FINITE GROUPS WITH LARGE CHERMAK–DELGADO LATTICES

Abstract

Given a finite group G, we denote by $L(G)$ the subgroup lattice of G and by ${\cal CD}(G)$ the Chermak–Delgado lattice of G. In this note, we determine the finite groups G such that $|{\cal CD}(G)|=|L(G)|-k$ , for $k=1,2$ .

1 Introduction

Let G be a finite group and $L(G)$ be the subgroup lattice of G. The Chermak–Delgado measure of a subgroup H of G is defined by

$$ \begin{align*} m_G(H)=|H||C_G(H)|. \end{align*} $$

Let

$$ \begin{align*} m^*(G)=\mathrm{\max}\{m_G(H)\mid H\leq G\} \quad\mbox{and}\quad {\cal CD}(G)=\{H\leq G\mid m_G(H)=m^*(G)\}. \end{align*} $$

Then the set ${\cal CD}(G)$ forms a modular, self-dual sublattice of $L(G)$ , which is called the Chermak–Delgado lattice of G. It was first introduced by Chermak and Delgado [4] and revisited by Isaacs [5]. In the last few years, there has been a growing interest in understanding this lattice (see [1–3, 6–8, 11–14]). We recall several important properties of the Chermak–Delgado measure:

• • if $H\leq G$ , then $m_G(H)\leq m_G(C_G(H))$ , and if the measures are equal, then $C_G(C_G(H))=H$ ;

• • if $H\in {\cal CD}(G)$ , then $C_G(H)\in {\cal CD}(G)$ and $C_G(C_G(H))=H$ ;

• • the maximal member M of ${\cal CD}(G)$ is characteristic and ${\cal CD}(M)={\cal CD}(G)$ ;

• • the minimal member $M(G)$ of ${\cal CD}(G)$ (called the Chermak–Delgado subgroup of G) is characteristic, abelian and contains $Z(G)$ .

In [12], the Chermak–Delgado measure of G has been seen as a function:

$$ \begin{align*} m_G:L(G)\longrightarrow\mathbb{N}^*,\quad H\mapsto m_G(H)\quad \mbox{for all } H\in L(G). \end{align*} $$

If G is nontrivial, then $m_G$ has at least two distinct values, or equivalently ${\cal CD}(G)\neq L(G)$ (see [11, Corollary 3]). This leads to the following natural question.

Question 1.1. How large can the lattice ${\cal CD}(G)$ be?

The dual problem of finding finite groups with small Chermak–Delgado lattices has been studied in [6, 7].

Our main result is stated as follows.

Theorem 1.2. Let G be a finite group. Then:

1. (a) $|{\cal CD}(G)|=|L(G)|-1$ if and only if $G\cong \mathbb {Z}_p$ or $G\cong Q_8$ ;

2. (b) $|{\cal CD}(G)|=|L(G)|-2$ if and only if $G\cong \mathbb {Z}_{p^2}$ .

For the proof of the above theorem, we need the following well-known result (see, for example, [10, Volume II, (4.4)]).

Theorem 1.3. A finite p-group has a unique subgroup of order p if and only if it is either cyclic or a generalised quaternion $2$ -group.

We recall that a generalised quaternion $2$ -group is a group of order $2^n$ for some positive integer $n\geq 3$ , defined by

$$ \begin{align*} Q_{2^n}=\langle a,b \mid a^{2^{n-2}} = b^2, a^{2^{n-1}}=1, b^{-1}ab=a^{-1}\rangle. \end{align*} $$

We also need the following theorem taken from [12].

Theorem 1.4. Let G be a finite group. For each prime p dividing the order of G and $P\in \mathrm {Syl}_p(G)$ , let $|Z(P)|=p^{n_p}$ . Then

(1.1) $$ \begin{align} |\mathrm{Im}(m_G)|\geq 1+\sum_{p}n_p. \end{align} $$

Finally, we indicate a natural open problem concerning the above study.

Open problem. Determine the finite groups G such that $|{\cal CD}(G)|=|L(G)|-k$ , where $k\geq 3$ .

Most of our notation is standard and will usually not be repeated here. Elementary notions and results on groups can be found in [5]. For subgroup lattice concepts, we refer to [9].

