A result on the sum of element orders of a finite group

TL;DR

This paper investigates the sum of element orders in finite groups, proving a modified version of Herzog's conjecture that relates this sum to the group's structure and solvability.

Contribution

It provides a proof for a modified Herzog's conjecture, establishing a new inequality involving the sum of element orders and group solvability.

Findings

Proves a modified version of Herzog's conjecture.

Establishes a bound on the sum of element orders for non-solvable groups.

Identifies the equality case as the group A_5.

Abstract

Let $G$ be a finite group and $\psi(G)=\sum_{g\in{G}}{o(g)}$. There are some results about the relation between $\psi(G)$ and the structure of $G$. For instance, it is proved that if $G$ is a group of order $n$ and $\psi(G)>\dfrac{211}{1617}\psi(C_n)$, then $G$ is solvable. Herzog {\it{et al.}} in [Herzog {\it{et al.}}, Two new criteria for solvability of finite groups, J. Algebra, 2018] put forward the following conjecture: \noindent{\bf Conjecture.} {\it {If $G$ is a non-solvable group of order $n$, then $${\psi(G)}\,{\leq}\,{{\dfrac{211}{1617}}{\psi(C_n)}}$$ with equality if and only if $G=A_5$. In particular, this inequality holds for all non-abelian simple groups.} } In this paper, we prove a modified version of Herzog's Conjecture.

Equations

$$\psi (G)\le \frac{211}{1617}\psi (C_n)$$

$$\psi (P)=\frac{p^{2r+1}+1}{p+1}=\frac{p|P{|}^2+1}{p+1}.$$

$$\psi (C_n)=\prod _{i=1^t}\psi (P_i)\ge \frac{2}{p+1}{n}^2.$$

$$[G:⟨x⟩]<\frac{s}{r}\cdot \frac{p_1+1}{p_1}\cdots \frac{p_k+1}{p_k}.$$

$$\psi (C_n)\ge \frac{5005}{1152}\frac{n^2}{p+1}.$$

$$\psi (P)\psi (G/P)\ge \psi (G)=\frac{211}{1617}\psi (C_n)=\frac{211}{1617}\psi (C_{|P|})\psi (C_{|G/P|}).$$

$$\psi (G/P)\ge \frac{211}{1617}\psi (C_{|G/P|}).$$

$$\psi (G)=\frac{211}{1617}\psi (C_n)>\frac{211}{1617}\cdot \frac{2^{2\alpha +1}}{3}\cdot \frac{3^{2\beta +1}}{4}\cdot \frac{5^{2\gamma +1}}{6}=\frac{211}{1617}\cdot \frac{5}{12}{n}^2.$$

$$18=[G:⟨x⟩]=[G:N_G(P_5)][N_G(P_5):⟨x⟩]=(1+5k)[N_G(P_5):⟨x⟩].$$

$$\frac{211}{1617}\cdot \frac{5}{12}{n}^2<\psi (G)\le \psi (G/H)|H{|}^2=1237(n/180{)}^2,$$

$$10=[G:⟨x⟩]=[G:N_G(P_3)][N_G(P_3):⟨x⟩]=(1+3k)[N_G(P_3):⟨x⟩].$$

$$12=(1+5k)[N_G(P_5):⟨x⟩].$$

$$\frac{211}{1617}\cdot \frac{5}{12}{n}^2<\psi (G)\le \psi (G/H)|H{|}^2\le 663(n/120{)}^2,$$

$$\frac{G}{G^{\prime }\cap Z(G)}\cong \frac{G^{\prime }Z(G)}{G^{\prime }\cap Z(G)}\cong \frac{G^{\prime }}{G^{\prime }\cap Z(G)}\times \frac{Z(G)}{G^{\prime }\cap Z(G)}\cong A_5\times C_m,$$

$$\psi (G)\le \psi (\frac{G}{G^{\prime }\cap Z(G)})|G^{\prime }\cap Z(G){|}^2\le \psi (A_5)\psi (C_m)|G^{\prime }\cap Z(G){|}^2=4\cdot 211\psi (C_m).$$

$$\psi (G)=\psi (A_5\times Z(G))$$

$$<211\cdot \frac{\psi (C_{2^{\alpha }})}{\psi (C_4)}\cdot \frac{\psi (C_{3^{\beta }})}{\psi (C_3)}\cdot \frac{\psi (C_{5^{\gamma }})}{\psi (C_5)}=\frac{211}{1617}\psi (C_n),$$

