Groups whose Chermak–Delgado lattice is a subgroup lattice of an abelian group

Abstract

The Chermak–Delgado lattice of a finite group G is a self-dual sublattice of the subgroup lattice of G. In this paper, we prove that, for any finite abelian group A, there exists a finite group G such that the Chermak–Delgado lattice of G is a subgroup lattice of A.

1 Introduction

Suppose that G is a finite group, and H is a subgroup of G. The Chermak–Delgado measure of H (in G) is denoted by $m_G(H)$ , and defined as $m_G(H)=|H|\cdot |C_G(H)|.$ The maximal Chermak–Delgado measure of G is denoted by $m^*(G)$ , and defined as

$$ \begin{align*}m^*(G)=\max\{ m_G(H)\mid H\le G\}.\end{align*} $$

Let

$$ \begin{align*}\mathcal{CD}(G)=\{ H\mid m_G(H)=m^*(G)\}.\end{align*} $$

Then the set $\mathcal {CD}(G)$ forms a sublattice of $\mathcal {L}(G)$ (the subgroup lattice of G), which is called the Chermak–Delgado lattice of G. It was first introduced by Chermak and Delgado [9], and revisited by Isaacs [12]. In the last years, there has been a growing interest in understanding this lattice (see, e.g., [1–11, 13–17, 19–22]).

A Chermak–Delgado lattice is always self-dual. So the question arises: Which types of self-dual lattices can be Chermak–Delgado lattices of finite groups? In [5], it is proved that, for any integer n, a chain of length n can be a Chermak–Delgado lattice of a finite p-group.

A quasi-antichain is a lattice consisting of a maximum, a minimum, and the atoms of the lattice. The width of a quasi-antichain is the number of atoms. For a positive integer $w\ge 3$ , a quasi-antichain of width w is denoted by $\mathcal {M}_{w}$ . In [6], it was proved that $\mathcal {M}_{w}$ can be a Chermak–Delgado lattice of a finite group if and only if $w=1+p^a$ for some positive integer a and some prime p.

An m-diamond is a lattice with subgroups in the configuration of an m-dimensional cube. A mixed n-string is a lattice with n components, adjoined end to end, so that the maximum of one component is identified with the minimum of the other component. The following theorem gives more self-dual lattices which can be Chermak–Delgado lattices of finite groups.

Theorem 1.1 [4]

If $\mathcal {L}$ is a Chermak–Delgado lattice of a finite p-group G such that both $G/Z(G)$ and $G'$ are elementary abelian, then so are $\mathcal {L}^+$ and $\mathcal {L}^{++}$ , where $\mathcal {L}^+$ is a mixed $3$ -string with center component isomorphic to $\mathcal {L}$ and the remaining components being m-diamonds, and $\mathcal {L}^{++}$ is a mixed $3$ -string with center component isomorphic to $\mathcal {L}$ and the remaining components being lattice isomorphic to $\mathcal {M}_{p+1}$ .

By [18, Theorem 8.1.4], $\mathcal {L}(A)$ is always self-dual for any finite abelian group A. If A is a cyclic p-group, then $\mathcal {L}(A)$ is chain, and hence can be a Chermak–Delgado lattice of a finite p-group. In [2], it is proved that, if A is an elementary abelian p-group, then $\mathcal {L}(A)$ can be a Chermak–Delgado lattice of a finite p-group. In this paper, we prove that, for any finite abelian group A, $\mathcal {L}(A)$ can be a Chermak–Delgado lattice of a finite group. The main results are the following theorems.

Theorem 1.2 For any finite abelian p-group A, there exists a finite p-group G such that $\mathcal {CD}(G)$ is isomorphic to $\mathcal {L}(A)$ .

Theorem 1.3 For any finite abelian group A, there exists a finite group G such that $\mathcal {CD}(G)$ is isomorphic to $\mathcal {L}(A)$ .

2 Preliminary

We gather next some basic properties of the Chermak–Delgado lattice, which will be used often throughout the paper without further reference.

Theorem 2.1 [9]

Suppose that G is a finite group and $H,K\in \mathcal {CD}(G)$ .

1. (1) $\langle H,K\rangle =HK$ . Hence, a Chermak–Delgado lattice is modular.

2. (2) $C_G(H\cap K)=C_G(H)C_G(K)$ .

