Sums of element orders in groups of odd order

TL;DR

This paper investigates the sum of element orders in finite groups of odd order, establishing bounds and characterizing groups that attain these bounds based on their prime factorization.

Contribution

It provides new bounds for the sum of element orders in non-cyclic odd order groups and characterizes the structure of groups that achieve these bounds.

Findings

For q=3, the ratio of sum of element orders to group order is at most 85/301.

For q>3, the ratio is bounded by a function involving the smallest prime greater than q.

Equality cases are explicitly characterized by specific group structures.

Abstract

Denote by $G$ a finite group and by $\psi(G)$ the sum of element orders in $G$. If $t$ is a positive integer, denote by $C_t$ the cyclic group of order $t$ and write $\psi(t)=\psi(C_t)$. In this paper we proved the following Theorem A: Let $G$ be a non-cyclic group of odd order $n=qm$, where $q$ is the smallest prime divisor of $n$ and $(m,q)=1$. Then the following statements hold. (1) If $q=3$, then $\frac {\psi(G)}{\psi(|G|)}\leq \frac {85}{301}$, and equality holds if and only if $n=3\cdot 7\cdot m_1$ with $(m_1,42)=1$ and $G=(C_7\rtimes C_3)\times C_{m_1}$, with $C_7\rtimes C_3$ non-abelian. (2) If $q>3$, then $\frac {\psi(G)}{\psi(|G|)}\leq \frac {p^4+p^3-p^2+1}{p^5+1}$, where $p$ is the smallest prime bigger than $q$ and equality holds if and only if $n=qp^2m_1$ with $(m_1,p!)=1$ and $G=C_q\times C_p\times C_p \times C_{m_1}$.

Equations

$$\psi (G)=\Sigma \{o(x)\mid x\in G\},$$

$$\psi (G)\le (\frac{1}{3}+\frac{2l}{3\psi (l)})\psi (C_n),$$

$$\frac{\psi (G)}{\psi (C_n)}\le \frac{q^4+q^3-q^2+1}{q^5+1}$$

$$f(x)=\frac{x^4+x^3-x^2+1}{x^5+1}\text{and}{g}_q(x)=\frac{x^2-x+1+x(q^2-q)}{(x^2-x+1)(q^2-q+1)}.$$

$$g_q(q_1)\le g_q(2q+1)\text{and}f(p)\ge f(2q-3).$$

$$\frac{(2q+1{)}^2-(2q+1)+1+(2q+1)(q^2-q)}{((2q+1{)}^2-(2q+1)+1)(q^2-q+1)}<\frac{(2q-3{)}^4+(2q-3{)}^3-(2q-3{)}^2+1}{(2q-3{)}^5+1}.$$

$$\frac{2q^3+3q^2+q+1}{4q^4-2q^3+3q^2+q+1}<\frac{(2q-3{)}^2+(2q-3)-1}{(2q-3{)}^3}$$

$$\frac{2q^3+3q^2+q+1}{4q^4-2q^3+3q^2+q+1}<\frac{4q^2-10q+5}{8q^3-36q^2+54q-27}.$$

$$16q^6-48q^5+8q^4+80q^3-63q^2+27q-27<16q^6-48q^5+52q^4-36q^3+9q^2-5q+5,$$

$$116q^3+32q<44q^4+72q^2+32,$$

$$\psi (G)<\psi (⟨a⟩\times ⟨b⟩)\psi (⟨y⟩).$$

$$\psi (⟨a⟩\times ⟨b⟩)=\psi (⟨b⟩)+p(\psi (⟨a⟩)-1)=p^2-p+1-p+p\psi (⟨a⟩)=(p-1{)}^2+p\psi (⟨a⟩).$$

$$\psi (G)=(p-1{)}^2\psi (Z)+p\psi (⟨a⟩)\psi (⟨y⟩)<((p-1{)}^2+p\psi (⟨a⟩))\psi (⟨y⟩)=\psi (⟨a⟩\times ⟨b⟩)\psi (⟨y⟩),$$

$$\frac{\psi (G)}{\psi (|G|)}\le \frac{p^2-p+1+p(q^2-q)}{(p^2-p+1)(q^2-q+1)}.$$

$$\frac{\psi(G}{\psi(|G|)}=\frac{\psi(Z)}{\psi(F)}+\frac{1}{\lambda(p^{\alpha})}\big{(}1-\frac{\psi(Z)}{\psi(F)}\big{)}\leq\frac{\psi(Z)}{\psi(F)}+\frac{1}{\lambda(p)}\big{(}1-\frac{\psi(Z)}{\psi(F)}\big{)}=\frac{\psi(Z)}{\psi(F)}\big{(}1-\frac{1}{\lambda(p)}\big{)}+\frac{1}{\lambda(p)}.$$

$$\frac{\psi (G)}{\psi (|G|)}$$

$$=\frac{(p^2-2p+1)+p(q^2-q+1)}{(q^2-q+1)(p^2-p+1)}=\frac{p^2-p+1+p(q^2-q)}{(q^2-q+1)(p^2-p+1)},$$

$$\frac{\psi (G)}{\psi (|G|)}\le \frac{p_2^2-p_2+1+6p_2}{7(p_2^2-p_2+1)}=g_3(p_2)\le g_3(7)=\frac{85}{301},$$

$$[G:⟨x⟩]<\frac{2(p^5+1)p_2}{(p^4+p^3-p^2+1)p}<2p_2.$$

$$\frac{\psi (G)}{\psi (|G|)}\le \frac{p_2^2-p_2+1+p_2(q^2-q)}{(p_2^2-p_2+1)(q^2-q+1)}=g_q(p_2)\le g_q(q_1)<f(p),$$

$$\frac{\psi (G)}{\psi (|G|)}\le \frac{p_2^2-p_2+1+6p_2}{7(p_2^2-p_2+1)}=g_3(p_2)\le g_3(q_1)=\frac{\psi (G)}{\psi (|G|)}.$$

$$\psi (G)<\psi (⟨a⟩\times ⟨b⟩)\psi (⟨y⟩)\le f(p_2)\psi (|G|)\le f(5)\psi (|G|)=\frac{121}{521}\psi (|G|)<\frac{85}{301}\psi (|G|).$$

