On Two Conjectures about the Sum of Element Orders

TL;DR

This paper proves a conjecture linking the sum of element orders to supersolvability in finite groups and provides a counterexample to a related conjecture about subgroup sums.

Contribution

It confirms a conjecture by Trnuceanu connecting the sum of element orders with supersolvability and demonstrates that another proposed inequality does not hold universally.

Findings

Proved that if mbda(G) exceeds a certain bound, G is supersolvable.

Provided a counterexample to the conjecture about mbda(G) for subgroups.

Established limitations of the conjectured inequality in group theory.

Abstract

Let $G$ be a finite group and $\psi(G) = \sum_{g \in G} o(g)$, where $o(g)$ denotes the order of $g \in G$. First, we prove that if $G$ is a group of order $n$ and $\psi(G) >31\psi(C_n)/77$, where $C_n$ is the cyclic group of order $n$, then $G$ is supersolvable. This proves a conjecture of M.~{T\u{a}rn\u{a}uceanu}. Moreover, M. Herzog, P. Longobardi and M. Maj put forward the following conjecture: If $H\leq G$, then $\psi(G) \leqslant \psi(H) |G:H|^2$. In the sequel, by an example we show that this conjecture is not satisfied in general.

Equations

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

$$\psi (C_n)\ge \frac{385}{96}\frac{n^2}{p+1}.$$

$$\frac{31}{77}\psi (C_{|G/Q|})\psi (C_{|Q|})=\frac{31}{77}\psi (C_n)<\psi (G)\le \psi (Q)\psi (G/Q)=\psi (C_{|Q|})\psi (G/Q).$$

$$\psi (G)>\frac{31}{77}\psi (C_n)\ge \frac{31}{77}\cdot \frac{385}{96}\frac{n^2}{p+1}.$$

$$[G:⟨x⟩]<\frac{77}{31}\cdot \frac{96}{385}(p+1)\le \frac{77}{31}\cdot \frac{96}{385}\cdot \frac{12}{11}p<0.68p<p,$$

$$\psi (G)>\frac{31}{77}\psi (C_{|G|})>\frac{31}{77}\cdot \frac{2^{2{\alpha }_1+1}}{2+1}\cdot \frac{3^{2{\alpha }_2+1}}{3+1}=\frac{31}{77}\cdot \frac{1}{2}{n}^2.$$

$$\frac{31}{77}\cdot \frac{1}{2}{n}^2<99\frac{n^2}{24^2},$$

$$\psi (G)>\frac{31}{77}\psi (C_4)\psi (C_{3^{\beta }})=\frac{31}{77}\cdot 11\cdot \frac{3^{2\beta +1}+1}{4}.$$

$$\frac{31}{77}\cdot 11\cdot \frac{3^{2\beta +1}+1}{4}<265\cdot 3^{2\beta -4}$$

$$\Rightarrow 31\cdot 3^5<28\cdot 265,$$

$$\psi (P)\psi (G/P)\ge \psi (G)=\frac{31}{77}\psi (C_n)=\frac{31}{77}\psi (C_{|P|})\psi (C_{|G/P|}).$$

$$\psi (G/P)\ge \frac{31}{77}\psi (C_{|G/P|}).$$

$$\psi (G/P)=\frac{31}{77}\psi (C_{|G/P|}).$$

$$[G:⟨x⟩]<\frac{77}{31}\cdot \frac{96}{385}(p+1)\le \frac{77}{31}\cdot \frac{96}{385}\cdot \frac{12}{11}p<0.68p<p,$$

$$f^{\prime }(1)=1,f^{\prime }(r)=\prod _{i=2}^r\frac{q_i}{q_i+1};$$

$$h^{\prime }(2)=3,h^{\prime }(s)=f^{\prime }(s-1)q_s.$$

$$\psi (C_n)\ge h^{\prime }(12)\frac{n^2}{p+1}.$$

$$\psi (G)>\frac{271}{3647}\psi (C_n)\ge \frac{271}{3647}\cdot h^{\prime }(12)\frac{n^2}{p+1}.$$

$$[G:⟨x⟩]<\frac{3647}{271}\cdot \frac{1}{h^{\prime }(12)}(p+1)\le \frac{3647}{271}\cdot \frac{1}{h^{\prime }(12)}\cdot \frac{38}{37}p<p,$$

