of n which is generated by a semi-primitive root modulo n and we have reviewed some properties, Finally we have also studied the solution of some modular arithmetic equations by using of semi-primitive roots. 1. Introduction Definition 1.1. Let n be a positive integer, all of the integer numbers between 1 and n which are coprime to n form a group with multiplication modulo n as the Z∗ binary operation, this group denoted by n and is called the multiplicative group of integers modulo n. the cardinality of this group denoted by '(n), where '(n) is the Euler's phi- function. 2 Z∗ '(n) ≡ Note 1.1. It is obvious that for each a n , Euler's theorem states that a 1(mod n), and it's order which is denoted by ordn(a), is the smallest positive integer number k such that ak ≡ 1(mod n). Therefore a is said to be a primitive root modulo n ( i.e. primitive root) if the order of a modulo n is equal to '(n). Z∗ Note 1.2. It is well known that n as a cyclic multiplicative group usually has a Z∗ k k primitive root, also the n is cyclic if and only if n is equals to 1; 2; 4; p or 2p , where p is an odd prime number and k ≥ 1. Definition 1.2. Let n be any positive integer number, a 2 Zn is denoted as a Z∗ semi-primitive root modulo n, if the set of H is a cyclic subgroup of n as the following: '(n) − ∗ f 2 2 1g ≤ Z H = 1; a; a ; :::; a n 2 Z∗ 2 2 Z∗ It is obvious that if a n be a primitive root , then a n is a semi-primitive root. 2010 Mathematics Subject Classification. 11A07, 11A05. Key words and phrases. Multiplicative group of integers modulo n, Primitive roots, Semi- primitive roots. ∗Department of Mathematics, Faculty of Science, University of Yasouj, Yasouj, IRAN. 1 104 IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 6 Issue 12, December 2019 ISSN (Online) 2348 – 7968 | Impact Factor (2019) – 6.248 www.ijiset.com 2 A. R. ALIZADEH M. Z∗ Also all of non-cyclic groups as the form of n which are contain some semi-primitive roots are classified by the following Theorem. Z∗ Theorem 1.1. Let n be the multiplicative group of integers modulo n that does Z∗ not possess any primitive roots, then n has a semi-primitive root if and only if n k k k1 k2 k1 k2 is equals to 2 (k > 2), 4p1 1, p1 p2 or 2p1 p2 , where p1 and p2 are odd prime k1 k2 ≥ integers satisfying ('(p1 );'(p2 )) and k1; k2 1. Proof. See reference [4] □ Note 1.3. It is shown that in [3] that 3 is a semi-primitive root modulo 2k for any Z∗ integer k greater than 2 and 2k , k > 2 can be represented as Z∗ { i k−2g 2k = 3 mod n : i = 1; :::; 2 : Z∗ ∼ × Theorem 1.2. Suppose n = C2 C'(n)=2, then there exists a semi-primitive root 2 Z∗ h n such that '(n) Z∗ = {hi mod n : i = 1; :::; g: n 2 Proof. See reference [4] □ 2 Z∗ Z∗ Definition 1.3. The a n is said to be a good semi-primitive root "GSP " if n can be expressed as Z∗ × − n =< h > < 1 > Z∗ Theorem 1.3. If n be a non-cyclic group contains some semi-primitive roots, Z∗ '(n) then n has exactly 2'( 2 ) incongruent "GSP " roots. Z∗ × − Proof. According by Theorem "1.2", n =< h > < 1 > for a semi-primitive 2 Z∗ 2 Z∗ i − i root h n. In other words, any element a n can be expressed a = h or h , '(n) where i = 1; 2; :::; 2 . For the case of a = hi, i ordn(h) '(n)=2 ordn(h ) = = (ordn(h); i) ('(n)=2; i) Where ordn(a) indicates the order of a modulo n. This implies that '(n) '(n) ord (hi) = () ( ; i) = 1: n 2 2 Since h is a "GP S" root modulo n, it is also clear that (hi)j =6 −1 for all integers '(n) Z∗ j. Therefore they said that there are '( 2 ) incongruent "GSP " roots in n in the form of hi. '(n) Now they shown that there are also '( 2 ) incongruent "GSP " roots modulo n Z∗ − i − i in n in the form of h . More precisely, h is a "GSP " roots modulo n if and '(n) only if ( 2 ; i) = 1. They first note that i − i i 2ordn(h ) ( '(n) ) ordn( h ) = [2; ordn(h )] = i = ( ) : (2; ordn(h )) '(n) '(n)=2 2 ; i 2; (('(n)=2;i);i) ( ) − i '(n) '(n) '(n)=2 Then ordn( h ) = 2 if and only if ( 2 ; i) 2; ('(n)=2;i) = 2. In other words, '(n) (1) ( '(n) ; i) = 1 and ( '(n) ; 2) = 2 or ord (−hi) = () 2 2 n 2 '(n) '(n) (2) ( 2 ; i) = 2 and ( 2 ; 2) = 1 105 IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 6 Issue 12, December 2019 ISSN (Online) 2348 – 7968 | Impact Factor (2019) – 6.248 www.ijiset.com FOUNDATIONS ON SEMI-PRIMITIVE ROOTS 3 '(n) Since 2 is even for the cases under their consideration, the first case is simply '(n) '(n) − i j ̸≡ − ( 2 ; i) = 1. If ( 2 ; i) = 1 , ( h ) 1 for all integers j. '(n) − i j ≡ − Suppose that ( 2 ; i) = 1 and ( h ) 1 for an integer j, then j must be odd integer , since h is a "GP S" root modulo n, which it gives that hij ≡ −1 , '(n) j equivalently 2 ij. '(n) '(n) '(n) j Since ( 2 ; i) = 1 and 2 is even, therefore 2 j and hence j must be an even integer, which is leading to contradiction. '(n) '(n) '(n) For the second case, if ( 2 ; i) = 2 and ( 4 ; 2) = 1, then 4 is an odd integer i '(n) '(n) 1 and i is an even integer. Then (−h ) 4 = (−1)(h 2 ) 2 ≡ −1 mod n, which means that any semi-primitive root modulo n in the second case is not a "GSP ". − i '(n) □ This implies that the h is a "GSP " root modulo n if and only if ( 2 ; i) = 1 Z∗ Note 1.4. By the style of proving in the last theorem, it is shown that if n is '(n) a noncyclic group possessing semi-primitive roots and 4 is an odd integer, then Z∗ '(n) '(n) n has '( 4 ) more semi-primitive roots in addition to 2'( 2 )"GSP " roots. Now we want to show that the way of finding a "GSP " root in a non-cyclic multiplicative group. Theorem 1.4. Let m1 and m2 be coprime integers possessing primitive roots a and b respectively. If ('(m1);'(m2)) = 2 , there exist some semi-primitive roots modulo n = m1m2 and the solution to the system of linear congruences: x ≡ a(mod m ) (1) 1 x ≡ −b(mod m2) is a "GSP" root modulo n. Proof. By Chinese Remainder Theorem it is ensured that the system in (1) has a unique answer modulo n, say x0. Now by the hypotheses since a and b are primitive roots and (m1; m2) = 1 , it concludes that ['(m1);'(m2)] ≡ x0 1(mod n) k ≡ k ≡ In addition the equation of x0 1(mod n) equivalent to a 1(mod m1) and k (−b) ≡ 1(mod m2). Since a is a primitive root modulo m1 and '(m1)jk and ('(m1);'(m2)) = 2 so k k the k must be an even integer. Therefore (−b) = b ≡ 1(mod m2), equivalently '(m2)jk. Hence: '(m )'(m ) '(n) ord (x ) = ['(m );'(m )] = 1 2 = n 0 1 2 2 2 This implies that x0 is a semi-primitive root modulo n. '(n)=4 ̸≡ − Now if x0 1(mod n) , then x0 will be a "GSP " root modulo n. '(n)=4 ≡ − ≡ For this, let x0 1(mod m1m2) , then by the first congruence x a(mod m1) we have '(m1) '(m2) '(m2) −1 ≡ (a 2 ) 2 ≡ (−1) 2 (mod m1) '(m2) and therefore 2 is an odd integer. In the meantime, the second congruence x ≡ −b(mod m2) gives '(m2) '(m1) '(m2) '(m1) '(m1) '(m1) −1 ≡ ((−1) 2 ) 2 (b 2 ) 2 ≡ (−1) 2 (−1) 2 (mod m2) 106 IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 6 Issue 12, December 2019 ISSN (Online) 2348 – 7968 | Impact Factor (2019) – 6.248 www.ijiset.com 4 A. R. ALIZADEH M. Which is a contradiction □ Z∗ By the last two Theorems and their proofs we conclude that, if n is a non-cyclic Z∗ '(n) group possessing semi-primitive roots then n has exactly 2'( 2 ) (i.e. '('(n))) incongruent "GSP " roots and also the number of semi-primitive roots is either '(n) '(n) '('(n)) or '('(n)) + '( 4 ) according as 4 is even or odd. 2. Some Properties of Semi-primitive Roots In this section, it has been produced some key results for the semi-primitive roots Z∗ and "GSP " roots in n At the first step for producing of the number of semi-primitive roots for different positive values of n, we define f 2 Z∗ j g S(n) = g n g is a semi-primitive root modulo n So ( '('(n)) + '( '(n) if '(n) is odd j j 4 4 S(n) = '(n) '('(n)) if 4 is even Where cardinality of S(n) is denoted by jS(n)j.