$$[G:⟨x⟩]<\frac{2p}{p+3}(p_2+1)<2p_2.$$

$$\frac{\psi (G)}{\psi (|G|)}\le \frac{p_2^2-p_2+1+p_2(q^2-q)}{(q^2-q+1)(p_2^2-p_2+1)}=g_q(p_2)\le g_q(p_1).$$

$$\psi (G)<\psi (⟨a⟩\times ⟨b⟩)\psi (⟨y⟩)\le f(p_2)\psi (|G|)\le f(p)\psi (|G|)=\psi (G).$$

$$\frac{\psi (G)}{\psi (|G|)}\le f(p),$$

$$\frac{\psi (G)}{\psi (|G|)}\le f(7)=\frac{337}{2101},$$

$$\frac{\psi (G)}{\psi (|G|)}\le \frac{85}{301},$$

Full text

Sums of element orders in groups of odd order

Marcel Herzog*, Patrizia Longobardi** and Mercede Maj**

*School of Mathematical Sciences

Tel-Aviv University

Ramat-Aviv, Tel-Aviv, Israel

**Dipartimento di Matematica

Università di Salerno

via Giovanni Paolo II, 132, 84084 Fisciano (Salerno), Italy

Denote by $G$ a finite group and by $\psi(G)$ the sum of element orders in $G$. If $t$ is a positive integer, denote by $C_{t}$ the cyclic group of order $t$ and write $\psi(t)=\psi(C_{t})$. In this paper we proved the following Theorem A: Let $G$ be a non-cyclic group of odd order $n=qm$, where $q$ is the smallest prime divisor of $n$ and $(m,q)=1$. Then the following statements hold. (1) If $q=3$, then $\frac{\psi(G)}{\psi(|G|)}\leq\frac{85}{301}$, and equality holds if and only if $n=3\cdot 7\cdot m_{1}$ with $(m_{1},42)=1$ and $G=(C_{7}\rtimes C_{3})\times C_{m_{1}}$, with $C_{7}\rtimes C_{3}$ non-abelian. (2) If $q>3$, then $\frac{\psi(G)}{\psi(|G|)}\leq\frac{p^{4}+p^{3}-p^{2}+1}{p^{5}+1}$, where $p$ is the smallest prime bigger than $q$ and equality holds if and only if $n=qp^{2}m_{1}$ with $(m_{1},p!)=1$ and $G=C_{q}\times C_{p}\times C_{p}\times C_{m_{1}}$.

KEYWORDS: Group element orders; Finite groups

MSC[2010] 20D60; 20E34

††support: This work was supported by the National Group for Algebraic and Geometric Structures, and their Applications (GNSAGA - INDAM), Italy. The first author is grateful to the Department of Mathematics of the University of Salerno for its hospitality and support, while this investigation was carried out.

1. Introduction

The problem of detecting structural properties of periodic groups by looking at the orders of their elements has been considered by various authors, from many different points of view. In [1], H. Amiri, S.M. Jafarian Amiri and I.M. Isaacs introduced the function

[TABLE]

where $o(x)$ denotes the order of the element $x$ of the finite group $G$. They proved the following theorem:

Theorem 1

Let $G$ be a group of order $n$, and denote by $C_{n}$ the cyclic group of order $n$. Then $\psi(G)\leq\psi(C_{n})$, with equality if and only if $G=C_{n}$.

Thus $C_{n}$ is the unique group of order $n$ with the largest value of $\psi(G)$ for groups of that order. Following this publication, in several recent papers it was proved that certain classes of finite groups $G$ can be characterized using the function $\psi(G)$ (see [2],[3],[5],[6],[8],[9],[10],[12],[14],[15] and [18]).

The problem of determining the second largest value of $\psi$ for groups of order $n$ and finding groups satisfying that condition was discussed in several papers, for example [4],[7], [14] and [17]. Recently S.M. Jafarian Amiri and M. Amiri in [7] (see also [4]) and R. Shen, G. Chen and C. Wu in [17] studied finite groups $G$ of order $n$ with the second largest value of $\psi(G)$, and obtained information about the structure of $G$ if $n=p_{1}^{\alpha_{1}}\dotsb p_{t}^{\alpha_{t}}$, where $p_{i}$ are primes satisfying $p_{1}<\dotsb<p_{t}$, $\alpha_{i}$ are positive integers and $\alpha_{1}>1$. Our aim is to continue this investigation in the case when $\alpha_{1}=1$.

In our paper [11] we obtained the following result which improves Theorem 1:

Theorem 2

If $G$ is a non-cyclic group of order $n$, then $\psi(G)\leq\frac{7}{11}\psi(C_{n})$. Moreover, this upper bound is best possible.

Recently we determined all groups $G$ of order $n$ satisfying $\psi(G)=\frac{7}{11}\psi(C_{n})$. In fact, we proved in [14] the following theorem:

Theorem 3

Let $G$ be a non-cyclic group of order $n$. Then $\psi(G)=\frac{7}{11}\psi(C_{n})$ if and only if $n=4m$ with $m$ an odd integer and $G=C_{2}\times C_{2}\times C_{m}$.

In the paper [13], we studied groups of order $n=p_{1}^{\alpha_{1}}\dotsb p_{t}^{\alpha_{t}}$ with $p_{i}$ and $\alpha_{i}$ as stated above, where $p_{1}=2$ and $\alpha_{1}=1$. We proved the following two theorems.

Theorem 4

Let $G$ be a non-cyclic group of order $n=2m$, where $m$ is an odd integer. Then $\psi(G)\leq\frac{13}{21}\psi(C_{n})$. Moreover, $\psi(G)=\frac{13}{21}\psi(C_{n})$ if and only if $G=S_{3}\times C_{n/6}$, where $n=6m_{1}$ with $(m_{1},6)=1$ and $S_{3}$ is the symmetric group on three letters.

Thus the groups $G=S_{3}\times C_{n/6}$, where $n=6m_{1}$ and $(m_{1},6)=1$, are the groups with the second largest value of $\psi$ for groups of order $2m$, with $m$ odd.

The second result in [13] is the following theorem.

Theorem 5

Let $\Delta_{n}$ be the set of non-cyclic groups of the fixed order $n=2m$, where $m$ is an odd integer, and suppose that $m=p_{1}^{\alpha_{1}}\dotsb p_{t}^{\alpha_{t}}$, where $p_{i}$ are distint primes and $\alpha_{i}$ are positive integers for all $i$. If $G\in\Delta_{n}$, then

[TABLE]

where $l=\min(p_{i}^{\alpha_{i}}\mid i\in\{1,\dots,t\})$ and $\psi(l)=\psi(C_{l})$.

