A minimal non-abelin group is a non-abelian group all of whose subgroups are abelian. Following lemma gives an important property of minimal non-abelian p-groups. It will be used in the proof of Lemma 2.2.
Lemma 2.1
Let G be a minimal non-abelian p- group. Then${\scriptstyle \frac{G}{Z(G)}}\cong {Z}_{p}\times {Z}_{p}$.
ProofSince 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 ${\scriptstyle \frac{G}{Z(G)}}\cong {Z}_{p}\times {Z}_{p}$ 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.
Lemma 2.2
Let G be a group. If G is non-abelian, then G contains a subgroup H such that either${\scriptstyle \frac{H}{Z(H)}}\cong {Z}_{r}^{t}\u22ca{Z}_{q}$or Z_{p} × Z_{p} where p, q and r are distinct primes.
ProofTowards 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.
Lemma 2.3
Let G be a finite group.
(1) For every proper subgroup H of G, we have cp(G) ≤ cp(H).
(2) Whenever N ⊴ G, we have$cp(G)\le cp({\scriptstyle \frac{G}{N}})$.
(3) For every section, X, of G, we have cp(G) ≤ cp(X).
ProofSee proof of Lemma 2 of [3].
Lemma 2.4
Let G be a non-abelian group.
(1) If${\scriptstyle \frac{G}{Z(G)}}\cong {Z}_{p}^{t}\u22ca{Z}_{q}$, then$cp(G)={\scriptstyle \frac{{p}^{t}+{q}^{2}-1}{{p}^{t}q}}$.
(2) If${\scriptstyle \frac{G}{Z(G)}}\cong {Z}_{p}\times {Z}_{p}$, then$cp(G)={\scriptstyle \frac{{p}^{2}+p-1}{{p}^{3}}}$.
ProofLet G be a non abelian group and ${\scriptstyle \frac{G}{Z(G)}}\cong {Z}_{p}^{t}\u22ca{Z}_{q}$. Since ${\scriptstyle \frac{H}{Z(H)}}$ 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
$$cp(G)=\frac{(\mid G\mid -\mid Z(G)\mid (\frac{\mid G\mid}{p}))+\frac{{\mid G\mid}^{2}}{{p}^{2}}}{{\mid G\mid}^{2}}=\frac{{p}^{2}+p-1}{{p}^{3}}.$$Now we recall a nilpotent number defined in [7].
Definition 2.5
A positive integer n is called a nilpotent number if every group of order n is nilpotent.
Now we are ready to prove Theorem 1.1.
Proof of Theorem 1.1It is clear that there is at least one non-abelian group G of order n such that p is the smallest prime divisor of $\mid {\scriptstyle \frac{G}{Z(G)}}\mid $. 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 $cp(G)={\scriptstyle \frac{{p}^{2}+p-1}{{p}^{3}}}$ but ${\scriptstyle \frac{G}{Z(G)}}$ be not isomorphic to Z_{p} × Z_{p}. Then by Lemmas 2.2, 2.3 and 2.4 we have a contradiction. Now if ${\scriptstyle \frac{G}{Z(G)}}\cong {Z}_{p}\times {Z}_{p}$, 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|s^{i} − 1 for some integer i. Also let
$$\mathrm{\Upsilon}=\{(r,s)\mid r\ne s\hspace{0.17em}\text{are\hspace{0.17em}primes},\hspace{0.17em}r\mid {s}^{i}-1\hspace{0.17em}\text{but\hspace{0.17em}}r\nmid {s}^{j}-1\hspace{0.17em}\text{for\hspace{0.17em}all\hspace{0.17em}integers\hspace{0.17em}}j<i\}.$$Put also $\mathrm{\Gamma}=\{{\scriptstyle \frac{{r}^{t}+{s}^{2}-1}{{r}^{t}{s}^{2}}}\mid (r,s)\in \mathrm{\Upsilon}\}$. We know by [7] that ϒ and so Γ are non-empty. Now we choose (p, q) ∈ ϒ such that ${\scriptstyle \frac{{q}^{\alpha}+{p}^{2}-1}{{q}^{\alpha}{p}^{2}}}$ is a maximum element of Γ. We now define a group H.
It is well known that $Aut({Z}_{q}^{i})\cong {PGL}_{i}(q)$ and so
$$\mid Aut({Z}_{q}^{i})\mid =({q}^{i}-q)\hspace{0.17em}({q}^{i}-q)\cdots ({q}^{i}-{q}^{i-1}).$$Therefore there is some $\theta \in Aut({Z}_{q}^{i})$ with |θ| = q. Let
$$\iota :{Z}_{p}\to Aut({Z}_{q}^{i})$$be a group homomorphism such that ι(a) = θ and Z_{p} = 〈a〉. Now ${Z}_{q}^{i}\u22ca{Z}_{p}$ is a semidirect group with respect to ι and we denote ${Z}_{q}^{i}\u22ca{Z}_{p}\times A$ by H in which A is abelian group with $\mid A\mid ={\scriptstyle \frac{n}{{q}^{i}p}}$. By Lemma 2.4, we have $cp(H)={\scriptstyle \frac{{q}^{i}+{p}^{2}-1}{{q}^{i}p}}$.
Now we prove that for all non-nilpotent groups G of order n, $cp(G)\le {\scriptstyle \frac{{p}^{i}+{q}^{2}-1}{{p}^{i}q}}$. 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 ${\scriptstyle \frac{G}{Z(G)}}$ 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 ${\scriptstyle \frac{K}{Z(K)}}\cong {Z}_{r}^{t}\u22ca{Z}_{s}$ where r, s are distinct primes and t is an positive integer. By Lemma 2.3, $cp(G)\le cp({\scriptstyle \frac{K}{Z(K)}})={\scriptstyle \frac{{r}^{t}+{s}^{2}-1}{{r}^{t}{s}^{2}}}$. 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.
Corollary 2.6
Let G be a non- nilpotent group of order n. Then$cp(G)\le {\scriptstyle \frac{p+3}{4p}}$and equality holds if and only if${\scriptstyle \frac{G}{Z(G)}}\cong {D}_{2p}$.