2 Proof of the main result

First of all, we solve the problem for generalised quaternion $2$ -groups.

Lemma 2.1. With the above notation:

1. (a) $|{\cal CD}(Q_{2^n})|=|L(Q_{2^n})|-1$ if and only if $n=3$ , that is, $G\cong Q_8$ ;

2. (b) $|{\cal CD}(Q_{2^n})|\neq |L(Q_{2^n})|-2$ for all $n\geq 3$ .

Proof. We easily obtain

$$ \begin{align*} m^*(Q_{2^n})=2^{2n-2} \quad\mbox{for all } n\geq 3, \end{align*} $$

and

$$ \begin{align*} {\cal CD}(Q_{2^n})=\begin{cases} \{Q_8, \langle a\rangle, \langle b\rangle, \langle ab\rangle, \langle a^2\rangle\}&\hbox{if } n=3\\ \{\langle a\rangle\}&\hbox{if } n\geq 4.\end{cases} \end{align*} $$

These lead immediately to the desired conclusions.

We are now able to prove our main result.

Proof of Theorem 1.2

We divide the proof into two parts corresponding to the two parts of the theorem.

Part (a). Since $|{\cal CD}(G)|=|L(G)|-1$ , we have ${\cal CD}(G)=L(G)\setminus \{H_0\}$ , where $H_0\leq G$ . We infer that $|\mathrm { Im}(m_G)|=2$ and so G is a p-group with $|Z(G)|=p$ by Theorem 1.4. Then $m_G(1)<m_G(Z(G))$ , implying that

$$ \begin{align*} H_0=1\quad \mbox{and}\quad m^*(G)=m_G(Z(G))=p^{n+1}, \end{align*} $$

where $|G|=p^n$ .

Assume that there exists $H\leq G$ with $|H|=p$ and $H\neq Z(G)$ . Then $H\notin {\cal CD}(G)$ , which shows that $H=1$ and this contradicts the hypothesis. Thus, G has a unique subgroup of order p and Theorem 1.3 leads to

(2.1) $$ \begin{align} G\cong\mathbb{Z}_{p^n} \quad\mbox{or}\quad G\cong Q_{2^n} \quad\mbox{for some } n\geq 3. \end{align} $$

In the first case, we easily get $n=1$ , that is, $G\cong \mathbb {Z}_p$ , while in the second one, we get $G\cong Q_8$ by Lemma 2.1(a).

Part (b). The condition $|{\cal CD}(G)|=|L(G)|-2$ means ${\cal CD}(G)=L(G)\setminus \{H_1,H_2\}$ , where $H_1,H_2\leq G$ . Then $|\mathrm {Im}(m_G)|\leq 3$ . Recall that we cannot have $|\mathrm {Im}(m_G)|=1$ .

If $|\mathrm {Im}(m_G)|=2$ , then

$$ \begin{align*} m_G(H_1)=m_G(H_2)\neq m^*(G) \end{align*} $$

and again G is a p-group with $|Z(G)|=p$ . It is clear that one of the two subgroups $H_1$ and $H_2$ must be trivial, say $H_1=1$ . Then G has at most two subgroups of order p, namely $Z(G)$ and possibly $H_2$ . This implies that it has exactly one subgroup of order p because the number of subgroups of order p in a finite p-group is congruent to $1 \ (\mathrm {mod}\, p)$ . Consequently, one obtains again (2.1). For $G\cong \mathbb {Z}_{p^n}$ , we easily get $n=2$ , that is, $G\cong \mathbb {Z}_{p^2}$ , while for $G\cong Q_{2^n}$ , we get no solution by Lemma 2.1(b).

If $|\mathrm {Im}(m_G)|=3$ , then $m_G(H_1)$ , $m_G(H_2)$ and $m^*(G)$ are distinct. Also, (1.1) becomes

$$ \begin{align*} 3\geq 1+\sum_{p}n_p. \end{align*} $$

Since $n_p\geq 1$ for all p, we have the two possibilities described in Cases 1 and 2.

Case 1: $|G|=p^n$ and $|Z(G)|\in \{p,p^2\}$ .