Full text

Conjecture. If $G$ is a non-solvable group of order $n$, then

[TABLE]

*with equality if and only if $G=A_{5}$. In particular, this inequality holds for all non-abelian simple groups. *

In this paper, we prove a modified version of Herzog’s Conjecture.

1. Introduction

Let $G$ be a finite group and $\pi(G)$ be the set of all prime divisors of $|G|$. Let $\psi(G)=\sum_{g\in{G}}{o(g)}$. In [isaacs], it is proved that cyclic groups are characterized by their orders and the sum of element orders. In fact, they proved that the maximum amount of this function occurs on cyclic groups among all groups of the same order. Many authors have investigated some other properties of this function (see [dr, 2011, upperbound, twocriteria, 2m, 2013, 2014, 2015]). The following conjecture was posed by Herzog et al. in [twocriteria] :

Conjecture. If $G$ is a non-solvable group of order $n$, then

[TABLE]

with equality if and only if $G=A_{5}$. In particular, this inequality holds for all non-abelian simple groups.

It is proved that if $G$ is a group of order $n$ and $\psi(G)>\dfrac{211}{1617}\psi(C_{n})$, then $G$ is solvable (see [dr],[twocriteria]). Baniasad Azad and Khosravi in [dr] gave some other groups the equality holds for. Actually, they showed that $A_{5}{\times}C_{m}$, where $(30,m)=1$, satisfies the equality. Here, we prove that these groups are the only groups the equality holds for. Thus, we prove a modified version of Conjecture 6 in [twocriteria] as follows.

Main Theorem. *Suppose that $G$ is a non-solvable group of order $n$ and $\psi(G)=\dfrac{211}{1617}\psi(C_{n})$. Then $G={A_{5}}\times{C_{m}}$, where $(30,m)=1$. *

2. Preliminary Results

Lemma 2.1**.**

[isaacs, Corollary B]* Let $P\in{{\rm Syl}_{p}(G)}$, and assume that $P\,{\lhd}\,G$ and that $P$ is cyclic. Then $\psi(G)\,\leq\,{\psi(P)}{\psi(G/P)}$, with equality if and only if $P$ is central in $G$.*

Lemma 2.2**.**

[twocriteria, Proposition 2.6]* Let $H$ be a normal subgroup of the finite group $G$. Then $\psi(G)\,\leq\,{\psi(G/H)}{|H|^{2}}$.*

Lemma 2.3**.**

[2011, Lemma 2.1]* If $G$ and $H$ are finite groups, then ${\psi(G{\times}H)}\,\leq\,{\psi(G)}{\psi(H)}$. Also $\psi(G{\times}H)={\psi(G)}{\psi(H)}$ if and only if ${\rm gcd}(|G|,|H|)=1$.*

Lemma 2.4**.**

[upperbound, Proposition 2.5]* Let $G$ be a finite group and suppose that there exists $x\in{G}$ such that $|G:\langle{x}\rangle|\,<\,2p$, where $p$ is the maximal prime divisor of $|G|$. Then one of the following holds:*

• (i)

$G$* has a normal cyclic Sylow $p$-subgroup,*

• (ii)

$G$* is solvable and $\langle{x}\rangle$ is a maximal subgroup of $G$ of index either $p$ or $p+1$.*

Lemma 2.5**.**

[upperbound, Lemma 2.9]**

• (1)

If $P$ is a cyclic group of order $p^{r}$ for some prime $p$, then

[TABLE]

• (2)

