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 [Reference Chermak and Delgado9], and revisited by Isaacs [Reference Isaacs12]. In the last years, there has been a growing interest in understanding this lattice (see, e.g., [Reference An1–Reference Glauberman11, Reference McCulloch13–Reference Morresi Zuccari, Russo and Scoppola17, Reference Tǎrnǎuceanu19–Reference Wilcox22]).
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 [Reference Brewster, Hauck and Wilcox5], 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 [Reference Brewster, Hauck and Wilcox6], 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 [Reference An, Brennan, Qu and Wilcox4]
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 [Reference Schmidt18, 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 [Reference An2], 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 [Reference Chermak and Delgado9]
Suppose that G is a finite group and
$H,K\in \mathcal {CD}(G)$
.
-
(1)
$\langle H,K\rangle =HK$
. Hence, a Chermak–Delgado lattice is modular. -
(2)
$C_G(H\cap K)=C_G(H)C_G(K)$
. -
(3)
$C_G(H)\in \mathcal {CD}(G)$
and
$C_G(C_G(H))=H$
. Hence, a Chermak–Delgado lattice is self-dual. -
(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) The minimal member of
$\mathcal {CD}(G)$
is characteristic, abelian, and contains
$Z(G)$
.
We also need the following lemmas.
Theorem 2.2 [Reference Brewster and Wilcox7, 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 [Reference An2, 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 [Reference Tǎrnǎuceanu20, 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) If
$H/Z(G)$
is cyclic, then
$m_G(H)<m_G(G)$
. -
(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,
$$ \begin{align*}\mathcal{CD}(G)=\mathcal{CD}(P_1)\times \cdots \times \mathcal{CD}(P_n)\cong \mathcal{L}(A_1)\times \cdots \times \mathcal{L}(A_n) =\mathcal{L}(A).\\[-33pt] \end{align*} $$
Acknowledgment
I cordially thank the referee for detailed reading and helpful comments, which helped me to improve the whole paper considerably.