3. (3) $C_G(H)\in \mathcal {CD}(G)$ and $C_G(C_G(H))=H$ . Hence, a Chermak–Delgado lattice is self-dual.

4. (4) Let M be the maximal member of $\mathcal {CD}(G)$ . Then M is characteristic in G and $\mathcal {CD}(M)=\mathcal {CD}(G)$ .

5. (5) The minimal member of $\mathcal {CD}(G)$ is characteristic, abelian, and contains $Z(G)$ .

We also need the following lemmas.

Theorem 2.2 [7, Theorem 2.9]

For any finite groups G and H, $\mathcal {CD}(G\times H)=\mathcal {CD}(G)\times \mathcal {CD}(H).$

Lemma 2.3 [2, Lemma 3.3]

Suppose that G is a finite group and $H\le G $ such that $G=HC_G(H)$ . If $H\in \mathcal {CD}(H)$ , then H is contained in the unique maximal member of $\mathcal {CD}(G)$ .

Lemma 2.4 [20, Lemma 5]

Let G be a finite p-group. Then $\mathcal {CD}(G)=[G/Z(G)]$ if and only if the interval $[G/Z(G)]$ of $\mathcal {L}(G)$ is modular and $G'$ is cyclic.

In this section, we prove that, for any finite abelian group A, $\mathcal {L}(A\times A)$ can be a Chermak–Delgado lattice of a finite group. Although this result can be deduced from our main theorem, the proof is independent and short.

Lemma 2.5 Let A be a finite abelian p-group. Then there exists a finite p-group G such that $\mathcal {CD}(G)$ is isomorphic to $\mathcal {L}(A\times A)$ .

Proof Assume that the type of A is $(p^{e_1},p^{e_2},\dots ,p^{e_m})$ , where $e_1\ge e_2\ge \dots \ge e_m$ . Let G be the group generated by $2m$ elements $x_1,\dots ,x_m$ , $y_1,\dots ,y_m$ subject to the defining relations:

$$ \begin{gather*} [x_i,x_j]=[y_i,y_j]=[x_i,y_j]=1 \text{ if }i\ne j, \\[3pt] x_i^{p^{e_i}}=y_i^{p^{e_i}}=z^{p^{e_1}}=1, [x_i,y_i]=z^{p^{e_1-e_i}}, [z,x_i]=[z,y_i]=1 \text{ for }1\le i\le m. \end{gather*} $$

Let $P_i=\langle x_i,y_i,z\rangle $ . Then $Z(P_i)=\langle z\rangle $ . Thus, G is also the central product of $P_i$ . It is easy to see that $G'=Z(G)=\langle z\rangle $ and $G/Z(G)\cong A\times A$ . By Lemma 2.4, $\mathcal {CD}(G)$ is just the interval $[G/Z(G)]$ . Hence, $\mathcal {CD}(G)\cong \mathcal {L}(G/Z(G))\cong \mathcal {L}(A\times A)$ .

Theorem 2.6 For any finite abelian group A, there exists a finite group G such that $\mathcal {CD}(G)$ is isomorphic to $\mathcal {L}(A\times A)$ .

Proof Let $A=A_1\times \cdots \times A_n$ , where $A_i$ is the Sylow $p_i$ -subgroup of A. By Lemma 2.5, there exist finite group $P_i$ such that $\mathcal {CD}(P_i)$ is isomorphic to $\mathcal {L}(A_i\times A_i)$ . Let $G=P_1\times \cdots \times P_n$ . By Theorem 2.2,

$$ \begin{align*} \mathcal{CD}(G) &= \mathcal{CD}(P_1)\times \cdots \times \mathcal{CD}(P_n) \\[3pt] &\cong \mathcal{L}(A_1\times A_1)\times \cdots \times \mathcal{L}(A_n\times A_n) \\[3pt] &= \mathcal{L}(A\times A).\\[-2.7pc] \end{align*} $$

3 The groups $G(p,e)$

For any prime p and an integer $e\ge 1$ , we use $G(p,e)$ to denote the finite p-group generated by three elements $x,y,w$ subject to the following defining relations:

• • $[x,y]=z_1$ , $[y,w]=z_2$ , $[w,x]=z_3$ ,

• • $x^{p^e}=y^{p^e}=w^{p^e}=z_1^{p^e}=z_2^{p^e}=z_3^{p^e}=1$ , and

• • $[z_i,x]=[z_i,y]=[z_i,w]=1$ for all $i=1,2,3$ .