Let ${p_{1}}<{p_{2}}<{\cdots}<{{p_{t}}=p}$ be the prime divisors of $n$ and denote the corresponding Sylow subgroups of $C_{n}$ by ${P_{1}},{P_{2}},{\cdots},P_{t}$. Then

[TABLE]

Lemma 2.6**.**

[twocriteria, Theorem 1]* Let $G$ be a finite group of order n containing a subgroup $A$ of prime power index $p^{s}$. Suppose that $A$ contains a normal cyclic subgroup $B$ satisfying the following condition: $A/B$ is a cyclic group of order $2^{r}$ for some non-negative integer $r$. Then $G$ is a solvable group.*

Lemma 2.7**.**

[Herstein, Theorem]* Let $G$ be a finite group, $A$ an Abelian subgroup of $G$. If $A$ is a maximal subgroup of $G$ then $G$ is solvable.*

Lemma 2.8**.**

[finiteisaacs, Lemma 9.1]* Let $G$ be a group, and suppose that $G/Z(G)$ is simple. Then $G/Z(G)$ is nonabelian, and $G^{\prime}$ is perfect. Also $G^{\prime}/Z(G^{\prime})$ is isomorphic to the simple group $G/Z(G)$.*

Lemma 2.9**.**

[finiteisaacs, Corollary 5.14]* Let $P\in{{\rm Syl}_{p}(G)}$, where $G$ is a finite group and $p$ is the smallest prime divisor of $|G|$, and assume that $P$ is cyclic. Then $G$ has a normal $p$-complement.*

Lemma 2.10**.**

[finiteisaacs, Theorem 2.20]* Let $A$ be a cyclic proper subgroup of a finite group $G$, and let $K={\rm core}_{G}(A)$. Then $|A:K|\,<\,|G:A|$, and in particular, if $|A|\,\geq\,|G:A|$, then $K\,>\,1$*

Lemma 2.11**.**

[hall, Theorem 3.1]* If $n=1+rp$, with $1\,<r<\,(p+3)/2$ there is not a group $G$ with $n$ Sylow $p$-subgroups unless $n=q^{t}$ where $q$ is a prime, or $r=(p-3)/2$ and $p\,>\,3$ is a Fermat prime.*

Lemma 2.12**.**

[hall, Theorem 3.2]* There is no group $G$ with $n_{3}=22$, with $n_{5}=21$, or with $n_{p}=1+3p$ for $p\,\geq\,7$.*

Lemma 2.13**.**

[dr, Lemma 2.1]* Let $G$ be a group of order $n={p_{1}}^{\alpha_{1}}\cdots{p_{k}}^{\alpha_{k}}$, where $p_{1},\cdots,p_{k}$ are distinct primes. Let $\psi(G)\,>\,{\dfrac{r}{s}{\psi(C_{n})}}$, for some integers $r,s$. Then there exists a cyclic subgroup $\langle{x}\rangle$ such that*

[TABLE]

Lemma 2.14**.**

[dr, Lemma 2.3]**

• (a)

Let $p\in\{2,3,5\}$ and $a>0$. Then $p^{2a}>{{\dfrac{13}{12}}{\psi(C_{p^{a}})}}$.

• (b)

Let $\pi(m)\subseteq\{2,3,5\}$ and $m\geq 2$. Then ${m^{2}}>{\dfrac{13}{12}{\psi(C_{m})}}$.

Lemma 2.15**.**

[dr, Lemma 2.4]* Let $n={p_{1}}^{\alpha_{1}}{p_{2}}^{\alpha_{2}}\cdots{p_{r}}^{\alpha_{r}}$ be a positive integer, where $p_{i}$ are primes, ${p_{1}}<{p_{2}}<\cdots<{p_{r}}=p$ and ${\alpha_{i}}>0$, for each $1\,{\leq}\,i\,{\leq}\,r$. If $p\geq{13}$, then*

[TABLE]

By the proof given in [dr] for Lemma 2.2, we have:

Lemma 2.16**.**

Let $p$ be a prime number and $a,b>0$. Then ${\psi(C_{p^{a+b}})}>{\psi(C_{p^{a}})}{\psi(C_{p^{b}})}$.

