CENTRAL PRODUCTS AND THE CHERMAK–DELGADO LATTICE

. The Chermak–Delgado lattice of a ﬁnite group is a modular, self-dual sublattice of the lattice of subgroups of G. We prove that the Chermak–Delgado lattice of a central product contains the product of the Chermak–Delgado lattices of the relevant central factors. Furthermore, we obtain information about heights of elements in the Chermak–Delgado lattice relative to their heights in the Chermak–Delgado lattices of central factors. We also explore how the central product can be used as a tool in investigating Chermak–Delgado lattices.


Introduction.
The Chermak-Delgado lattice consists of subgroups of a finite group that have maximal Chermak-Delgado measure.Due to the many unique properties of the Chermak-Delgado lattice, it has attracted attention from researchers interested in lattice theory, general finite group theory, and centralizers of groups.
Originally defined by Chermak and Delgado [8], the Chermak-Delgado lattice for a finite group G is defined using the so-called Chermak-Delgado measure m which takes subgroups of G to positive integers via the formula It is quite interesting, and perhaps counter-intuitive that the subgroups with maximal value of m in a group form a sublattice of the subgroup lattice of G. Recall that the subgroup lattice of G is the poset of subgroups of G with the operations of meet and join defined by subgroup intersection and subgroup generated by respectively, i.e., H ∧ K = H ∩ K and H ∨ K = ⟨H, K⟩.Surprisingly, when H and K are in the Chermak-Delgado lattice, we have that ⟨H, K⟩ = HK, i.e., the set theoretic product is actually a subgroup.Furthermore if H is in the Chermak-Delgado lattice then C G (H) is in the Chermak-Delgado lattice.In addition, we know that for all H in the Chermak-Delgado lattice of G, the subgroup H must contain Z(G), the center of G.
The many properties we presented in this paragraph can be found in Isaacs [12,Section 1.G].Because of these properties as well as current research about the Chermak-Delgado lattice for various groups, questions about the Chermak-Delgado lattice make for good research projects that can be accessible to students [16,7].We write m * (G) for the maximum value that m takes on a finite group G, and CD(G) for the Chermak-Delgado lattice of G. Hence for a subgroup H of G we have that Recent work on the Chermak-Delgado lattice can be broadly classified as coming in two broad themes.While not exhaustive, we list some references below.
(2) Showing properties of the Chermak-Delgado lattice in general: the Chermak-Delgado lattice of a direct product is the direct product of the Chermak-Delgado lattices of the factors [4].All subgroups of the Chermak-Delgado lattice of a finite group are subnormal in that group with subnormal depth bounded by their relevant position within the lattice [4,9].Most recently, the question was asked about how many groups are not contained in the Chermak-Delgado lattice [10].
Related to the direct product is the central product.Recall that a group G is a central product of two of its subgroups A and B if G = AB and [A, B] = 1, i.e., ab = ba for all a ∈ A and b ∈ B.
We prove that the Chermak-Delgado lattice of a central product contains the product of the Chermak-Delgado lattices of the relevant central factors.
Theorem A. Let G be a finite group and A, B ≤ G such that G is a central product of A and B. Then CD(A) ⋅ CD(B) ⊆ CD(G).Furthermore, the top (resp.bottom) of CD(G) is equal to the product of the tops (resp.bottoms) of CD(A) and CD(B).
Hence we have the following corollary.
In addition, we have the following structural information about the Chermak-Delgado lattices of central products regarding the heights and depths of elements in the product.
Recall that the depth of H ∈ CD(G) is the length of a maximal chain where T G is the top element; the height of H ∈ CD(G) is the length of a maximal chain where B G is the bottom element.The height of CD(G) is the height of the top element (or, equivalently, the depth of the bottom element).These quantities are well-defined since CD(G) is a modular lattice, and so all of the maximal chains between two fixed elements are of the same length.In Section 3 we show how the central product can be used to prove results about the Chermak-Delgado lattice.

Central Products and the Chermak-Delgado Lattice.
A group G is a central product of A and B, which are subgroups of G, if G = AB and [A, B] = 1.Note for every group G we have the central product G = GZ(G).Also note that if G = AB is a central product, then both A and B are normal subgroups of G and that A ∩ B ⊆ Z(G).Also for any X 1 ≤ A and for any X 2 ≤ B, . This last observation is almost enough to prove Theorem A.
, and we have that CD(A) ⋅ CD(B) ⊆ CD(G), and also As it stands, Proposition 1 proves Theorem A modulo the existence of a subgroup of the form HK ∈ CD(G).While Theorem A tells us what such a subgroup could be, i.e., take the product of the tops of CD(A) and CD(B), it is difficult to directly show that this product is in CD(G).Instead, we utilize some lemmata comparing information about how the Chermak-Delgado measure behaves across various subgroups.
The following lemma is found in Issacs [12,Lemma 1.43].
The following lemma generalizes a result of An [1,Lemma 3.2].
Moreover, equality holds if and only if Proof.Note that , where equality holds if and only if where equality holds if and only if The last lemma we will use to prove Theorem A is one step below that of a central product.Here the group G will be equal to HC G (X) where And so We can now prove Theorem A which states that for a finite group G = AB a central product, we have CD(A) ⋅ CD(G) ⊆ CD(G), and furthermore we have that the top (resp.bottom) of CD(G) is equal to the product of the tops (resp.bottoms) of CD(A) and CD(B).
Proof of Theorem A. We write T A , T B , T G for the top elements of the Chermak-Delgado lattices of A, B and G respectively.Similarly B A , B B , B G refer to the bottom elements of these lattices.
For any X 1 ≤ A, G = AC G (X 1 ), and for any X 2 ≤ B, G = BC G (X 2 ), and so Lemma 3 applies for any X 1 ∈ CD(A) and any X 2 ∈ CD(B), with any Y ∈ CD(G).
By Lemma 3, T A T G ∈ CD(G), and so Thus, T A T B = T G , and so We apply Proposition 1 to complete the proof.
We provide two examples of equality in Theorem A. Proposition 2 appears in [4].We shall prove Proposition 3 in a moment, but first a few facts regarding central products and group products in general.

