A cyclic group C ( n ) is an elementary commutative group, and if n = p (prime), C ( p ) is known as one of the classifications of finite simple groups.

Every element in the cyclic group C ( n ) = { g 0 ( = e ) , g , g 2 , ⋯ , g n − 1 } is generated by a generator g. Euler’s totient function φ ( n ) is defined by

φ ( n ) = | { 1 ≤ k ≤ n | gcd ( k , n ) = 1 } | . (1)

Euler’s totient function φ ( n ) plays an intrinsically important role in the public key cipher RSA, which is essential in electronic commerce [1] .

The average probability of an element being a generator has not been discussed before. Several analytic properties of φ ( n ) have been investigated for a long time (e.g., [2] [3] ). However, it seems that some issues still remain unresolved.

Dirichlet [4] considered the mean values of sequences of integer values analytically; however, their understanding can be somewhat challenging because of their unfamiliarity.

In this paper, we derive the average probability of an element being a generator using the studies by Dirichlet [4] and Dirichlet and Dedekind [5] .

Throughout this paper, for a real number t, [t] denotes the integer part of t.

As for the possibility that two arbitrary natural numbers are coprime, the following result is mentioned in [6] . We prove the result for the sake of convenience.

Lemma 1. The probability that two arbitrary natural numbers are coprime is 6 π 2 .

Proof: Since the probability that two arbitrary natural numbers have a prime p as a common divisor is 1 − 1 p 2 , the probability that two arbitrary natural numbers are coprime is ∏ p : prime ( 1 − 1 p 2 ) . Noting that by the Euler product formula

∏ p : prime ( 1 / ( 1 − 1 p 2 ) ) = ∑ n = 1 ∞ 1 n 2 = ζ ( 2 ) ,

it holds that

∏ p : prime ( 1 − 1 p 2 ) = ζ ( 2 ) − 1 = 6 π 2 . (2)

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We also mention the following result.

Theorem 2. Choose two arbitrary natural numbers a and b, and consider an arithmetic progression { a , a + b , a + 2 b , ⋯ } . Then, the probability that the arithmetic progression includes an infinite number of primes is 6 π 2 .

Proof: The proof follows from Lemma 1 and Dirichlet’s theorem on arithmetic progressions [7] . □

In general, the cyclic group C ( n ) = { g 0 ( = e ) , g , g 2 , ⋯ , g n − 1 } generated by a generator g has generators g i , i ∈ S ( n ) ⊂ { 1 , 2 , ⋯ , n − 1 } . Then, C ( n ) is expressed as C ( n ) = { g 0 ( = e ) , g i , g 2 i , ⋯ , g ( n − 1 ) i } .

As for | S ( n ) | , which is the number of generators of the cyclic group C ( n ) , we prove the following lemma.

Lemma 3. | S ( n ) | = φ ( n ) .

Proof Let g be a generator. If g k is a generator, it follows that g = ( g k ) z = g k z , namely, g k z − 1 = e .

As we can write k z − 1 = q n + r , r < n ,

g k z − 1 = g q n + r = ( g n ) q ⋅ g r = g r = e .

As r = 0 because r < n ,

k z = 1 , mod n. (3)

Equation (3) implies that the Diophantine equation k z + n u = 1 has integer solutions z and u, which means k and n are coprimes by Bezout’s lemma. The converse is obvious.

Therefore, the theorem holds from the definition (1) of Euler’s totient function φ ( n ) . □

Consider P ( x ) = φ ( x ) x for x (integer). We can see that P ( 1 ) = 1 , P ( p ) = p − 1 p for p: prime, and P ( x ) > x − 1 2 for x > 6 , P ( x ) > x − 1 3 for x > 30 (see [8] ).

We can define E ( P ( X ) ) , the average probability of P ( x ) as follows.

E ( P ( X ) ) = lim x → ∞ 1 x ⋅ ∑ i = 1 x P ( i ) = lim x → ∞ 1 x ⋅ ∑ i = 1 x | S ( i ) | i = lim x → ∞ 1 x ⋅ ∑ i = 1 x φ ( i ) i .

Let ψ ( x ) = ∑ i = 1 x φ ( i ) .

Then, the following result holds.

Theorem 4. E ( P ( X ) ) = 6 π 2 = 0.6079271 ⋯ .

Proof: First, we derive φ ( n ) along the lines of Dirichlet [4] . It is well-known that

∑ δ | n φ ( δ ) = n (4)

for example, in [5] . Summing up both sides for n , n − 1 , ⋯ , 1 ,

∑ s = 1 n [ n s ] φ ( s ) = 1 2 n 2 + 1 2 n . (5)

For [ n s ] = t , it follows that

( ψ ( [ n t ] ) − ψ ( [ n t + 1 ] ) ) t = ( φ ( [ n t + 1 ] + 1 ) + ⋯ + φ ( [ n t ] ) ) t

because [ n t + 1 ] < s ≤ [ n t ] . We regard the right hand side as φ ( [ n t ] ) t if [ n t + 1 ] + 1 = [ n t ] , and as 0 if [ n t + 1 ] = [ n t ] .