Moreover, $G\in\Delta_{n}$ satisfies $\psi(G)=(\frac{1}{3}+\frac{2l}{3\psi(l)})\psi(C_{n})$ if and only if $G=D_{2l}\times C_{n/2l}$, where $D_{2l}$ denotes the dihedral group of order $2l$.

Thus if $n=2m$, with $m$ odd, then $D_{2l}\times C_{n/2l}$ is the unique group of order $n$ with the second largest value of $\psi$ for groups of that order. In particular, if $l=3$, then $D_{2l}=S_{3}$ and $\frac{1}{3}+\frac{2l}{3\psi(l)}=\frac{1}{3}+\frac{6}{3\cdot 7}=\frac{13}{21}$, in agreement with Theorem 4.

Using Theorem 2, Theorem 3, Theorem 4 and results from [7] and [17], we could obtain the following result:

Theorem 6

Let $G$ be a non-cyclic group of even order $n=2^{\alpha}m$, with $m$ an odd integer. Then the following statements hold.

where $Q_{8}$ is the quaternion group of order $8$ and $G_{1}=\langle a,b\mid a^{2^{\alpha}}=b^{3}=1,b^{a}=b^{-1}\rangle$.

In particular, these upper bounds are best possible.

In the paper [14], we started the investigation in the odd case. By Theorem 3 in [11] we have

Theorem 7

If $G$ is a non-cyclic group of order $n$ and if $q$ denotes the smallest prime divisor of $n$, then $\frac{\psi(G)}{\psi(C_{n})}<\frac{1}{q-1}$.

Theorem 7 implies the following corollary.

Corollary 8

If $G$ is a non-cyclic group of odd order $n$, then $\frac{\psi(G)}{\psi(C_{n})}<\frac{1}{2}$.

In [14] we proved the following theorem, which yields a better result in the odd case.

Theorem 9

If $G$ is a non-cyclic group of order $n$ and if $q$ denotes the smallest prime divisor of $n$, then

[TABLE]

and the equality holds if and only if $n=q^{2}k$ with $(k,q!)=1$ and $G=C_{q}\times C_{q}\times C_{k}$.

As shown in Lemma 2.1 of Section 2, the function $f(x)=\frac{x^{4}+x^{3}-x^{2}+1}{x^{5}+1}$ is strongly decreasing for $x\geq 2$. Notice that $f(2)=\frac{7}{11}$ and $f(3)=\frac{25}{61}$. Thus Theorem 9 implies the following result.

Theorem 10

If $G$ is a non-cyclic group of order $n$, then the following statements hold.

Demonstration Proof

Let $q$ denote the smallest prime divisor of $n$. If $n$ is even, then $q=2$ and by Theorem 9 $\frac{\psi(G)}{\psi(C_{n})}\leq f(2)$, with equality as required. If $n$ is odd, then $q\geq 3$ and by Theorem 9 $\frac{\psi(G)}{\psi(C_{n})}\leq f(q)$. It follows by Lemma 2.1 that $f(q)\leq f(3)$, so $\frac{\psi(G)}{\psi(C_{n})}\leq f(3)=\frac{25}{61}$, as required. The condition for equality is again as required. ∎

Hence if $G$ is a non-cyclic group of odd order, then $\frac{\psi(G)}{\psi(C_{n})}\leq\frac{25}{61}<\frac{1}{2}$ and the upper bound $\frac{25}{61}$ is best possible.

In this paper we are looking for a result similar to Theorem 5 for groups of odd order $n$. Denote by $q$ the smallest prime divisor of $n$. As in the case when $q=2$, we start with groups of order $n=qm$, where $q>2$ and $(q,m)=1$. We denote by $C_{7}\rtimes C_{3}$ the non-abelian group of order $21$. We prove the following theorem:

Theorem A

Let $G$ be a non-cyclic group of odd order $n=qm$, where $q$ is the smallest prime divisor of $n$ and $(m,q)=1$. Then the following statements hold.

Theorem A implies the following corollary.

Corollary B

Let $G$ be a non-cyclic group of odd order $n=qm$, where $q$ is the smallest prime divisor of $n$ and $(m,q)=1$. Then the following statements hold.

Finally, Theorem A and results of [7], [14] and [17] imply the following corollary.

Corollary C

Let $G$ be a non-cyclic group of odd order $n=mq^{\alpha}$, where $q$ is the smallest prime divisor of $n$, $\alpha$ is a positive integer and $(m,q)=1$. Moreover, let $p$ be the smallest prime bigger than $q$. Then the following statements hold.

These upper bounds are best possible.

The proof of Theorem A will be achieved in two steps. The first step is

Proposition D

Let $G$ be a non-cyclic group of odd order $n=qm$, where $q$ is the smallest prime divisor of $n$ and $(m,q)=1$. Then the following statements hold.

The second step is

Proposition E

Let $|G|=n=qm$, where $q$ is the smallest prime divisor of $n$ and $(m,q)=1$. Then the following statements hold.

It is clear that Theorem A follows from Propositions D and E. These propositions will be proved in Sections 3 and 4, respectively. Section 2 will be devoted to preliminary results and Corollary B will be proved in Section 5.

Our notation will be the usual one (see for example [16]). In particular, $\varphi(n)$ denotes the Euler’s function on $n$. If $k$ is a positive integer, we write $\psi(k)=\psi(C_{k})$, and, following [17], $\lambda(k)=\frac{\psi(k)}{k}$. Finally, if $X$ is a subset of $G$, we shall define $\psi(X)=\Sigma_{x\in X}o(x)$.

2. Preliminary results

First we define two functions of real variables which will play important roles in our proofs.

Definition Definition

Let $q>1$ be a positive integer. Then

[TABLE]

Our first result is the following lemma.

Lemma 2.1

The functions $f$ and $g_{q}$ are strongly decreasing for $x\geq 2$.