3. Main Results

In this section we prove the modified version of Herzog’s Conjecture.

Lemma 3.1**.**

Let $G$ be a non-solvable group of order n and $p\mid{n}$. Furthermore, $\psi(G)={\dfrac{211}{1617}{\psi(C_{n})}}$ and $P\in{{\rm Syl}_{p}(G)}$ is cyclic and normal in $G$. Then $P$ has a normal $p$-complement $K$ in $G$, where $\psi(K)=\dfrac{211}{1617}\psi(C_{|K|})$.

Proof.

By Lemma 2.1, we have

[TABLE]

Therefore

[TABLE]

Now by the non-solvability of $G$ and the main result of [dr], we conclude that $\psi(G/P)=\dfrac{211}{1617}{\psi(C_{|G/P|})}$. It follows that the equality holds in (1). Thus $P\,\leq\,Z(G)$ by Lemma 2.1. Therefore by Burnside’s normal $p$-complement theorem, there exists $K\,\unlhd\,G$ such that $G=P\times{K}$. ∎

Theorem 3.2**.**

Suppose that $G$ is a non-solvable group of order $n$, and $p=5$ is the largest prime divisor of $n$. Furthermore, $\psi(G)=\dfrac{211}{1617}\psi(C_{n})$. Then $G=A_{5}$.

Proof.

By the assumption $n=2^{\alpha}3^{\beta}5^{\gamma}$. We note that $G$ has no normal Sylow subgroup. By Lemma 2.5, we have

[TABLE]

Therefore, there exists $x\in{G}$ such that $[G:\langle{x}\rangle]<19$.

If $[G:\langle{x}\rangle]<10$, then by the non-solvability of $G$ and Lemma 2.4, $G$ has a normal Sylow $5$-subgroup, which is a contradiction. Hence $10\leq[G:\langle{x}\rangle]<19$. Now we consider the following cases:

• (a)

Let $[G:\langle{x}\rangle]=18$.

Then there exists $P_{5}\in{{\rm Syl}_{5}(G)}$ such that $P_{5}\,\leq\,\langle{x}\rangle$. Therefore

[TABLE]

Hence $[G:N_{G}(P_{5})]=6$ and $[N_{G}(P_{5}):\langle{x}\rangle]=3$. Let $H={\rm core}_{G}(\langle{x}\rangle)$. Now by Lemma 2.10, $[\langle{x}\rangle{:H}]<[G:\langle{x}\rangle]=18$. If $5\nmid{[\langle{x}\rangle:H]}$, then $H$ contains a Sylow $5$-subgroup of $G$, say $P_{5}$, implying $P_{5}{\unlhd}G$, a contradiction. Hence $5\mid{[\langle{x}\rangle{:H}]}$. By the non-solvability of $G/H$, $[\langle{x}\rangle{:H}]=10$ and $|G/H|=180$. Since $A_{5}{\times}C_{3}$ is the only non-solvable group of order $180$, we conclude that $G/H{\cong}A_{5}{\times}C_{3}$ and $\psi(G/H)=1237$. Now using Lemma 2.2 and (2), we have

[TABLE]

which is a contradiction.

• (b)

Let $[G:\langle{x}\rangle]=16$.

By Lemma 2.6, we conclude that G is solvable contradicting our hypothesis.

• (c)

Let $[G:\langle{x}\rangle]=15$.

Similar to the previous cases, $\langle{x}\rangle$ contains a Sylow $2$-subgroup of $G$, say $P_{2}$. Using Lemma 2.9, $G$ has a normal $2$-complement, say $N$. Hence by the Feit-Thompson theorem, $N$ is solvable contradicting the non-solvability of $G$.

• (d)

Let $[G:\langle{x}\rangle]=10$.