In this section, we prove that the Chermak–Delgado lattice of $G(p,e)$ is isomorphic to a subgroup lattice of a cyclic group of order $p^e$ . This group will be used to construct an example in the proof of Theorem 1.2. Let $G=G(p,e)$ . Then it is easy to check the following results:

• • $d(G)=3$ , $\exp (G)=p^e$ , $Z(G)=G'=\langle z_1,z_2,z_3\rangle $ , and

• • $|Z(G)|=p^{3e}$ , $|G/Z(G)|=p^{3e}$ , $m_G(G)=m_G(Z(G))=p^{9e}$ .

Lemma 3.1 Assume that $G=G(p,e)$ and $Z(G)<H<G$ .

1. (1) If $H/Z(G)$ is cyclic, then $m_G(H)<m_G(G)$ .

2. (2) If $H/Z(G)$ is not cyclic, then $m_G(H)\le m_G(G)$ , where “ $=$ ” holds if and only if the type of $H/Z(G)$ is $(p^{e_1},p^{e_1},p^{e_1})$ for some $1\le e_1<e$ .

Proof (1) Let $H=\langle h, Z(G)\rangle $ and $H/Z(G)$ be of order $p^{e_1}$ . Then we may let

$$ \begin{align*}h=x^{k_1p^{e-e_1}}y^{k_2p^{e-e_1}}w^{k_3p^{e-e_1}},\end{align*} $$

where $p\nmid k_i$ for some i. Without loss of generality, we may assume that $p\nmid k_1$ . Replacing x with $x^{k_1}y^{k_2}w^{k_3}$ , we have $h=x^{p^{e-e_1}}$ . It is easy to check that $C_G(H)=\langle x, y^{p^{e_1}}, w^{p^{e_1}}\rangle Z(G)$ . Since $|C_G(H)/Z(G)|=p^{3e-2e_1}$ ,

$$ \begin{align*}|H/Z(G)|\cdot |C_G(H)/Z(G)|=p^{3e-e_1}<p^{3e}=|G/Z(G)|.\end{align*} $$

Hence, $m_G(H)=|H|\cdot |C_G(H)|<|G|\cdot |Z(G)|=m_G(G)$ .

(2) Let $H=\langle h_1,h_2, h_3\rangle Z(G)$ and $H/Z(G)$ be of type $(p^{e_1},p^{e_2},p^{e_3})$ , where $e_1\ge e_2\ge e_3\ge 0$ . Since $H/Z(G)$ is not cyclic, $e_2\ge 1$ . By a similar argument as (1), we may assume that $h_1=x^{p^{e-e_1}}$ . We may let

$$ \begin{align*}h_2=x^{k_1p^{e-e_2}}y^{k_2p^{e-e_2}}w^{k_3p^{e-e_2}},\end{align*} $$

where $p\nmid k_i$ for some $2\le i\le 3$ . Without loss of generality, we may assume that $p\nmid k_2$ . Replacing y with $x^{k_1}y^{k_2}w^{k_3}$ , we have $h_2=y^{p^{e-e_2}}$ . It is easy to check that

$$ \begin{align*}C_G(H)=C_G(h_1)\cap C_G(h_2)=\langle x^{p^{e_2}}, y^{p^{e_1}},w^{p^{e_1}}\rangle Z(G).\end{align*} $$

Since $|H/Z(G)|=p^{e_1+e_2+e_3}$ and $|C_G(H)/Z(G)|=p^{3e-e_2-2e_1}$ ,

$$ \begin{align*}|H/Z(G)|\cdot |C_G(H)/Z(G)|=p^{3e+e_3-e_1}\le p^{3e}=|G/Z(G)|,\end{align*} $$

where “ $=$ ” holds if and only if $e_3=e_1$ . Hence, $m_G(H)=|H|\cdot |C_G(H)|\le |G|\cdot |Z(G)|=m_G(G)$ , where “ $=$ ” holds if and only if $e_1=e_2=e_3$ .