Obviously, if G is abelian, we get $G\cong \mathbb {Z}_{p^2}$ . Assume that G is not abelian. Since $m_G(1)<m_G(Z(G))=m_G(G)$ , we infer that one of the two subgroups $H_1$ and $H_2$ is trivial and that

$$ \begin{align*} m^*(G)=m_G(Z(G))=m_G(G). \end{align*} $$

If $|Z(G)|=p$ , then G has a unique subgroup of order p and so it is a generalised quaternion $2$ -group, contradicting Lemma 2.1(b). The same can also be said when $|Z(G)|=p^2$ because all subgroups of order p of G are outside of ${\cal CD}(G)$ .

Case 2: $|G|=p^nq^m$ and the Sylow p-subgroups and q-subgroups of G have centres of orders p and q, respectively.

Let P be a Sylow p-subgroup and Q be a Sylow q-subgroup of G. Since $P\subseteq C_G(Z(P))$ , we have

$$ \begin{align*} m_G(Z(P))=p\,|C_G(Z(P))|=p^{n+1}q^x \quad\mbox{for some } x \mbox{ with } 0\leq x\leq m, \end{align*} $$

and similarly,

$$ \begin{align*} m_G(Z(Q))=p^yq^{m+1} \quad\mbox{for some } y \mbox{ with } 0\leq y\leq n. \end{align*} $$

Also,

$$ \begin{align*} m_G(1)=p^nq^m \quad\mbox{and}\quad m_G(G)=p^nq^m|Z(G)|. \end{align*} $$

We observe that the measures $m_G(Z(P))$ , $m_G(Z(Q))$ and $m_G(1)$ are distinct and consequently they are all possible measures of the subgroups of G. We distinguish two subcases.

Subcase 2.1: $Z(G)=1$ .

Then $m^*(G)=m_G(1)=m_G(G)$ . Indeed, if $m^*(G)=m_G(Z(P))$ , then $1$ , G and $Z(Q)$ will be outside of ${\cal CD}(G)$ , and this contradicts the hypothesis. In the same way, we cannot have $m^*(G)=m_G(Z(Q))$ . Since $m_G(P)$ is divisible by $p^{n+1}$ and $m_G(Q)$ is divisible by $q^{m+1}$ , we infer that $m_G(P)=m_G(Z(P))$ and $m_G(Q)=m_G(Z(Q))$ . Thus, $P,Z(P),Q,Z(Q)\notin {\cal CD}(G)$ and our hypothesis implies that $P=Z(P)$ and $Q=Z(Q)$ , that is, G is a nonabelian group of order $pq$ . Assume that $p<q$ . Then ${\cal CD}(G)$ consists of the unique subgroup of order q of G and therefore we obtain $|L(G)|=3$ , and this contradicts the hypothesis.

Subcase 2.2: $Z(G)\neq 1$ .

Then $m_G(1)<m_G(G)$ , which shows that $m_G(G)$ equals either $m_G(Z(P))$ or $m_G(Z(Q))$ . Assume that $m_G(G)=m_G(Z(P))$ . Then $x=m$ and $|Z(G)|=p$ , implying that

(2.2) $$ \begin{align} Z(G)=Z(P). \end{align} $$

Note that we cannot have $m^*(G)=m_G(Z(Q))$ because in this case, $1$ , $Z(G)$ and G will be outside of ${\cal CD}(G)$ , and this contradicts the hypothesis. Consequently,

$$ \begin{align*} m^*(G)=m_G(Z(P))=m_G(P)=m_G(Z(G))=m_G(G). \end{align*} $$

It follows that $1$ , $Z(Q)$ and Q are not contained in ${\cal CD}(G)$ , which leads to $Q=Z(Q)$ . In other words, ${\cal CD}(G)=L(G)\setminus \{1,Q\}$ . Thus, ${\cal CD}(G)$ is the lattice interval

$$ \begin{align*} [G/Z(G)]=\{H\in L(G)\mid Z(G)\leq H\leq G\} \end{align*} $$

and [11, Corollary 2] shows that G is nilpotent. Then $G=P\times Q$ and it follows that $Z(G)=Z(P)\times Q$ , which contradicts (2.2).

This completes the proof.