There exists $P_{3}\in{{\rm Syl}_{3}(G)}$ such that $P_{3}\leq{\langle{x}\rangle}$. Then $\langle{x}\rangle\,{\leq}\;{N_{G}(P_{3})}$ and we have

[TABLE]

Then $N_{G}(P_{3})=\langle{x}\rangle$. We claim that $N_{G}(P_{3})$ is a maximal subgroup of $G$. Suppose that $N_{G}(P_{3})<L\leq{G}$. Hence $N_{L}(P_{3})=N_{G}(P_{3})$ and $10=[G:L][L:N_{L}(P_{3})]$ implying $[G:L]=1$. Now Lemma 2.7 implies that $G$ is solvable, a contradiction.

• (e)

Let $[G:\langle{x}\rangle]=12$.

There exists $P_{5}\in{{\rm Syl}_{5}(G)}$ such that $P_{5}\leq{\langle{x}\rangle}$. As we argued, we have

[TABLE]

Thus $[G:N_{G}(P_{5})]=6$ and $[N_{G}(P_{5}):\langle{x}\rangle]=2$. Let $H={\rm core}_{G}(\langle{x}\rangle)$. Hence by Lemma 2.10, $[\langle{x}\rangle:H]<[G:\langle{x}\rangle]=12$.

Similarly to Case (a), $5\mid[\langle{x}\rangle:H]$, and so $[\langle{x}\rangle:H]=5$ or $10$. Now we consider each case:

$\bullet$ Let $[\langle{x}\rangle:H]=10$.

In this case, $|G/H|=120$ and by the non-solvability of $G/H$, we have $\psi(G/H)\leq{663}$. By Lemma 2.2 and (2), we conclude that:

[TABLE]

which is a contradiction.

$\bullet$ Let $[\langle{x}\rangle:H]=5$.

In this case, $|G/H|=60$, and so $G/H{\cong}A_{5}$. As a result, $H$ is a maximal normal subgroup of $G$. On the other hand, $H$ is cyclic, so $H\,\leq\,C_{G}(H)$. Then $H=C_{G}(H)$ or $G=C_{G}(H)$. The Normalizer-Centralizer Theorem implies that $G=C_{G}(H)$, and so $H\,\leq\,{Z(G)}$. Therefore $H=Z(G)$ and $G/Z(G){\cong}A_{5}$. Hence ${G^{\prime}Z(G)/Z(G)}\cong{A_{5}}$ implying $G=G^{\prime}Z(G)$, and we have $\dfrac{G^{\prime}}{G^{\prime}{\cap}Z(G)}{\cong}A_{5}$. On the other hand, Lemma 2.8 implies that $G^{\prime}$ is perfect and $G^{\prime}/Z(G^{\prime})\cong{A_{5}}$. By the fact that the Schur multiplier of $A_{5}$ is equal to 2, we have $G^{\prime}\cong{{\rm SL}(2,5)}$ or $G^{\prime}\cong{A_{5}}$.

If $G^{\prime}{\cong}{\rm SL}(2,5)$, then $|G^{\prime}{\cap}Z(G)|=2$ and we have

[TABLE]

where $m=n/120$. So by Lemmas 2.2 and 2.3, we conclude that

[TABLE]

Now by (2) and some simple calculations, we obtain $72000m^{2}<77616\psi(C_{m})<78000\psi(C_{m})$. Hence $12m^{2}<13\psi(C_{m})$. By Lemma 2.14, $m=1$, $n=120$ and $G={\rm SL}(2,5)$. It means that, $663=\psi(G)=\dfrac{211}{1617}\psi(C_{120})>824$, a contradiction.

If $G^{\prime}\cong{A_{5}}$, then $|G^{\prime}{\cap}Z(G)|=1$ and $G={G^{\prime}}\times{Z(G)}\cong{A_{5}}\times{Z(G)}$.

If $1\neq Z(G)$, then $|Z(G)|={2^{\alpha-2}}{3^{\beta-1}}{5^{\gamma-1}}$ and by Lemmas 2.3 and 2.16, we have

[TABLE]

which is a contradiction. That is why, we must have $Z(G)=1$ and $G={A_{5}}$.

∎