Theorem 3.2 Let $G=G(p,e)$ . Then $G\in \mathcal {CD}(G)$ and $\mathcal {CD}(G)$ is a chain of length e. Moreover, $H\in \mathcal {CD}(G)$ if and only if $H=\langle x^{p^{e-e_1}},y^{p^{e-e_1}},w^{p^{e-e_1}}\rangle Z(G)$ for some $0\le e_1\le e$ .

Proof By Lemma 3.1, $m^*(G)=m_G(G)=p^{9e}$ , and $H\in \mathcal {CD}(G)$ if and only if the type of $H/Z(G)$ is $(p^{e_1},p^{e_1},p^{e_1})$ for some $0\le e_1\le e$ . Hence, all elements of $\mathcal {CD}(G)$ are $\langle x^{p^{e-e_1}},y^{p^{e-e_1}},w^{p^{e-e_1}}\rangle Z(G)$ where $0\le e_1\le e$ .

4 The proof of main results

For any prime p and an abelian p-group A with type $(p^{e_1},p^{e_2},\dots ,p^{e_m})$ , where $e_1\ge e_2\ge \cdots \ge e_m$ , we use $G_A$ to denote the finite p-group generated by $3m$ elements $x_1,\dots ,x_m$ , $y_1,\dots ,y_m$ , $w_1,\dots ,w_m$ subject to the following defining relations:

• • $x_i^{p^{e_i}}=y_i^{p^{e_i}}=w_i^{p^{e_i}}=z_1^{p^{e_1}}=z_2^{p^{e_1}}=z_3^{p^{e_1}}=1$ for $1\le i\le m$ ,

• • $[x_i,x_j]=[y_i,y_j]=[w_i,w_j]=[x_i,y_j]=[y_i,w_j]=[w_i,x_j]=1$ if $i\ne j$ ,

• • $[x_i,y_i]=z_1^{p^{e_1-e_i}}$ , $[y_i,w_i]=z_2^{p^{e_1-e_i}}$ , $[w_i,x_i]=z_3^{p^{e_1-e_i}}$ for $1\le i\le m$ , and

• • $[z_j,x_i]=[z_j,y_i]=[z_j,w_i]=1$ for $1\le i\le m$ and $j=1,2,3$ .

In this section, we require the following notation and straightforward results for a finite p-group $G=G_A$ .

• • $Z(G)=G'=\langle z_1,z_2,z_3\rangle $ is of order $p^{3e_1}$ .

• • Let $P_i=\langle x_i,y_i,w_i\rangle $ for $1\le i\le m$ . Then $P_i\cong G(p,e_i)$ , $|P_iZ(G)/Z(G)|=p^{3e_i}$ , and G is the central product $P_1*P_2*\cdots *P_m$ .

• • Let $X=\langle x_1,x_2,\dots , x_m\rangle $ , $Y=\langle y_1,y_2,\dots ,y_m\rangle $ , and $W=\langle w_1,w_2,\dots ,w_m\rangle $ . Then $X\cong Y\cong W\cong A$ .

• • Let $n=e_1+e_2+\cdots +e_m$ . Then $|A|=p^n$ , $|G/Z(G)|=p^{3n}$ , $|G|=p^{3n+3e_1}$ , and $m_{G}(G)=p^{3n+6e_1}$ .

• • Let $\alpha ,\beta ,\gamma $ be isomorphisms from A to $X,Y,W$ , respectively, such that $x_i^{\alpha ^{-1}}=y_i^{\beta ^{-1}}=w_i^{\gamma ^{-1}}$ for all $1\le i\le m$ .

• • For $a\in A$ , let $a^\varphi =\langle a^\alpha , a^\beta , a^\gamma \rangle Z(G)$ .

• • For $B\le A$ , let $B^\varphi =\langle B^\alpha ,B^\beta ,B^\gamma \rangle Z(G)=\prod _{b\in B}b^\varphi $ .

The proof of Theorem 1.2

Assume that the type of A is $(p^{e_1},p^{e_2},\dots ,p^{e_m})$ , where $e_1\ge e_2\ge \cdots \ge e_m$ . Let $G=G_A$ . We will prove $\mathcal {CD}(G)\cong \mathcal {L}(A)$ in six steps.

(1) $G\in \mathcal {CD}(G)$ and $m^*(G)=p^{3n+6e_1}$ .