Therefore, (4) turns out to be

∑ s = 1 n [ n s ] φ ( s ) = ∑ s = 1 n ψ ( [ n s ] ) . (6)

Hence,

∑ s = 1 n ψ ( [ n s ] ) = 1 2 n 2 + 1 2 n . (7)

We put

ψ ( n ) = 3 π 2 n 2 + ζ χ ( n ) , ∃ ζ   for   n . (8)

By (7),

ψ ( n ) = − ∑ s = 2 n ψ ( [ n s ] ) + 1 2 n 2 + 1 2 n . (9)

By (8),

− ∑ s = 2 n ψ ( [ n s ] ) = − ∑ s = 2 n ( 3 π 2 [ n s ] 2 + ζ χ ( [ n s ] ) ) = − 3 π 2 ∑ s = 2 n ( n s − ε ) 2 − ∑ s = 2 ∞ ζ χ ( [ n s ] ) ,     0 ≤ ε < 1 = − 3 n 2 π 2 ∑ s = 2 n 1 s 2 + 6 n π 2 ∑ s = 2 n ε s − 3 π 2 ∑ s = 2 n ε 2 − ∑ s = 2 ∞ ζ χ ( [ n s ] ) = − 3 n 2 π 2 ( π 2 6 − 1 + τ ) + 6 n π 2 ∑ s = 2 n ε s − 3 π 2 ∑ s = 2 n ε 2 − ∑ s = 2 ∞ ζ χ ( [ n s ] ) = − 1 2 n 2 + 3 π 2 n 2 + P n log n + B ∑ s = 2 ∞ χ ( n s ) ,     ∃ P , ∃ B

Substituting this back into (9),

ψ ( n ) = 3 π 2 n 2 + P n log n + A ∑ s = 2 ∞ χ ( n s ) , ∃ P , ∃ A (10)

From (8) and (10),

ζ χ ( n ) = P n log n + A ∑ s = 2 ∞ χ ( n s )

ζ = P n log n χ ( n ) + A χ ( n ) ∑ s = 2 ∞ χ ( n s ) .

We can write

χ ( n ) = n δ ,

where δ is a constant satisfying

∑ s = 2 n 1 s δ < ∑ s = 2 ∞ 1 s δ = q , 1 < δ < 2 .

Hence,

ζ = P n log n χ ( n ) + A n δ ∑ s = 2 ∞ ( n s ) δ = P n log n n δ + A q

for ∀ k > 0 , P n log n n δ < k , n ≥ N ,

Which implies

A ′ < k + A q , ∃ A ′ .

Especially, if we use δ satisfying

∑ s = 2 ∞ 1 s δ = 1 , (11)

we obtain

A ′ < A , n ≫ N .

Therefore, it follows that

ψ ( n ) = 3 π 2 n 2 + A ′ n δ , 1 < δ < 2 .

Thus,

φ ( x ) = ψ ( x ) − ψ ( x − 1 ) = 3 π 2 ( x 2 − ( x − 1 ) 2 ) + A ′ ( x δ − ( x − 1 ) δ ) = 6 π 2 x + o ( x ) ,     x : integer

this is because ( x − 1 ) δ = x δ + δ x δ − 1 ( − 1 ) + ⋯ by the Taylor expansion.

Hence,

E ( P ( X ) ) = lim x → ∞ 1 x ⋅ ∑ i = 1 x φ ( i ) i = 6 π 2 + lim x → ∞ 1 x ⋅ x ⋅ o ( 1 ) = 6 π 2 . (12)

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We conducted an elementary computational experiment, in which we computed 1 x ⋅ ∑ i = 1 x φ ( i ) i , x = 1 , ⋯ , 120 using Python (Figure 1). Figure 1 shows the validity of the result.

In this study, we have proved that the average probability of an element being a generator in the cyclic group is 6 / π 2 . It is interesting that this value is equal to the value of Lemma 1. This is evident in the following discussion.

Consider ( j , k ) , j = 1 , ⋯ , x ; k = 1 , ⋯ , x . Note that ( j , j ) is coprime for j = 1 and not coprime for j = 2 , ⋯ , x . Then,

Pr { ( j , k )   are   coprime   integers | j = 1 , ⋯ , x ; k = 1 , ⋯ , x } = ∑ k = 1 x Pr { ( ( j , k ) , j ≤ k )   arecoprimeintegers | j = 1 , ⋯ , k }       + ∑ j = 1 x Pr { ( ( j , k ) , j ≥ k )   arecoprimeintegers | k = 1 , ⋯ , j }       − ∑ j = 1 x Pr { ( j , j )   arecoprime   integers }

= 1 2 ( 1 + 1 x ) ⋅ 1 x ⋅ ∑ k = 1 x φ ( k ) k + 1 2 ( 1 + 1 x ) ⋅ 1 x ⋅ ∑ j = 1 x φ ( j ) j − 1 x 2 = 1 x ⋅ ∑ i = 1 x φ ( i ) i + 1 x 2 ⋅ ∑ i = 1 x φ ( i ) i − 1 x 2 → x → ∞ lim x → ∞ 1 x ⋅ ∑ i = 1 x φ ( i ) i = E ( P ( X ) )

We would like to further clarify the asymptotic property of P ( x ) = φ ( x ) / x itself, which seems like φ ( x ) / x → P 6 / π 2 when x → ∞ , in our future work.

The author declares no conflicts of interest regarding the publication of this paper.

Tanaka, Y. (2023) Average Probability of an Element Being a Generator in the Cyclic Group. American Journal of Computational Mathematics, 13, 230-235. https://doi.org/10.4236/ajcm.2023.132012