Demonstration Proof

We have $g_{q}^{\prime}(x)=\frac{(q^{2}-q)(1-x^{2})}{(q^{2}-q+1)(x^{2}-x+1)^{2}}<0$ for $x\geq 2$. Furthermore, if $x\geq 2$, then $f(x)=\frac{x^{3}-x+1}{x^{4}-x^{3}+x^{2}-x+1}$ and $f^{\prime}(x)=\frac{-xh(x)}{(x^{4}-x^{3}+x^{2}-x+1)^{2}}$, where $h(x)=x^{5}-4x^{3}+8x^{2}-7x+2>0$. Hence also $f^{\prime}(x)<0$ for $x\geq 2$. ∎

In sequel, we shall often use without reference the fact that if $m$ and $n$ are integers satisfying $m>n\geq 2$, then $f(m)<f(n)$ and $g_{q}(m)<g_{q}(n)$.

Our second result is the following important proposition.

Proposition 2.2

Let $q\geq 3$ be a prime. Denote by $p$ the smallest prime bigger than $q$ and by $q_{1}$ the smallest prime congruent to $1\ (mod\ q)$. Then the following statements hold.

Demonstration Proof

(1) If $q=3$, then $p=5$ and $q_{1}=7$. Thus $g_{3}(q_{1})=\frac{7^{2}-7+1+7(3^{2}-3)}{(7^{2}-7+1)(3^{2}-3+1)}=\frac{85}{301}$ and $f(p)=\frac{5^{4}+5^{3}-5^{2}+1}{5^{5}+1}=\frac{121}{521}<\frac{85}{301}$, as required.

(2) By our assumption $q>3$. Clearly $q_{1}\geq 2q+1$ and by the Bertrand’s conjecture $p\leq 2q-3$. As shown in Lemma 2.1, the functions $g_{q}(x)$ and $f(x)$ are both decreasing for $x\geq 2$, which implies that

[TABLE]

Therefore it suffices to prove that $g_{q}(2q+1)<f(2q-3)$. So we need to prove that

[TABLE]

Since $\frac{a+1}{b+1}>\frac{a}{b}$ if $b>a$, it suffices to prove that

[TABLE]

or

[TABLE]

This inequality is equivalent to the inequality

[TABLE]

or

[TABLE]

which is certainly true if $q>3$.

The proof of the proposition is now complete. ∎

In the next lemma, we collect some basic information about the function $\psi$.

Lemma 2.3

Let $G$ denote a finite group, $p,p_{i}$ denote primes and $n,m,\alpha_{i}$ denote positive integers. Then the following statements hold.

We conclude this section with the following two lemmas which follow easily from previous results in [13].

Lemma 2.4

Let $G=(\langle a\rangle\times\langle b\rangle)\rtimes\langle y\rangle$, where $p$ is an odd prime number and $o(a)=p^{\alpha-1}$ for some integer $\alpha>1$, $o(b)=p$, $(o(y),p)=1$, $a^{y}=a$ and $b^{y}=b^{r}$ for some integer $r$ not congruent to $1$ modulo $p$. Then

[TABLE]

Demonstration Proof

By Theorem 3.2 of [13] we have $\psi(G)=(p-1)^{2}\psi(Z)+p\psi(\langle a\rangle)\psi(\langle y\rangle)$, where $Z=C_{\langle y\rangle}(\langle b\rangle)$. On the other hand,

[TABLE]

Thus $\psi(\langle a\rangle\times\langle b\rangle)\psi(\langle y\rangle)=((p-1)^{2}+p\psi(\langle a\rangle))\psi(\langle y\rangle)$ and $Z\subset\langle y\rangle$ implies that $\psi(Z)<\psi(\langle y\rangle)$. Hence

[TABLE]

as required. ∎

Lemma 2.5

Let $G$ be a non-cyclic group and $G=P\rtimes\langle y\rangle$, where $P$ is a non-trivial cyclic $p$-subgroup of $G$ for some prime $p$ and $(|P|,|\langle y\rangle|)=1$. If $q$ is the smallest prime divisor of $|\langle y\rangle|$, then

[TABLE]

Demonstration Proof

Write $F=\langle y\rangle$ and let $Z=C_{F}(P)$. Then, by Lemma 2.3(6), we have $\psi(G)=\psi(P)\psi(Z)+|P|(\psi(F)-\psi(Z))$. Thus $\frac{\psi(G)}{\psi(|G|)}=\frac{\psi(G)}{\psi(P)\psi(F)}=\frac{\psi(Z)}{\psi(F)}+\frac{|P|}{\psi(|P|)}(1-\frac{\psi(Z)}{\psi(F)})$.

Write $|P|=p^{\alpha}$, with $\alpha$ a positive integer, and if $k$ is a positive integer, let $\lambda(k)=\frac{\psi(k)}{k}$. By Lemma 2.8 of [13] we have $\lambda(p)\leq\lambda(p^{\alpha})$, so $\frac{1}{\lambda(p^{\alpha})}\leq\frac{1}{\lambda(p)}$. Thus we have

[TABLE]

Arguing as in the proof of Proposition 6 of [14], we have $\frac{\psi(Z)}{\psi(F)}\leq\frac{1}{q^{2}-q+1}$. Hence

[TABLE]

as required. ∎

3. Proof of Proposition D

Demonstration Proof

We assume that $G$ is a non-cyclic group of odd order $|G|=n=qm$ with $(m,q)=1$. This implies that $G=M\rtimes C$, where $C\simeq C_{q}$ and $M$ is its normal complement.

First assume that $q=3$. Then $|G|=3m$ with $(3,m)=1$ and suppose that $\frac{\psi(G)}{\psi(|G|)}>\frac{85}{301}$. Our aim is to reach a contradiction. We argue by induction on $|G|$.

If $G=M\times C$, then $\psi(G)=\psi(M)\psi(C)$, where $M$ is non-cyclic and $3$ does not divide $|M|$. By Theorem 4 of [14] and Lemma 2.1 we have $\psi(M)\leq\frac{5^{4}+5^{3}-5^{2}+1}{5^{5}+1}\psi(|M|)=\frac{121}{521}\psi(|M|)$. Hence $\psi(G)\leq\frac{121}{521}\psi(|G|)<\frac{85}{301}\psi(|G|)$, a contradiction. Therefore $C$ is not normal in $G$.

Let $p_{2}$ be the largest prime divisor of $n$. Clearly $p_{2}\geq 5$. By Lemma 2.3(4), we have $\psi(|G|)\geq\frac{3}{p_{2}+1}n^{2}$, so $\psi(G)>\frac{85}{301}\psi(|G|)\geq\frac{255}{301(p_{2}+1)}n^{2}$. Hence there exists $x\in G$ such that $o(x)>\frac{255}{301(p_{2}+1)}n$, which implies that $[G:\langle x\rangle]<\frac{301}{255}(p_{2}+1)<2p_{2}$.