$$\psi (G)>\frac{271}{3647}\psi (C_{|G|})>\frac{271}{3647}\cdot \frac{3^{2{\alpha }_1+1}}{3+1}\cdot \frac{5^{2{\alpha }_2+1}}{5+1}=\frac{271}{3647}\cdot \frac{5}{8}{n}^2.$$

$$\frac{271}{3647}\cdot \frac{5}{8}{n}^2<2197(n/225{)}^2,$$

$$\frac{271}{3647}\cdot \frac{5}{8}<3771(n/375{)}^2.$$

$$\psi (G)=\psi (\mathrm{SmallGroup}(32,7))\psi (C_m)=167\psi (C_m)\psi (H)|G:H{|}^2=156\psi (C_m).$$

Full text

First, we prove that if $G$ is a group of order $n$ and $\psi(G)>31\psi(C_{n})/77$, where $C_{n}$ is the cyclic group of order $n$, then $G$ is supersolvable. This proves a conjecture of M. Tărnăuceanu.

Moreover, M. Herzog, P. Longobardi and M. Maj put forward the following conjecture: If $H\leq G$, then $\psi(G)\leqslant\psi(H)|G:H|^{2}$. In the sequel, by an example we show that this conjecture is not satisfied in general.

1. Introduction

In this paper all groups are finite. The cyclic group of order $n$ is denoted by $C_{n}$. Let $\psi(G)=\sum_{g\in G}o(g)$, the sum of element orders in a group $G$. The function $\psi(G)$ was introduced by Amiri, Jafarian and Isaacs [2].

Many authors try to get some relations between the structure of the group $G$ and $\psi(G)$ (see [7, 8, 9, 11]). Recently it is proved that if $\psi(G)>\frac{211}{1617}\psi(C_{n})$, where $G$ is a finite group of order $n$, then $G$ is solvable [4, 6].

From the observation that $A_{4}$ satisfies $\psi(A_{4})=31$ and $\psi(C_{12})=77$, in [12] M. Tărnăuceanu put forward the following conjecture:

[T]-Conjecture [12, Conjecture 1.5] If $G$ is a group of order $n$ and $\psi(G)>\frac{31}{77}\psi(C_{n})$, then $G$ is supersolvable.

First, in this paper we prove the validity of this conjecture. Nevertheless, for groups of odd order, it is possible to prove a stronger result:

Theorem 1.1**.**

Suppose that $n=|G|$ is odd. If $\psi(G)>\frac{271}{3647}\psi(C_{n})$, then $G$ is supersolvable.

On the other hand, in [6] there exist three conjectures about $\psi(G)$. The validity of Conjecture 5 was proved in [4] and the validity of Conjecture 6 was proved in [3]. Moreover the following conjecture was posed in that paper.

[HLM]-Conjecture [6, Conjecture 7] If $H\leq G$, then $\psi(G)\leqslant\psi(H)|G:H|^{2}$.

Finally, in the sequel of this paper we give an example which shows that this conjecture is not satisfied in general and it seems that we need extra conditions on $G$ or $H$ to have this result.

For the proof of these results, we need the following lemmas.

Lemma 1.2**.**

[2, Corollary B]** Let $P\in{\rm Syl}_{p}(G),$ and assume that $P\unlhd G$ 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 1.3**.**

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

Lemma 1.4**.**

[1, 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 $\gcd(|G|,|H|)=1$.

Lemma 1.5**.**

[5, 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 1.6**.**

[10, Theorem 2.20]** (Lucchini) 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 $|A|>|G:A|$ , then $K>1$.

Lemma 1.7**.**

[4, 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 1.8**.**

[6, Lemmas 2.4 and 2.5]** 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 $p\geq 11$, then

[TABLE]

2. Supersolvability

2.1. Proof of [T]-Conjecture

Let $G$ be a group of order $n={p_{1}}^{\alpha_{1}}{p_{2}}^{\alpha_{2}}\cdots{p_{r}}^{\alpha_{r}}$, where $p_{1},p_{2},\dots,p_{r}$ are primes, $p_{1}<p_{2}<\cdots<p_{r}=p$ such that $\alpha_{i}>0$, for each $1\leq i\leq r$. By assumption, $\frac{31}{77}\psi(C_{n})<\psi(G)$. By induction on $|\pi(G)|$, we prove that $G$ is a supersolvable group.

