Projective Geometry Structure of zn* (A Case of n=pqrs)

Projective Geometry Structure of z_n^* (A Case of n=pqrs)

Felix Komu¹, Benard M. Kivunge², Fredrick O. Nyamwala³

^1,3Moi University

²Kenyatta University

DOI: https://doi.org/10.51244/IJRSI.2025.12020031

Received: 21 January 2025; Review: 29 January 2025; Accepted: 31 January 2025; Published: 05 March 2025

ABSTRACT

The concept of projective geometry has been studied by a number of mathematicians, though the initial focus was on Euclidean and non Euclidean geometries on the relationship between lines and points on a 3D projective space. This paper will focus on the dempotent elements in Z^∗, for n=pqrs, specifically focusing on the triples, fano planes and order 15 projective structures. We shall count the number of the triples, fano planes and order 15 projective structures, and establish the relationship between them. It extends the results of projective geometry structure of Z_n*,n=pqr.

Key Phrases: Primes, Triple Systems, Fano Planes.

INTRODUCTION

Projective geometry structure of various permutation groups has been also computed. Projective geometry structure for the group Z_n*,n=pqr, has been computed This paper seeks to extent the established results for the group  We shall investigate the relationship between order 2 elements in the ring of units modulo , the triples, fano planes and order 15 geometric structures, n= pqrs.

Preliminaries

Fano plane

A fano plane is the smallest finite projective plane of order 2, containing 7 points and 7 lines that involves the set of integers and the modulus operations.

Triple system

A triple system in Z_n* is denoted by (a₁, a₂, a₃) where there exists k_i > 1, i = 1, 2, 3, such that a² ≡ 1(mod n) with a₁a₂ ≡ a₃(mod n), a₁a₃ ≡ a₂ (mod n) and a₂a₃ ≡ a₁ (mod n).

Euler phi function

Let a be an elements in Z^∗. Then Euler’s phi function φ(a) denotes the number of positive integers ≤ a and relatively prime to a.

Given that a is a natural number with (a, n) = 1, a^φ(n) ≡ 1 (mod n).

Chinese Remainder Theorem

Let n₁, n₂, . . . , n₃ be pairwise co-prime positive integers and x₁, x₂, . . . , x_k be arbitrary integers. The system of simultaneous congruence

a ≡ x₁(mod n₁)

a≡x₂ (modn₂)

a ≡ x_k(mod n_k)

has a unique solution modulo n = n₁n₂ . . . n_k

Corollary (Chinese remainder theorem)

If n is an odd number with n =p_1^(k_1 ) p_2^(k_2 )…p_r^(k_r ) where r is the number of distinct primes of n and k_i > 0 for 1 ≤ i ≤ r, then the equation x² ≡ 1 (mod n) has exactly 2^r distinct solutions (mod n).

Proof :

Suppose x^2≡1(modp_1^(k_1 ) p_2^(k_2 )…p_r^(k_r )), then p_1^(k_1 ) p_2^(k_2 )…(p_r^(k_r ))⁄x^2 -1. But since p_i^’ are distinct primes, then p_1^(k_1 ) p_2^(k_2 )…(p_r^(k_r ))⁄x^2 -1 only happens iff p_i^(k_i ) / x^2-1, for all 1≤i≤r. But each of the congruences has two solutions , i.e x=±1 (modp_i^(k_i )).

For each i, 1 ≤ i ≤ r choose y_i = ±1 and utilize the linear congruence’s system;

x≡y_1^(k_1 )
x≡y_2^(k_2 )
.
.
.
.
.
.
.
x≡y_r^(k_r )                                                                    .

By the Chinese Remainder Theorem, this system has a unique solution  modp_1^(k_1 ) p_2^(k_2 )…p_r^(k_r ).

Since we have two choices for each y_i (namely ±1), and we have r congruence’s, then the possible choices for y₁ , … , y_r are 2^r.

Assuming that the 2^r choices of x are not distinct (mod n), i.e. x₁ ≡ x₂(mod n), then

x1 ≡ x2(modp_i^(k_i )) for all i. However, any two values of x are not congruent p_i^(k_i ) for at least one i. Therefore, the above system of linear congruence’s has 2r distinct solutions x(mod n). Any of the 2r choices satisfies x2 ≡ 1 (modp_i^(k_i )) for 1 ≤ i ≤ r. Hence, there are 2r distinct solutions to x2 ≡ 1(mod n)

Lemma

For x² ≡ 1(mod n) and y² ≡ 1(mod n), we have (xy)² ≡ 1(mod n)

Proof:

(xy)² = x²y² ≡ 1 ∗ 1(mod n) = 1(mod n)

Lemma

Given x(xy) = x²y = 1 ∗ y ≡ y(mod n) and y(xy) = xy² = x ∗ 1 ≡ x(mod n), then xy ≡ ±1(mod n) Hence, (xy)² ≡ 1(mod n) if x² ≡ 1(mod n) and y² ≡ 1(mod n)

Proof:

From x² ≡ 1(mod n) and y² ≡ 1(mod n) we have, x ≡ ±1(mod n) and y ≡ ±1(mod n).