Suppose, first, that $p_{2}$ does not divide $[G:\langle x\rangle]$. Then $\langle x\rangle$ contains a cyclic Sylow $p_{2}$-subgroup $P$ of $G$. Since $\langle x\rangle\leq N_{G}(P)$, it follows that $[G:N_{G}(P)]<2p_{2}$. But $[G:N_{G}(P)]=1+kp_{2}$ for some non-negative integer $k$, so $1+kp_{2}<2p_{2}$ and since $G$ is of odd order, $[G:N_{G}(P)]=1+p_{2}$ is impossible. Hence $[G:N_{G}(P)]=1$ and $P$ is a normal cyclic Sylow $p_{2}$-subgroup of $G$. Now Lemma 2.3(5) and our assumptions imply that $\psi(P)\psi(G/P)\geq\psi(G)>\frac{85}{301}\psi(P)\psi(|G/P|)$. Thus $\psi(G/P)>\frac{85}{301}\psi(|G/P|)$. Since $3$ divides $|G/P|$, it follows by the induction hypothesis that $G/P$ is cyclic. Write $G=P\rtimes F$, where $F$ is isomorphic to $G/P$. Then $F$ is cyclic and $3$ divides $|F|$. Since $C$ is not normal in $G$, $3$ does not divide $C_{F}(P)$ and since $P$ is cyclic, it follows by Lemma 2.3(7) that $p_{2}$ is congruent to $1$ (mod $3$). Hence $p_{2}\geq 7$ and $g_{3}(p_{2})\leq g_{3}(7)$. Now Lemma 2.5 implies that

[TABLE]

a contradiction.

Now suppose that $p_{2}$ does divide $[G:\langle x\rangle]$. Then $[G:\langle x\rangle]<2p_{2}$ implies that $[G:\langle x\rangle]=p_{2}$. By Theorem 3.1 of [13], $G=P\rtimes F$, where $P$ is a normal Sylow $p_{2}$-subgroup of $G$, $F$ is a cyclic subgroup of $G$, and either $P$ is cyclic, or $G$ is nilpotent, or $G=(\langle a\rangle\times\langle b\rangle)\rtimes\langle y\rangle$, where $o(a)=p_{2}^{\alpha-1}$ for some integer $\alpha>1$, $o(b)=p_{2}$, $(o(y),p_{2})=1$, $a^{y}=a$ and $b^{y}=b^{r}$ for some integer $r$ not congruent to $1$ (mod $p_{2}$). If $P$ is cyclic, then we get a contradiction arguing as in the previous paragraph. If $G$ is nilpotent, then $C$ is normal in $G$, a contradiction. Finally, in the last case, it follows by Lemma 2.4 that $\psi(G)<\psi(\langle a\rangle\times\langle b\rangle)\psi(\langle y\rangle)$. Applying Theorem 9 to $\langle a\rangle\times\langle b\rangle$ we obtain $\psi(G)<\frac{p_{2}^{4}+p_{2}^{3}-p_{2}^{2}+1}{p_{2}^{5}+1}\psi(|\langle a\rangle\times\langle b\rangle|)\psi(\langle y\rangle)$. Now $\psi(|\langle a\rangle\times\langle b\rangle|)\psi(\langle y\rangle)=\psi(|G|)$ and since $p_{2}\geq 5$, Lemma 2.1 implies that $\frac{p_{2}^{4}+p_{2}^{3}-p_{2}^{2}+1}{p_{2}^{5}+1}\leq\frac{5^{4}+5^{3}-5^{2}+1}{5^{5}+1}=\frac{121}{521}$. Thus $\psi(G)<\frac{121}{521}\psi(|G|)<\frac{85}{301}\psi(|G|)$, a contradiction. The proof in the case when $q=3$ is now complete.

We assume, now, that $q>3$. Then $|G|=qm$ with $(q,m)=1$ and suppose that $\frac{\psi(G)}{\psi(|G|)}>\frac{p^{4}+p^{3}-p^{2}+1}{p^{5}+1}$, where $p$ is the smallest prime larger than $q$. Denote by $q_{1}$ the smallest prime congruent to $1$ modulo $q$. Our aim is to reach a contradiction. We argue by induction on $|G|$.

If $G=M\times C$, then $\psi(G)=\psi(M)\psi(C)$, where $M$ is non-cyclic and $q$ does not divide $|M|$. Therefore $\frac{\psi(G)}{\psi(|G|)}=\frac{\psi(M)}{\psi(|M|)}\frac{\psi(C)}{\psi(|C|)}=\frac{\psi(M)}{\psi(|M|)}$ and by Theorem 9 $\frac{\psi(M)}{\psi(|M|)}\leq\frac{q_{2}^{4}+q_{2}^{3}-q_{2}^{2}+1}{q_{2}^{5}+1}=f(q_{2})$, where $q_{2}$ is the smallest prime divisor of $|M|$. Obviously $q_{2}\geq p$, so $f(q_{2})\leq f(p)$ and $\frac{\psi(G)}{\psi(|G|)}=\frac{\psi(M)}{\psi(|M|)}\leq\frac{p^{4}+p^{3}-p^{2}+1}{p^{5}+1}$, a contradiction. Therefore $C$ is not normal in $G$.

Let $p_{2}$ be the largest prime divisor of $n$. Then, by Lemma 2.3(4), we have $\psi(|G|)\geq\frac{q}{p_{2}+1}n^{2}>\frac{q-1}{p_{2}}n^{2}$, so $\psi(G)>\frac{p^{4}+p^{3}-p^{2}+1}{p^{5}+1}\psi(|G|)>\frac{(p^{4}+p^{3}-p^{2}+1)(q-1)}{(p^{5}+1)p_{2}}n^{2}$. Hence there exists $x\in G$ such that $o(x)>\frac{(p^{4}+p^{3}-p^{2}+1)(q-1)}{(p^{5}+1)p_{2}}n$, which implies that $[G:\langle x\rangle]<\frac{(p^{5}+1)p_{2}}{(p^{4}+p^{3}-p^{2}+1)(q-1)}$. By Bertrand’s conjecture $p\leq 2q-2$, so $q-1\geq\frac{p}{2}$ and

[TABLE]