If $|\pi(G)|=1$, then $G$ is a $p$-group and so $G$ is supersolvable. Assume that $|\pi(G)|\geq 2$ and the theorem holds for each group $H$ such that $|\pi(H)|<|\pi(G)|$. Now we consider the following two cases:

Case (I) If $G$ has a normal cyclic Sylow subgroup $Q$, then by Lemma 1.2 we have $\psi(G)\leq\psi(Q)\psi(G/Q)$. Using Lemma 1.4 and the assumptions we have

[TABLE]

Therefore $\frac{31}{77}\psi(C_{|G/Q|})<\psi(G/Q)$ and $|\pi(G/Q)|<|\pi(G)|$. By the inductive hypothesis, $G/Q$ is supersolvable and so $G$ is a supersolvable group.

Case (II) Let $G$ have no normal cyclic Sylow subgroup.

If $p\geq 11$, then by Lemma 1.8, we have

[TABLE]

Thus there exists $x\in G$ such that $o(x)>\frac{31}{77}\cdot\frac{385}{96}\frac{n}{p+1}$. Therefore

[TABLE]

which is a contradiction, by Lemma 1.5. Therefore $\pi(G)\subseteq\{2,3,5,7\}$, where $2\leq|\pi(G)|\leq 4$. Now we consider the following cases:

Case 1. Let $\pi(G)=\{2,3\}$. Then $|G|=2^{\alpha_{1}}3^{\alpha_{2}}$. In this case we have

[TABLE]

It follows that there exists $x\in G$ such that $o(x)>\frac{31}{77}\cdot\frac{1}{2}n$. We conclude that $|G:\langle x\rangle|<\frac{77}{31}\cdot 2<4.97$. By Lemma 1.5, we have $[G:\langle x\rangle]=3$ or $4$. Let $H={\rm core}_{G}(\langle x\rangle)$.

• •

Let $[G:\langle x\rangle]=3$. By Lemma 1.6, $[\langle x\rangle:H]<[G:\langle x\rangle]=3$. Therefore $G/H$ is supersolvable and so $G$ is a supersolvable group.

• •

Let $[G:\langle x\rangle]=4$. By Lemma 1.6, $[\langle x\rangle:H]<[G:\langle x\rangle]=4$. If $G/H$ is a supersolvable group, then we get the result. Let $G/H$ be non-supersolvable. Therefore $[\langle{x}\rangle:H]=3$, $|G/H|=12$ and so $G/H\cong A_{4}$.

$\blacktriangleright$ If $2$ divides $|H|$, then there exists a characteristic subgroup $M$ in $H$ such that $|H:M|=2$. Thus $G/M$ is a non-supersolvable group of order $24$. Therefore using the list of such groups ($\rm{SL}(2,3),S_{4},C_{2}\times A_{4}$) and their $\psi$-values ($99,67,87$), we have $\psi(G/M)\leq 99$. By Lemma 1.3 we have $\psi(G)\leq\psi(G/M)|M|^{2}\leq 99(n/24)^{2}$. Using (1), $\psi(G)>\frac{31}{77}\cdot\frac{1}{2}n^{2}$. Therefore

[TABLE]

which is a contradiction.

$\blacktriangleright$ If $2\nmid|H|$, then $|G|=2^{2}3^{\beta}$. Let $\beta\geq 2$. Then there exists a characteristic subgroup $M$ in $H$ such that $|H:M|=3$. Therefore $M\unlhd G$ and $|G/M|=36$. Thus $G/M$ is a non-supersolvable group of order $36$. Therefore using the list of such groups ($(C_{2}\times C_{2}):C_{9},(C_{3}\times C_{3}):C_{4},C_{3}\times A_{4}$) and their $\psi$-values ($265,115,121$), we have $\psi(G/M)\leq 265$. By Lemma 1.3 we have $\psi(G)\leq\psi(G/M)|M|^{2}\leq 265(n/36)^{2}=265\cdot 3^{2\beta-4}$. On the other hand, we have $\psi(G)>\frac{31}{77}\psi(C_{n})$. Thus

[TABLE]

Therefore

[TABLE]

which is a contradiction. Thus $\beta=1$ and so $|H|=1$ and $G\cong A_{4}$, which is a contradiction, since $\psi(A_{4})=31$.