Hence, xy ≡ (±1)(±1)(mod n) = (±1)(mod n)

MAIN RESULTS

Theorem

Consider the set Z^∗ n = pqrs, where p, q, r, s are distinct odd primes. The equation x² ≡ 1(mod n) has 16 distinct solutions.

Proof:

By The Chinese Remainder Theorem, there are 2⁴ = 16 distinct solutions to the equation x² ≡ 1(mod n), for n = pqrs, with p, q, r, s being odd prime numbers and p > q > r > s > 2. These solutions are of the form:

Case A: We have two trivial solutions that correspond to the following two items;

x ≡ 1(mod p) ≡ 1(mod q) ≡ 1(mod r) ≡ 1(mod s)

This is the unit solution, and

x ≡ −1(mod p) ≡ −1(mod q) ≡ −1(mod r) ≡ −1(mod ) s)

Case B: The non-trivial solutions correspond to the following;

x ≡ 1(mod p) ≡ −1(mod q) ≡ −1(mod r) ≡ 1(mod s)

x ≡ 1(mod p) ≡ 1(mod q) ≡ −1(mod r) ≡ −1(mod s)

x ≡ 1(mod p) ≡ −1(mod q) ≡ 1(mod r) ≡ −1(mod s)

x ≡ −1(mod p) ≡ 1(mod q) ≡ 1(mod r) ≡ 1(mod s)

x ≡ −1(mod p) ≡ −1(mod q) ≡ −1(mod r) ≡ 1(mod s)

x ≡ −1(mod p) ≡ 1(mod q) ≡ −1(mod r) ≡ −1(mod s)

x ≡ 1(mod p) ≡ 1(mod q) ≡ −1(mod r) ≡ 1(mod s)

x ≡ −1(mod p) ≡ −1(mod q) ≡ 1(mod r) ≡ −1(mod s)

x ≡ 1(mod p) ≡ −1(mod q) ≡ 1(mod r) ≡ 1(mod s)

x ≡ 1(mod p) ≡ 1(mod q) ≡ 1(mod r) ≡ −1(mod s)

x ≡ 1(mod p) ≡ −1(mod q) ≡ −1(mod r) ≡ −1(mod s)

x ≡ −1(mod p) ≡ 1(mod q) ≡ −1(mod r) ≡ 1(mod s)

 x ≡ −1(mod p) ≡ −1(mod q) ≡ 1(mod r) ≡ 1(mod s)

x ≡ −1(mod p) ≡ 1(mod q) ≡ 1(mod r) ≡ −1(mod s)

Table 1: Non-unit idempotents in Z_n, for n = pqrs

p	Q	R	R	N	values of x satisfying x2 ≡ 1(mod n)
3	5	7	11	1155	34	76	274	386	419	461	496	659	694	736
3	5	7	13	1365	64	118	209	274	391	454	664	701	911	974
3	5	7	17	1785	169	239	356	526	596	664	764	1021	1121	1189
3	5	11	13	2145	131	274	571	584	716	859	989	1156	1286	1429
3	5	11	17	2805	254	494	749	934	1121	1189	1376	1429	1616	1684
3	11	13	17	3315	664	766	781	1106	1429	1444	1546	1769	1871	1886

values of x satisfying x2 ≡ 1(mod n)
769	881	1079	1121	1154
1091	1156	1184	1301	1364
1258	1429	1546	1616	1784
1561	1574	1871	2014	2144
1871	2056	2311	2551	2804
2209	2549	2549	2651	3314

Example

Consider the 15 non unit solutions Z^∗, n = 1155,, the triples are given by 34, 76, 274, 386, 419, 461, 496,659, 694, 736, 769, 881, 1079, 1121 and 1154. From these solutions, we make the following observations;

The 15 non unit solutions of the group generate 35 triples as follows. These triples are computed 9mod 1155).

34 ∗ 76 ≡ 274

34 ∗ 386 ≡ 419

34 ∗ 694 ≡ 496

34 ∗ 736 ≡ 769

34 ∗ 881 ≡ 1079

34 ∗ 1121 ≡ 1154

76 ∗ 386 ≡ 461

76 ∗ 419 ≡ 659

76 ∗ 496 ≡ 736

76 ∗ 694 ≡ 769

76 ∗ 881 ≡ 1121

76 ∗ 1079 ≡ 1154

274 ∗ 386 ≡ 659

274 ∗ 419 ≡ 461

274 ∗ 496 ≡ 769

274 ∗ 694 ≡ 736

274 ∗ 881 ≡ 1154

274 ∗ 1079 ≡ 1121

386 ∗ 496 ≡ 881

386 ∗ 694 ≡ 1079

386 ∗ 736 ≡ 1121

386 ∗ 769 ≡ 1154

419 ∗ 694 ≡ 881

419 ∗ 736 ≡ 1154

419 ∗ 769 ≡ 1121

461 ∗ 694 ≡ 1154

461 ∗ 736 ≡ 881

461 ∗ 769 ≡ 1079

496 ∗ 659 ≡ 1154

419 ∗ 496 ≡ 1079

496 ∗ 461 ≡ 1121

659 ∗ 694 ≡ 1121

659 ∗ 769 ≡ 881

659∗ 736 ≡ 1079

34∗ 461 ≡ 659

That;