First, suppose that $p_{2}$ does not divide $[G:\langle x\rangle]$. Then $\langle x\rangle$ contains a cyclic Sylow $p_{2}$-subgroup $P$ of $G$ and it follows, as in the $q=3$ case, that $P$ is a normal subgroup of $G$. Now Lemma 2.3(5) and our assumptions imply that $\psi(P)\psi(G/P)\geq\psi(G)>\frac{p^{4}+p^{3}-p^{2}+1}{p^{5}+1}\psi(P)\psi(|G/P|)$. Thus $\psi(G/P)>\frac{p^{4}+p^{3}-p^{2}+1}{p^{5}+1}\psi(|G/P|)$. Since $q$ divides $|G/P|$, it follows by the inductive hypothesis that $G/P$ is cyclic. Write $G=P\rtimes F$, where $F$ is isomorphic to $G/P$. Then $F$ is a cyclic group and $q$ divides $|F|$. As shown in the $q=3$ case, $p_{2}$ is congruent to $1$ (mod $q$), implying that $p_{2}\geq q_{1}$ and $g_{q}(p_{2})\leq g_{q}(q_{1})$. Since by Proposition 2.2(1) $g_{q}(q_{1})<f(p)$, applying Lemma 2.5 we obtain

[TABLE]

a contradiction.

Now suppose that $p_{2}$ does divide $[G:\langle x\rangle]$. Since $[G:\langle x\rangle]<2p_{2}$, it follows that $[G:\langle x\rangle]=p_{2}$. Then, by Theorem 3.1 of [13], $G=P\rtimes F$, where $P$ is a normal Sylow $p_{2}$ subgroup of $G$, $F$ is a cyclic subgroup of $G$, and either $P$ is cyclic, or $G$ is nilpotent, or $G=(\langle a\rangle\times\langle b\rangle)\rtimes\langle y\rangle$, where $o(a)=p_{2}^{\alpha-1}$ for some integer $\alpha>1$, $o(b)=p_{2}$, $(o(y),p_{2})=1$, $a^{y}=a$ and $b^{y}=b^{r}$ for some integer $r$ not congruent to $1$ (mod $p_{2}$). If $P$ is cyclic, then we get a contradiction arguing as in the previous paragraph. If $G$ is nilpotent, then $C$ is normal in $G$, a contradiction. Finally, in the last case, it follows by Lemma 2.4 that $\psi(G)<\psi(\langle a\rangle\times\langle b\rangle)\psi(\langle y\rangle)$. Applying Theorem 9 to $\langle a\rangle\times\langle b\rangle$ we obtain $\psi(G)<\frac{p_{2}^{4}+p_{2}^{3}-p_{2}^{2}+1}{p_{2}^{5}+1}\psi(|\langle a\rangle\times\langle b\rangle|)\psi(\langle y\rangle)$. Now $\psi(|\langle a\rangle\times\langle b\rangle|)\psi(\langle y\rangle)=\psi(|G))$ and $p_{2}\geq p$ implies that $\frac{p_{2}^{4}+p_{2}^{3}-p_{2}^{2}+1}{p_{2}^{5}+1}\leq\frac{p^{4}+p^{3}-p^{2}+1}{p^{5}+1}$. It follows that $\psi(G)<\frac{p^{4}+p^{3}-p^{2}+1}{p^{5}+1}\psi(|G|)$, a contradiction.

The proof of Proposition D is now complete.

3. Proof of Proposition E

Demonstration Proof

Let $q_{1}$ be the smallest prime congruent to $1$ (mod $q$). Recall that $p$ is the smallest prime larger than $q$, $f(x)=\frac{x^{4}+x^{3}-x^{2}+1}{x^{5}+1}$ and $g_{q}(x)=\frac{x^{2}-x+1+x(q^{2}-q)}{(x^{2}-x+1)(q^{2}-q+1)}$.

First assume that $q=3$.

Then $q_{1}=7$ and $p=5$. First we show that if $G=C_{k}\times(C_{7}\rtimes C_{3})$ where $(k,42)=1$, then $\frac{\psi(G)}{\psi(|G|)}=g_{3}(7)=\frac{85}{301}$. Indeed, $\psi(G)=\psi(C_{k})\psi(C_{7}\rtimes C_{3})=\psi(C_{k})(7^{2}-7+1+14\cdot 3)=\psi(C_{k})85$, so $\frac{\psi(G)}{\psi(|G|)}=\frac{\psi(C_{k})85}{\psi(C_{k})\psi(C_{7})\psi(C_{3})}=\frac{85}{301}$, as claimed.

Now we turn to the ”only if” part. So suppose that $q=3$, $|G|=n=3m$, $m$ is odd, $(m,3)=1$ and $\frac{\psi(G)}{\psi(|G|)}=g_{3}(7)=\frac{85}{301}$. Then $G$ is non-cyclic and $G=M\rtimes C$, where $C\simeq C_{3}$ and $3$ does not divide $|M|$. In particular, the least prime divisor $r$ of $|M|$ satisfies $r\geq 5$. Our aim is to show that $G=C_{k}\times(C_{7}\rtimes C_{3})$ where $(k,42)=1$. We argue by induction on $|G|$.

If $G=M\times C$, then $M$ is non-cyclic and $\psi(G)=\psi(M)\psi(C)$. Moreover, since $r\geq 5$, Theorem 9 and Lemma 2.1 imply that $\psi(M)\leq\frac{(5^{3}-5+1)(5+1)}{5^{5}+1}\psi(|M|)=\frac{121}{521}\psi(|M|)$. Hence $\frac{\psi(G)}{\psi(|G|)}=\frac{\psi(M)\psi(C)}{\psi(|M|)\psi(C)}\leq\frac{121}{521}<\frac{85}{301}$, a contradiction. Therefore $C$ is not normal in $G$.

Let $p_{2}$ be the largest prime divisor of $|G|$ and let $P$ be a Sylow $p_{2}$-subgroup of $G$. Then, by Lemma 2.3(4), we have $\psi(|G|)\geq\frac{3}{p_{2}+1}n^{2}$ and our assumptions imply that $\psi(G)=\frac{85}{301}\psi(|G|)\geq\frac{255}{301(p_{2}+1)}n^{2}$. Hence there exists $x\in G$ such that $o(x)>\frac{255}{301(p_{2}+1)}n$, which implies that $[G:\langle x\rangle]<\frac{301}{255}(p_{2}+1)<2p_{2}$.

