Since G is non-abelian, there exist two elements x and y such that xy ≠ yx. Therefore C_G(x) and C_G(y) are two distinct centralizers of G. By hypothesis these centralizers are abelian and have index p. On the other hand if t ∈ (C_G(x) ∩ C_G(y)) Z(G) then C_G(t) = C_G(x) = C_G(y). This is a contradiction. Thus C_G(x) ∩ C_G(y) = Z(G) and so GZ(G)≅Zp×Zp as a desired.

Using the above lemma and Theorem 10.1.7 of [8] and Exercise 9.1.11 of [8], we can say that every non-abelian group contains either a minimal non-abelian p-subgroup or a subgroup with Frobenius central factor. The following lemma play an important role in the proof of our main theorem.

Towards contradiction, let G be a counter-example of minimal order. If H is a non-abelian subgroup of G, then it, and consequently G, has a subgroup with the desired structure. So we may assume that G is a minimal non-abelian group. Using Theorem 10.1.7 of [8] and Exercise 9.1.11 of [8] and Lemma 2.1 we get a contradiction.

The following lemma, which will be used in prove of Theorem 1.1, gives some relations between commutativity degree of certain groups and their sections and subgroups.

See proof of Lemma 2 of [3].

Let G be a non abelian group and GZ(G)≅Zpt⋊Zq. Since HZ(H) has an abelian subgroup of index q, Theorem A of [2] tells us that Ḡ is a Frobenius group. Now, all centralizers of Ḡ are either of order q or p^t. By counting all of them and using the definition we get the result in first case.

Now let G be a group with central factor isomorphic to Z_p × Z_p. It is clear that all centralizers of noncentral elements of G are abelian of index p. So as a result we have

Now we recall a nilpotent number defined in [7].

It is clear that there is at least one non-abelian group G of order n such that p is the smallest prime divisor of ∣GZ(G)∣. We denote some p-sylow subgroup of G by P. First we suppose that P is non-abelian and thus all Sylow p-subgroups P of G are non-abelian. By Lemma 2.2, P contains some minimal non abelian subgroup. Therefore by Lemma 2.3 and Lemma 2.4 we have the result in first case. Now let cp(G)=p2+p-1p3 but GZ(G) be not isomorphic to Z_p × Z_p. Then by Lemmas 2.2, 2.3 and 2.4 we have a contradiction. Now if GZ(G)≅Zp×Zp, then Lemma 2.4 gives the result in first case.

Now let G be a non-nilpotent group. Thus n is not a nilpotent number and so there are positive integers r and s such that r|sⁱ − 1 for some integer i. Also let

Put also Γ={rt+s2-1rts2∣(r,s)∈ϒ}. We know by [7] that ϒ and so Γ are non-empty. Now we choose (p, q) ∈ ϒ such that qα+p2-1qαp2 is a maximum element of Γ. We now define a group H.

It is well known that Aut(Zqi)≅PGLi(q) and so

Therefore there is some θ∈Aut(Zqi) with |θ| = q. Let

be a group homomorphism such that ι(a) = θ and Z_p = 〈a〉. Now Zqi⋊Zp is a semidirect group with respect to ι and we denote Zqi⋊Zp×A by H in which A is abelian group with ∣A∣=nqip. By Lemma 2.4, we have cp(H)=qi+p2-1qip.

Now we prove that for all non-nilpotent groups G of order n, cp(G)≤pi+q2-1piq. If all proper subgroups of G are nilpotent, then using Theorem 10.1.7 of [8] and Exercise 9.1.11 of [8] one can see that GZ(G) is Frobenius group whose Frobenius kernel is elementary abelian and whose Frobenius complement is of prime order. Now assume that G contains at least one proper non-nilpotent subgroup. Therefore by Lemma 2.2, G has a proper subgroup whose central factor is a Frobenius group with elementary abelian Frobenius kernel and cyclic Frobenius complement of prime order. Anyway, we can assume that K is a subgroup of G (perhaps G itself ) such that KZ(K)≅Zrt⋊Zs where r, s are distinct primes and t is an positive integer. By Lemma 2.3, cp(G)≤cp(KZ(K))=rt+s2-1rts2. But by choosing (p, q) ∈ ϒ, we have cp(G) ≤ cp(H) and proof is complete.

In the above theorem, if |G| is even, then we have the following.