34 ≡ 1(mod 3) ≡ −1(mod 5) ≡ −1(mod 7) ≡ 1(mod 11)

76 ≡ 1(mod 3) ≡ 1(mod 5) ≡ −1(mod 7) ≡ −1(mod 11)

274 ≡ 1(mod 3) ≡ −1(mod 5) ≡ 1(mod 7) ≡ −1(mod 11)

386 ≡ −1(mod 3) ≡ 1(mod 5) ≡ 1(mod 7) ≡ 1(mod 11)

419 ≡ −1(mod 3) ≡ −1(mod 5) ≡ −1(mod 7) ≡ 1(mod 11)

461 ≡ −1(mod 3) ≡ 1(mod 5) ≡ −1(mod 7) ≡ −1(mod 11)

496 ≡ 1(mod 3) ≡ 1(mod 5) ≡ −1(mod 7) ≡ 1(mod 11)

659 ≡ −1(mod 3) ≡ −1(mod 5) ≡ 1(mod 7) ≡ −1(mod 11)

694 ≡ 1(mod 3) ≡ −1(mod 5) ≡ 1(mod 7) ≡ 1(mod 11)

736 ≡ 1(mod 3) ≡ 1(mod 5) ≡ 1(mod 7) ≡ −1(mod 11)

769 ≡ 1(mod 3) ≡ −1(mod 5) ≡ −1(mod 7) ≡ −1(mod 11)

881 ≡ −1(mod 3) ≡ 1(mod 5) ≡ −1(mod 7) ≡ 1(mod 11)

1079 ≡ −1(mod 3) ≡ −1(mod 5) ≡ 1(mod 7) ≡ 1(mod 11)

1121 ≡ −1(mod 3) ≡ 1(mod 5) ≡ 1(mod 7) ≡ −1(mod 11)

1154 ≡ −1(mod 3) ≡ −1(mod 5) ≡ −1(mod 7) ≡ −1(mod 11)

This establishes theorem 3.3

To get a fano plane, we fix a triple, we premultiply the three points of the triple with a fourth point, we shall generate three more points. In total we shall have 7 points which forms a fano plane. Consider the 35 triples of Z^∗, n= 1155. Repeating this procedure for all the 35 triples, we generate 15 fano planes as listed as follows;

34, 76, 274, 386, 419, 461, 659

34, 76, 274, 496, 694, 736, 769

34, 76, 274, 881, 1079, 1121, 1154

34, 386, 419, 496, 694, 881, 1079

34, 386, 419, 736, 769, 1121, 1154

34, 461, 659, 496, 694, 1121, 1154

34, 461, 659, 736, 769, 881, 1079

76, 386, 461, 496, 736, 881, 1121

76, 386, 461, 694, 769, 1079, 1154

76, 419, 659, 694, 769, 881, 1121

76, 419, 496, 659, 736, 1079, 1154

1274, 386, 496, 659, 769, 881, 1154

274, 386, 659, 694, 736, 1079, 1121

274, 419, 461, 694, 736, 881, 1154

274, 419, 461, 496, 769, 1079, 1121

Fitting triples into a Fano plane:

The first fano plane can be drawn as follows;

All the other fano planes can be drawn in a similar manner.

Remarks

Every triple will appear in three distinct fano planes.

Order 15 Projective geometry structure of Z^∗, n= pqrs

To get the order 15 geometric structure, we fix a fano plane, and pick a distinct eighth element and premultiply it with all the elements of the fixed fano plane, we get the required geometric structure.

Example

Consider the 15 non unit solutions of Z^∗, n= 1155 given by 34, 76, 274, 386, 419, 461, 496, 659,

694, 736, 769, 881, 1079, 1121 and 1154. We have already seen that Z^∗, n= 1155 has 35 triples and 15 fano planes. Consider the first fano plane given by 34,76,274,386,419,461,659. If we fix these elements and premultiply all by 694, we shall get the elements 34, 76, 274, 386, 419, 461,659 ,694,694,736,769,881,1079,1121,1154. Fixing any other fano plane and pre multiplying its elements with a distinct element from, we shall get the same result. Therefore, we have only one order 15 geometric structure given as follows; 34,76,274,386,419,461,659 ,694 ,694,736,769,881,1079,1121,1154

We can make the following observations;

Each fano plane appears once in the order 15 geometric structures

The order 15 projective geometric structure is made up of the 15 fano planes and 35 triples

CONCLUSIONS

We have established that Z^∗, n= pqrs has 15 non unit solutions. These non unit solutions generate 35 triples. The 35 triples generate 15 fano planes which in turn generate only one order 15 projective geometric stricture.

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