Case 2. Let $\pi(G)=\{2,3,5\}$. Then by Lemma 1.7, there exists $x\in G$ such that $[G:\langle x\rangle]<\frac{77}{31}\cdot\frac{12}{5}<5.97$. By Lemma 1.5, we have $[G:\langle x\rangle]=5$. By Lemma 1.6, $[\langle x\rangle:{\rm core}_{G}(\langle x\rangle)]<[G:\langle x\rangle]=5$. We have $|G/{\rm core}_{G}(\langle x\rangle)|=5,10,15$ or $20$. Therefore $G/{\rm core}_{G}(\langle x\rangle)$ is supersolvable and so $G$ is a supersolvable group.

Case 3. Let $\pi(G)=\{2,3,5,7\}$. Then by Lemma 1.7, there exists $x\in G$ such that $[G:\langle x\rangle]<\frac{77}{31}\cdot\frac{96}{35}<6.9<7$, which is impossible.

Case 4. Let $\pi(G)=\{2,5\}$ or $\pi(G)=\{3,5\}$. Then by Lemma 1.7, there exists $x\in G$ such that $[G:\langle x\rangle]<5$, which is a contradiction

Case 5. Let $\pi(G)=\{2,7\}$, $\pi(G)=\{3,7\}$, $\pi(G)=\{5,7\}$, $\pi(G)=\{2,3,7\}$ or $\pi(G)=\{2,5,7\}$. Then by Lemma 1.7, there exists $x\in G$ such that $[G:\langle x\rangle]<7$ and we get a contradiction.

The proof is now complete.

2.2. Equality condition of [T]-Conjecture

Lemma 2.1**.**

Let $G$ be a non-supersolvable group of order $n$ and $p\mid n$. Furthermore, let $\psi(G)=\frac{31}{77}\psi(C_{n})$ and $P\in{\rm Syl}_{p}(G)$ is cyclic and normal in $G$. Then $P$ has a normal $p$-complement $K$ in $G$, where $\psi(K)=\frac{31}{77}\psi(C_{|K|})$.

Proof.

By Lemma 1.2, we have

[TABLE]

Therefore

[TABLE]

Now by the non-supersolvability of G and the above result, we conclude that

[TABLE]

It follows that the equality holds in (2). Thus $P\leq Z(G)$, by Lemma 1.2. Therefore by Burnside’s normal $p$-complement theorem, there exists $K\unlhd G$ such that $G=P\times K.$ ∎

Theorem 2.2**.**

Suppose that $G$ is a non-supersolvable group of order $n$ such that $\psi(G)=\frac{31}{77}\psi(C_{n})$. Then $G=A_{4}\times C_{m}$, where $(6,m)=1$.

Proof.

Let $p$ be the largest prime divisor of $n$. We prove the result by induction on $p$. By the assumption, $|\pi(G)|\geq 2$, and so $p\geq 3$.

1. (1)

Let $p=3$.

We note that $G$ has no normal cyclic Sylow subgroup. There exists $x\in G$ such that $|G:\langle x\rangle|<\frac{77}{31}\cdot 2<4.97$. By Lemma 1.5, we have $[G:\langle x\rangle]=3$ or $4$. If $[G:\langle x\rangle]=3$, then $G$ is a supersolvable group, which is a contradiction. Therefore $[G:\langle x\rangle]=4$. We have $G/{\rm core}_{G}(\langle x\rangle)\cong A_{4}$, since $G$ is non-supersolvable. If ${\rm core}_{G}(\langle x\rangle)\neq 1$, then similarly to Case 1 in the proof of [T]-Conjecture we get a contradiction. Therefore ${\rm core}_{G}(\langle x\rangle)=1$ and so $G\cong A_{4}$. 2. (2)

Let $p=5$.

1. (a)

Let $\pi(G)=\{2,3,5\}$.

Then by Lemma 1.7, there exists $x\in G$ such that $[G:\langle x\rangle]<\frac{77}{31}\cdot\frac{12}{5}<5.97<10$. By Lemma 1.5, $G$ has a normal cyclic Sylow $5$-subgroup or $[G:\langle x\rangle]=5$.