By Theorem 3.2, $P_i\in \mathcal {CD}(P_i)$ . Since $G=P_iC_{G}(P_i)$ , by Lemma 2.3, $P_i$ is contained in the unique maximal member of $\mathcal {CD}(G)$ . Hence, G is the unique maximal member of $\mathcal {CD}(G)$ and $m^*(G)=m_{G}(G)=p^{3n+6e_1}$ .

(2) For any $a\in A$ , there exists a subgroup $C_a$ of A such that $C_X(a^\beta )=C_X(a^\gamma )=(C_a)^\alpha $ , $C_Y(a^\alpha )=C_Y(a^\gamma )=(C_a)^\beta $ , and $C_W(a^\alpha )=C_W(a^\beta )=(C_a)^\gamma $ .

Notice that for $x\in X$ , $[x,a^\beta ]=1$ if and only if $[x,a^\gamma ]=1$ . We have $C_X(a^\beta )=C_X(a^\gamma )$ . Let $C_a=(C_X(a^\beta ))^{\alpha ^{-1}}$ . Then $C_X(a^\beta )=C_X(a^\gamma )=(C_a)^\alpha $ . Notice that for $c\in A$ , $[c^\alpha ,a^\gamma ]=1$ if and only if $[c^\beta ,a^\gamma ]=1$ . We have

$$ \begin{align*}c\in C_a\Longleftrightarrow c^\alpha\in C_X(a^\gamma)\Longleftrightarrow c^\beta\in C_Y(a^\gamma).\end{align*} $$

It follows that $C_Y(a^\gamma )=(C_a)^\beta $ . By the symmetry, the conclusions hold.

(3) $C_G(a^\varphi )=(C_a)^\varphi $ and $a^\varphi \in \mathcal {CD}(G)$ .

Suppose that a is of order $p^t$ . Then $|a^\varphi /Z(G)|=p^{3t}$ . Since $[a^\alpha ,G]\le \langle z_1^{p^{e_1-t}},z_3^{p^{e_1-t}}\rangle $ , the length of the conjugacy class of $a^\alpha $ does not exceed $p^{2t}$ . Hence, $|C_{G}(a^\alpha )|\ge p^{3n+3e_1-2t}$ and $|C_{G}(a^\alpha )/Z(G)|\ge p^{3n-2t}$ . Notice that

$$ \begin{align*}C_{G}(a^\alpha)/Z(G)=XZ(G)/Z(G)\times C_Y(a^\alpha)Z(G)/Z(G)\times C_W(a^\alpha)Z(G)/Z(G),\end{align*} $$

$|XZ(G)/Z(G)|=|X|=p^n$ , and by (2),

$$ \begin{align*}|C_a|=|C_Y(a^\alpha)|=|C_W(a^\alpha)|=|C_Y(a^\alpha)Z(G)/Z(G)|=|C_W(a^\alpha)Z(G)/Z(G)|.\end{align*} $$

We have $|C_a|\ge p^{n-t}$ . Hence, $|(C_a)^\varphi /Z(G)|\ge p^{3n-3t}$ . By (2), $(C_a)^\varphi \le C_G(a^\varphi )$ . Hence,

$$ \begin{align*}|a^\varphi/Z(G)|\cdot |C_{G}(a^\varphi)/Z(G)|\ge |a^\varphi/Z(G)|\cdot |(C_a)^\varphi/Z(G)|\ge p^{3n}=|G/Z(G)|.\end{align*} $$

It follows that

$$ \begin{align*}m_{G}(a^\varphi)=|a^\varphi|\cdot |C_{G}(a^\varphi)|\ge |G|\cdot |Z(G)|=m^*(G).\end{align*} $$

Thus, “ $=$ ” holds, $C_G(a^\varphi )=(C_a)^\varphi $ , and $a^\varphi \in \mathcal {CD}(G)$ .

(4) For any $B\le A$ , $B^\varphi \in \mathcal {CD}(G)$ and there exists a subgroup $C_B$ of A such that $C_G(B^\varphi )=(C_B)^\varphi $ . Moreover, $|B|\cdot |C_B|=p^n$ .