First, suppose that $p_{2}$ does not divide $[G:\langle x\rangle]$. Then $\langle x\rangle$ contains a cyclic Sylow $p_{2}$-subgroup $P$ of $G$ and as shown in the proof of Proposition D, $P$ is a normal cyclic Sylow subgroup of $G$. Now Lemma 2.3(5) and our assumption imply that $\psi(P)\psi(G/P)\geq\psi(G)=\frac{85}{301}\psi(P)\psi(|G/P|)$, with equality if and only if $P$ is central in $G$. Hence $\psi(G/P)\geq\frac{85}{301}\psi(|G/P|)$. Notice that $|G/P|=3m_{1}$, with $m_{1}$ odd and $(m_{1},3)=1$.

If $G/P$ is non-cyclic, then by Proposition D the above inequality implies that $\psi(G/P)=\frac{85}{301}\psi(|G/P|)$. Hence $\psi(P)\psi(G/P)=\psi(G)$ and by Lemma 2.3(5) $P$ is central in $G$. Thus $G=P\times F$, where $F\simeq G/P$. Moreover, by induction $G/P=C_{k}\times(C_{7}\rtimes C_{3})$, where $(k,42)=1$, and since $F\simeq G/P$, we may also write $F=C_{k}\times(C_{7}\rtimes C_{3})$, where $(k,42)=1$. Since $(|P|,|G/P|)=1$, it follows that $(|P|,42k)=1$ and $G=C_{|P|}\times F=C_{k|P|}\times(C_{7}\rtimes C_{3})$, where $(k|P|,42)=1$, as required.

So assume that $G/P$ is cyclic. Write $G=P\rtimes F$, where $F$ is isomorphic to $G/P$. Then $F$ is cyclic and $3$ divides $|F|$. Since $C$ is not normal in $G$, $3$ does not divide $C_{F}(P)$. Therefore, by Lemma 2.3(7), $C_{3}$ acts fixed-point-freely on $P$, implying that $p_{2}$ is congruent to $1$ (mod $3$). Hence $p_{2}\geq q_{1}$ and $g_{3}(p_{2})\leq g_{3}(q_{1})$. Applying Lemma 2.5 we obtain

[TABLE]

Hence $p_{2}=q_{1}=7$ and $\frac{\psi(G)}{\psi(|G|)}=\frac{p_{2}^{2}-p_{2}+1+6p_{2}}{7(p_{2}^{2}-p_{2}+1)}$. The second equality implies, by the proof of Lemma 2.5, that $|P|=p_{2}=7$. Since $p_{2}$ is the largest prime divisor of $|G|$, it follows that $G=C_{7}\rtimes C_{3}C_{5^{c}}$ for some non-negative integer $c$. Since no $5$-element can act fixed -point-freely on $C_{7}$, it follows by Lemma 2.3(7) that $C_{5^{c}}$ acts trivially on $C_{7}$. Since $3$ does not divide $C_{F}(P)$, it follows that $G=C_{5^{c}}\times(C_{7}\rtimes C_{3})$, as required.

Now suppose that $p_{2}$ does divide $[G:\langle x\rangle]$. Since $[G:\langle x\rangle]<2p_{2}$, it follows that $[G:\langle x\rangle]=p_{2}$. Then, by Theorem 3.1 of [13], $G=P\rtimes F$, where $P$ is a normal Sylow $p_{2}$-subgroup of $G$, $F$ is a cyclic subgroup of $G$, and either $P$ is cyclic, or $G$ is nilpotent, or $G=(\langle a\rangle\times\langle b\rangle)\rtimes\langle y\rangle$, where $o(a)=p_{2}^{\alpha-1}$ for some integer $\alpha>1$, $o(b)=p_{2}$, $(o(y),p_{2})=1$, $a^{y}=a$ and $b^{y}=b^{r}$ for some integer $r$ not congruent to $1$ (mod $p_{2}$). Since $G/P$ is cyclic, the previous arguments imply that if $P$ is cyclic, then $G=C_{5^{c}}\times(C_{7}\rtimes C_{3})$, where $c$ is a non-negative integer, as required. Since $C$ is not normal in $G$, $G$ is not nilpotent. Finally, since $p_{2}\geq 5$, in the last case we reach the following contradiction, using Lemma 2.4, Theorem 9 and Lemma 2.1:

[TABLE]

The proof in the case $q=3$ is now complete.

Now assume that $q>3$.

If $G=C_{k}\times C_{q}\times C_{p}\times C_{p}$ with $(k,p!)=1$, then $\psi(G)=\psi(C_{k})\psi(C_{q})\psi(C_{p}\times C_{p})=\psi(C_{k})\psi(C_{q})(\frac{p^{4}+p^{3}-p^{2}+1}{p+1})$ and $\frac{\psi(G)}{\psi(|G|)}=(\frac{p^{4}+p^{3}-p^{2}+1}{p+1})\frac{1}{\psi(C_{p^{2}})}=\frac{p^{4}+p^{3}-p^{2}+1}{p^{5}+1}$, as claimed.

Now we turn to the ”only if” part. So suppose that $q>3$, $|G|=n=qm$, $m$ is odd, $(m,q)=1$ and $\frac{\psi(G)}{\psi(|G|)}=\frac{p^{4}+p^{3}-p^{2}+1}{p^{5}+1}=f(p)$. Then $G$ is non-cyclic and $G=M\rtimes C$, where $C\simeq C_{q}$ and $q$ does not divide $|M|$. Denote by $r$ the least prime divisor of $|M|$. Clearly $r\geq p$. Our aim is to show that $G=C_{k}\times C_{q}\times C_{p}\times C_{p}$ with $(k,p!)=1$. We argue by induction on $|G|$.

If $G=M\times C$, then $M$ is non-cyclic and $\psi(G)=\psi(M)\psi(C)$. Moreover, since $r\geq p$, Theorem 9 and Lemma 2.1 imply that $\psi(M)\leq f(r)\psi(|M|)\leq f(p)\psi(|M|)$. Hence $\frac{\psi(G)}{\psi(|G|)}=\frac{\psi(M)\psi(C)}{\psi(|M|)\psi(C)}\leq f(p)$. But by our assumptions $\frac{\psi(G)}{\psi(|G|)}=f(p)$, so $r=p$ and $\frac{\psi(M)}{\psi(|M|)}=f(p)=f(r)$. Therefore, by Theorem 9, $M=C_{p}\times C_{p}\times C_{k}$, with $(k,p!)=1$ and hence $G=C_{q}\times C_{p}\times C_{p}\times C_{k}$ with $(k,p!)=1$, as required.