$\bullet$ Let $G$ has a normal cyclic Sylow $5$-subgroup, say $P_{5}$. Using Lemma 2.1, there exists $K\unlhd G$ such that $G=P_{5}\times K$ and $\psi(K)=\frac{31}{77}\psi(C_{|K|})$. We notice that $K$ is non-supersolvable and $\pi(K)=\{2,3\}$. By the above case, we have $K=A_{4}$ and $G=A_{4}\times P_{5}$.

$\bullet$ Let $[G:\langle x\rangle]=5$. Then $|G/{\rm core}_{G}(\langle x\rangle)|=5,10,15$ or $20$. Therefore $G/{\rm core}_{G}(\langle x\rangle)$ is supersolvable and so $G$ is a supersolvable group, which is a contradiction. 2. (b)

Let $\pi(G)=\{2,5\}$ or $\pi(G)=\{3,5\}$. Then there exists $x\in G$ such that $[G:\langle x\rangle]<5$. By Lemma 1.5, $G$ has a normal cyclic Sylow $5$-subgroup. Thus $G$ is a supersolvable group, which is a contradiction. 3. (3)

Let $p=7$.

Then there exists $x\in G$ such that $[G:\langle x\rangle]<7$. By Lemma 1.5, $G$ has a normal cyclic Sylow $7$-subgroup. Using Lemma 2.1, there exists $K\unlhd G$ such that $G=P_{7}\times K$ and $\psi(K)=\frac{31}{77}\psi(C_{|K|})$. We notice that $K$ is non-supersolvable and $\pi(K)=\{2,3\}$ or $\{2,3,5\}$, therefore by the above cases we have $G\cong P_{7}\times A_{4}$ or $P_{7}\times P_{5}\times A_{4}$, respectively and we get the result. 4. (4)

Let $p\geq 11$. Then there exists $x\in G$ such that

[TABLE]

By Lemma 1.5, $G$ has a normal cyclic Sylow $p$-subgroup, say $P$. Using Lemma 2.1, there exists $K\unlhd G$ such that $G=P\times K$ and $\psi(K)=\frac{31}{77}\psi(C_{|K|})$. Then $K$ is non-supersolvable and $\pi(K)=\pi(G)\setminus\{p\}$. Therefore using the above discussion and the induction hypothesis we get the result.

∎

2.3. Supersolvability for group of odd order

Notation 2.3**.**

Let $\{q_{1},q_{2},q_{3},\cdots\}$ be the set of all primes in an increasing order: $2=q_{1}<q_{2}<q_{3}<\cdots$. Let also $q_{0}=1$. If $r,s$ are two positive integers, we define the functions $f^{\prime}(r)$ and $h^{\prime}(s)$ as follows:

[TABLE]

[TABLE]

In the sequel, let $A:={\rm SmallGroup}(75,2)=(C_{5}\times C_{5}):C_{3}$. We know that $\psi(A)=271$ and $\psi(C_{75})=3647$. Similarly to the proof of Lemmas 2.4 and 2.5 in [6] we get that:

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, $2<p_{1}<p_{2}<\cdots<p_{r}=p$ and $\alpha_{i}>0$, for each $1\leq i\leq r$. If $p\geq 37=q_{12}$, then

[TABLE]

The proof of Theorem 1.1 is very similar to the proof of [T]-conjecture, and so we remove the details of the proof.

Proof of Theorem 1.1.

We prove the result by induction on $|\pi(G)|$. The assertion is trivial if $G$ is a $p$-group, and let the result holds for each group $H$, where $|\pi(H)|<|\pi(G)|$ (exactly similar to the proof of [T]-Conjecture).

We consider the following two cases:

Case(I) If $G$ has a normal cyclic Sylow subgroup $Q$, then we get the result.

Case (II) Let $G$ have no normal cyclic Sylow subgroup.

Let $p$ be the largest prime divisor of $|G|$. If $p\geq 37$, then by Lemma 2.4, we have

[TABLE]

Thus there exists $x\in G$ such that $o(x)>\frac{271}{3647}\cdot h^{\prime}(12)\frac{n}{p+1}$. Therefore

[TABLE]

which is a contradiction.