Let $C_B=\bigcap _{b\in B} C_b$ . Since $B^\varphi =\prod _{b\in B}b^\varphi $ , $B^\varphi \in \mathcal {CD}(G)$ and

$$ \begin{align*}C_G(B^\varphi)=\bigcap_{b\in B}C_G(b^\varphi)=\bigcap_{b\in B}(C_b)^\varphi=(C_B)^\varphi.\end{align*} $$

Since $|B^\varphi /Z(G)|=|B|^3$ and $|(C_B)^\varphi /Z(G)|=|C_B|^3$ , we have

$$ \begin{align*}|B|^3\cdot |C_B|^3=|B^\varphi/Z(G)|\cdot |(C_B)^\varphi/Z(G)|=|G/Z(G)|=p^{3n}.\end{align*} $$

Hence, $|B|\cdot |C_B|=p^n$ .

(5) If $K\in \mathcal {CD}(G)$ , then there exists a subgroup B of A such that $K=B^\varphi $ .

Let $H=C_G(K)$ . Then $H\in \mathcal {CD}(G)$ and $K=C_G(H)$ . Let

$$ \begin{align*}B_1=\{ a\in A\mid \mbox{there exist}\ y\in Y, w\in W, \ \mbox{and}\ z\in Z(G)\ \mbox{such that}\ a^\alpha ywz\in H \},\end{align*} $$

$$ \begin{align*}B_2=\{ a\in A\mid \mbox{there exist}\ x\in X, w\in W, \ \mbox{and}\ z\in Z(G)\ \mbox{such that}\ xa^\beta wz\in H \},\end{align*} $$

$$ \begin{align*}B_3=\{ a\in A\mid \mbox{there exist}\ x\in X, y\in Y, \ \mbox{and}\ z\in Z(G)\ \mbox{such that}\ xya^\gamma z\in H \}.\end{align*} $$

Then $B_1$ , $B_2$ , and $B_3$ are subgroups of A and $|H/Z(G)|\le |B_1|\cdot |B_2|\cdot |B_3|$ . By (2),

$$ \begin{align*} C_X(H)\le C_X(B_2^\beta)=(C_{B_2})^\alpha. \end{align*} $$

Hence, $|C_X(H)|\le |C_{B_2}|$ . Similarly, $|C_Y(H)|\le |C_{B_3}|$ and $|C_W(H)|\le |C_{B_1}|$ . It follows that

$$ \begin{align*}|H/Z(G)|\cdot |K/Z(G)|\le |B_1|\cdot |B_2|\cdot |B_3|\cdot |C_{B_2}|\cdot |C_{B_3}|\cdot |C_{B_1}|=p^{3n}=|G/Z(G)|.\end{align*} $$

Since $H\in \mathcal {CD}(G)$ , “ $=$ ” holds. Hence,

$$ \begin{align*}K=C_G(H)=\langle (C_{B_2})^\alpha, (C_{B_3})^\beta, (C_{B_1})^\gamma\rangle Z(G)\end{align*} $$

and

$$ \begin{align*} C_X(H)=(C_{B_2})^\alpha, C_Y(H)=(C_{B_3})^\beta, \text{ and }C_W(H)=(C_{B_1})^\gamma. \end{align*} $$

By the symmetry, we also have

$$ \begin{align*} C_X(H)=(C_{B_3})^\alpha, C_Y(H)=(C_{B_1})^\beta, \text{ and } C_W(H)=(C_{B_2})^\gamma. \end{align*} $$

It follows that $C_{B_1}=C_{B_2}=C_{B_3}$ . Let $B=C_{B_1}$ . Then $K=C_G(H)=B^\varphi $ .

(6) $\mathcal {CD}(G)$ is isomorphic to $\mathcal {L}(A)$ .

It is a direct result of (4) and (5).

The proof of Theorem 1.3

Let $A=A_1\times \cdots \times A_n$ , where $A_i$ is the Sylow $p_i$ -subgroup of A. By Theorem 1.2, there exist finite groups $P_i$ such that $\mathcal {CD}(P_i)$ is isomorphic to $\mathcal {L}(A_i)$ . Let $G=P_1\times \cdots \times P_n$ . By Theorem 2.2,

$$ \begin{align*}\mathcal{CD}(G)=\mathcal{CD}(P_1)\times \cdots \times \mathcal{CD}(P_n)\cong \mathcal{L}(A_1)\times \cdots \times \mathcal{L}(A_n) =\mathcal{L}(A).\\[-33pt] \end{align*} $$