Suppose, now, that $C$ is not normal in $G$.

Let $p_{2}$ be the largest prime divisor of $|G|$ and let $P$ be a Sylow $p_{2}$-subgroup of $G$. Then, by Lemma 2.3(4), we have $\psi(|G|)\geq\frac{q}{p_{2}+1}n^{2}$ and our assumptions imply that $\psi(G)=f(p)\psi(|G|)\geq\frac{f(p)q}{p_{2}+1}n^{2}$. Hence there exists $x\in G$ such that $o(x)>\frac{f(p)q}{p_{2}+1}n$, which implies that $[G:\langle x\rangle]<\frac{p_{2}+1}{f(p)q}$. Now by the Bertrand’s conjecture we have $p\leq 2q-3$, so $q\geq\frac{p+3}{2}$. Since $f(p)=\frac{p^{4}+p^{3}-p^{2}+1}{p^{5}+1}>\frac{1}{p}$, it follows that $f(p)q>\frac{p+3}{2p}$ and as $p_{2}\geq p$, we get

[TABLE]

First, suppose that $p_{2}$ does not divide $[G:\langle x\rangle]$. Then $\langle x\rangle$ contains a cyclic Sylow $p_{2}$-subgroup $P$ of $G$. As shown in the proof of Proposition D, $P$ is a normal cyclic Sylow subgroup of $G$. Now Lemma 2.3(5) and our assumptions imply that $\psi(P)\psi(G/P)\geq\psi(G)=f(p)\psi(P)\psi(|G/P|)$. Hence $\psi(G/P)\geq f(p)\psi(|G/P|)$. Notice that $|G/P|=qm_{1}$, with $m_{1}$ odd and $(m_{1},q)=1$.

Suppose, first, that $G/P$ is non-cyclic. Then, by Proposition D, the above inequality implies that $\psi(G/P)=f(p)\psi(|G/P|)$. Hence $\psi(P)\psi(G/P)=\psi(G)$ and by Lemma 2.3(5) $P$ is central in $G$. Thus $G=P\times F$, where $F\simeq G/P$. Moreover, by induction, $G/P=C_{q}\times C_{p}\times C_{p}\times C_{k_{1}}$ with $(k_{1},p!)=1$ and since $F\simeq G/P$, we may also write $F=C_{q}\times C_{p}\times C_{p}\times C_{k_{1}}$ with $(k_{1},p!)=1$. As $(|P|,|G/P|)=1$, we have $(|P|,k_{1}pq)=1$. Hence $(|P|,k_{1})=1$ and as $p_{2}>p$, also $(|P|,p!)=1$. Thus $G=P\times F=C_{k_{1}|P|}\times C_{q}\times C_{p}\times C_{p}$ with $(k_{1}|P|,p!)=1$, as required.

Suppose, next, that $G/P$ is cyclic. Write $G=P\rtimes F$, where $F$ is isomorphic to $G/P$. Then $F$ is cyclic and $q$ divides $|F|$. Since $C$ is not normal in $G$, $q$ does not divide $C_{F}(P)$. Therefore, by Lemma 2.3(7), $C_{q}$ acts fixed-point-freely on $P$, implying that $p_{2}$ is congruent to $1$ (mod $q$). Hence $p_{2}\geq p_{1}$ and $g_{q}(p_{2})\leq g_{q}(p_{1})$. Applying Lemma 2.5 we obtain

[TABLE]

By Proposition 2.2(1) $g_{q}(p_{1})<f(p)$, so $\frac{\psi(G)}{\psi(|G|)}<f(p)$, in contradiction to our assumptions.

Now suppose that $p_{2}$ does divide $[G:\langle x\rangle]$. Since $[G:\langle x\rangle]<2p_{2}$, it follows that $[G:\langle x\rangle]=p_{2}$. Then, by Theorem 3.1 of [13], $G=P\rtimes F$, where $P$ is a normal Sylow $p_{2}$-subgroup of $G$, $F$ is a cyclic subgroup of $G$, and either $P$ is cyclic, or $G$ is nilpotent, or $G=(\langle a\rangle\times\langle b\rangle)\rtimes\langle y\rangle$, where $o(a)=p_{2}^{\alpha-1}$ for some integer $\alpha>1$, $o(b)=p_{2}$, $(o(y),p_{2})=1$, $a^{y}=a$ and $b^{y}=b^{r}$ for some integer $r$ not congruent to $1$ (mod $p_{2}$).

If $P$ is cyclic, then $G=P\rtimes F$, where both $P$ and $F$ are cyclic, and we reach a contradiction as above. Since $C$ is not normal in $G$, $G$ is not nilpotent. Finally, consider the last case. Since $p_{2}\geq p$, we have $f(p_{2})\leq f(p)$. Now, applying Lemma 2.4 and Theorem 9, we reach the following contradiction:

[TABLE]

The proof of Proposition E is now complete. ∎

5. Proof of Corollary B

Demonstration Proof

Recall that $f(x)=\frac{x^{4}+x^{3}-x^{2}+1}{x^{5}+1}$. Let $G$ be a non-cyclic group of odd order $n=qm$, where $q$ is the minimum prime divisor of $n$ and $(m,q)=1$.

Suppose, first, that $q>3$. Then, by Theorem A,

[TABLE]

where $p$ is the smallest prime bigger than $q$. Since $q>3$, it follows that $q\geq 5$, $p\geq 7$ and by Lemma 2.1 $f(p)\leq f(7)=\frac{337}{2101}$. Hence if $q>3$, then

[TABLE]

and by Proposition E equality holds if and only if $n=5\cdot 7^{2}\cdot m_{1}$ with $(m_{1},7!)=1$ and $G=C_{5}\times C_{7}\times C_{7}\times C_{m_{1}}$, as required.

Suppose, now, that $q=3$. Then by Theorem A

[TABLE]

with equality if and only if $n=3\cdot 7\cdot m_{1}$ with $(m_{1},42)=1$ and $G=(C_{7}\rtimes C_{3})\times C_{m_{1}}$.

Since $\frac{337}{2101}<\frac{85}{301}$, the result concerning groups with $q\geq 3$ follows immediately.

∎