Therefore $\pi(G)\subseteq\{3,5,7,\cdots,31\}$ and so $2\leq|\pi(G)|\leq 10$. Now we consider the following cases:

Case 1. Let $p=31$, i.e. $\{31\}\subseteq\pi(G)\subseteq\{3,5,7,\cdots,31\}$.

Using Lemma 1.7, there exists $x\in G$ such that $[G:\langle x\rangle]<2p=62$. By Lemma 1.5, we have $[G:\langle x\rangle]=31$. Using Lemma 1.6, $[\langle x\rangle:{\rm core}_{G}(\langle x\rangle)]<[G:\langle x\rangle]=31$. We see that $G/{\rm core}_{G}(\langle x\rangle)$ is supersolvable and so $G$ is a supersolvable group.

Case 2. Let $p\in\{17,19,23,29\}$. Similarly to Case 1 we obtain that $G$ is a supersolvable group.

Therefore $\pi(G)\subseteq\{3,5,7,11,13\}$, where $2\leq|\pi(G)|\leq 5$.

Case 3. Let $p=13$. If $5\nmid|G|$, then by Lemma 1.7, there exists $x\in G$ such that $[G:\langle x\rangle]<25$. Thus by Lemma 1.5 we have $[G:\langle x\rangle]=13$, and we get the result similarly to the above. If $5\in\pi(G)$, by Lemma 1.7, there exists $x\in G$ such that $[G:\langle x\rangle]<29$, and by Lemma 1.5 we have $[G:\langle x\rangle]=13$ or $27$.

If $[G:\langle x\rangle]=13$, then by Lemma 1.6 we have $|G/{\rm core}_{G}(\langle x\rangle)|=13m$, where $m\in\{3,9,5,7,11\}$. Therefore $G/{\rm core}_{G}(\langle x\rangle)$ is a supersolvable group and so $G$ is a supersolvable group.

If $[G:\langle x\rangle]=27$, then by Lemma 1.6 we have $|\langle x\rangle:{\rm core}_{G}(\langle x\rangle)|<27$. If $13\mid|\langle x\rangle:{\rm core}_{G}(\langle x\rangle)|$, then $|G/{\rm core}_{G}(\langle x\rangle)|=13\cdot 27$. Let $P_{5}\in{\rm Syl}_{5}({\rm core}_{G}(\langle x\rangle))$. We have $P_{5}\in{\rm Syl}_{5}(G)$ and $P_{5}\unlhd G$, which is a contradiction. Therefore $13\nmid|\langle x\rangle:{\rm core}_{G}(\langle x\rangle)|$. Let $P_{13}\in{\rm Syl}_{13}({\rm core}_{G}(\langle x\rangle))$. We have $P_{13}\in{\rm Syl}_{13}(G)$ and $P_{13}\unlhd G$, which is a contradiction.

Case 4. Let $p=11$.

$\bullet$ If $\pi(G)=\{3,5,7,11\},$ then there exists $x\in G$ such that $[G:\langle x\rangle]<26$. Therefore $[G:\langle x\rangle]=11$ or $25$.

$\blacktriangleright$ If $[G:\langle x\rangle]=11$, then we have the result.

$\blacktriangleright$ Let $[G:\langle x\rangle]=25$. Then $|G/{\rm core}_{G}\langle x\rangle|=25m$, where $m\in\{3,5,7,9,11,15,21\}$. If $m=11$, then we get the result. Otherwise, $G$ has a Sylow $11$-subgroup which is cyclic and normal, and this is impossible.

$\bullet$ If $\pi(G)\subsetneq\{3,5,7,11\}$, then there exists $x\in G$ such that $[G:\langle x\rangle]<24$. Therefore $[G:\langle x\rangle]=11$ and so we have the result.

Case 5. Let $p=7$.

$\bullet$ If $\pi(G)=\{3,7\}$ or $\{5,7\}$, then easily we can see that $G$ is supersolvable.

$\bullet$ If $\pi(G)=\{3,5,7\}$, then there exists $x\in G$ such that $[G:\langle x\rangle]<25$. Therefore $[G:\langle x\rangle]=7,15$ or $21$.

$\blacktriangleright$ If $[G:\langle x\rangle]=7$, then we get the result.

$\blacktriangleright$ Let $[G:\langle x\rangle]=15$. Then $|G/{\rm core}_{G}(\langle x\rangle)|=15m$, where $m\in\{3,5,7,9\}$. If $m=7$, then we get the result. Otherwise, the Sylow $7$-subgroup of $G$ is cyclic and normal, which is impossible.

$\blacktriangleright$ Let $[G:\langle x\rangle]=21$. Then $|G/{\rm core}_{G}\langle x\rangle|=21m$, where $m\in\{3,5,7,9,15\}$. If $m\in\{5,15\}$, then we get the result. Otherwise, the Sylow $5$-subgroup of $G$ is cyclic and normal, which is impossible.

Case 6. Let $\pi(G)=\{3,5\}$. Then $|G|=3^{\alpha_{1}}5^{\alpha_{2}}$. In this case we have

[TABLE]

Hence there exists $x\in G$ such that $|G:\langle x\rangle|<22$. By Lemma 1.5, we have $[G:\langle x\rangle]=5$ or $15$. If $[G:\langle x\rangle]=5$, then we have the result. Let $[G:\langle x\rangle]=15$ and $H={\rm core}_{G}(\langle x\rangle)$. By Lemma 1.6, $[\langle x\rangle:H]<[G:\langle x\rangle]=15$. If $G/H$ is a supersolvable group, then we get the result. Let $G/H$ be non-supersolvable. Therefore $[\langle{x}\rangle:H]=5$ and $|G/H|=75$. Thus $G/H\cong A=(C_{5}\times C_{5}):C_{3}$.

$\blacktriangleright$ If $3$ divides $|H|$, then there exists a characteristic subgroup $M$ in $H$ such that $|H:M|=3$. Therefore $M\unlhd G$ and $|G/M|=225$. Thus $G/H$ is a non-supersolvable group of order $225$. Therefore using the list of such groups ($(C_{5}\times C_{5}):C_{9},C_{3}\times((C_{5}\times C_{5}):C_{3})$) and their $\psi$-values ($2197,1297$), we have $\psi(G/H)\leq 2197$. By Lemma 1.3 we have $\psi(G)\leq\psi(G/H)|H|^{2}\leq 2197(n/225)^{2}$. Using (3), $\psi(G)>\frac{271}{3647}\cdot\frac{271}{3647}\cdot\dfrac{5}{8}n^{2}$. Therefore

[TABLE]

which is a contradiction.

$\blacktriangleright$ If $3\nmid|H|$, then $|G|=3\cdot 5^{\beta}$. Let $\beta>2$. Then there exists a characteristic subgroup $M$ in $H$ such that $|H:M|=5$. Therefore $M\unlhd G$ and $|G/M|=375$. Thus $G/H$ is a non-supersolvable group of order $375$. Therefore using the list of such groups ($((C_{5}\times C_{5}):C_{5}):C_{3},C_{5}\times((C_{5}\times C_{5}):C_{3})$) and their $\psi$-values ($3771,3771$), we have $\psi(G/H)=3771$. By Lemma 1.3 we have $\psi(G)\leq\psi(G/H)|H|^{2}\leq 3771(n/375)^{2}$. Using (3), we have

[TABLE]

which is a contradiction. Thus $\beta=2$ and so $G\cong A=(C_{5}\times C_{5}):C_{3}$, which is a contradiction. ∎

3. Counterexample for [HLM]-Conjecture

Using the notation in GAP, let $G={\rm SmallGroup}(32,7)\times C_{m}\cong((C_{8}:C_{2}):C_{2})\times C_{m}$, where $(m,2)=1$. We know that ${\rm SmallGroup}(32,7)$ has a maximal subgroup $M\cong C_{2}\times D_{8}$. Therefore $G$ has a maximal subgroup $H=M\times C_{m}\cong C_{2}\times D_{8}\times C_{m}$. Using GAP we have $\psi({\rm SmallGroup}(32,7))=167$ and $\psi(M)=39$ and so $\psi(M)|G:M|^{2}=39\cdot 4=156$. Now note that

[TABLE]

Therefore this example shows that [HLM]-Conjecture is not satisfied in general. Obviously this is a natural question to ask with which extra conditions the validity of [HLM]-Conjecture is obtained.