and similarly
Note that Lemma 4 applies for any finite central product G = AB with H 1 , H 2 ∈ CD(A) and K 1 , K 2 ∈ CD(B), since A∩B ≤ Z(A) and A∩B ≤ Z(B) and all subgroups in the Chermak-Delgado lattice contain the center.It follows that for any finite central product G = AB, we have that the sum of the heights of CD(A) and CD(B) cannot exceed the height of CD(G).For if CD(A) has height n, and say , and if CD(B) has height m, and say Proof.We prove the assertion for π A (U ), and the assertion for π B (U ) is similar.
Clearly π A (U ) ⊆ A. We show that π A (U ) is a subgroup of A. Note that π A (U ) is nonempty as 1 ∈ π A (U ).Let x, y ∈ π A (U ).We show that xy −1 ∈ π A (U ).
So there exists u ∈ U and v ∈ U so that u = xb 1 and v = yb We now prove Proposition 3 which states that for a finite group G = AB with B ≤ Z(G), we have that Proof of Proposition 3. We have that G = AB is a central product, and since B is abelian, Let U ∈ CD(G).Then Z(G) ≤ U , and so B ≤ U , and so by Lemma 6, We say that a central product G = AB is proper if Z(G) < A < G and Z(G) < B < G, and in such case we say that the group G admits a proper central product.
Note that if a group G = AB is a central product with one of A or B abelian, say B, then B ≤ Z(G), and we are in the situation of Proposition 3.
Given a finite group G which admits a proper central product, one wonders if CD(G) is equal to the subgroup collection of G that is generated by all CD(X) ⋅ CD(Y ) where G = XY a proper central product.The answer is "no" in general.Proof.One direction is true by Lemma 8. Suppose that every atom in CD(G) is abelian, and let G = AB be an arbitrary central product.Since G ∈ CD(G), by Corollary 2 we have that A ∈ CD(A) and B ∈ CD(B).Now if one of CD(A) or CD(B) has height 0, say CD(A), then CD(A) = {A} and so A is abelian, and the result follows.
Otherwise, since CD(G) has height 2 or 3, we have that at least one of CD(A) or CD(B) has height 1, say CD(A).So CD(A) = {A, Z(A)}.By Theorem B we have that AZ(B) has height 1 in CD(G), and thus is a nonabelian atom in CD(G), a contradiction.
The fun stops after height 3. Construct the central product G = Q 8 * Q 8 , which is of course proper.Note that CD(G) has height 4, and all of the atoms have order 4 so they are all abelian.
3. Investigating the Chermak-Delgado Lattice via the Central Product.
In this section we show how using the central product and the results from Section 2 we can obtain new proofs of some results about Chermak-Delgado lattice.
Let L(G) denote the lattice of all subgroups of a group G.Given a sublattice C of L(G), and given Lemma 10.Let G be a finite group and let We establish that for any X with Z The following corollary appears in An [1,Theorem 3.4] and [2,Theorem 4.4].However, as stated at the start of the section, our goal is to show the power of the central product when investigating the Chermak-Delgado lattice.Of note, our proof utilizes Lemma 10 and Proposition 1 together with some standard results about the Chermak-Delgado lattice.Of note, Corollary 3 can be used to argue that certain groups never appear in a Chermak-Delgado lattice for any group G.For example, the alternating group A 4 can never be in a Chermak-Delgado lattice since A 4 ∉ CD(A 4 ).
Taking the central product approach also allows us to prove a result of Tȃrnȃuceanu Theorem B. If G is a central product of A and B, then the height of the Chermak-Delgado lattice of G is equal to the sum of the heights of the Cheramk-Delgado lattices of A and B respectively.Moreover, an element HK ∈ CD(G) with A ∩ B ≤ H ≤ A and A ∩ B ≤ K ≤ B has height (resp.depth) equal to the sum of the heights (resp.depths) of H ∈ CD(A) and K ∈ CD(B).

Proposition 2 .Proposition 3 .
Suppose that a finite group G = A × B is a direct product.Then CD(G) = CD(A) ⋅ CD(B).Suppose that a finite group G = AB with B ≤ Z(G).Then CD(G) = CD(A) ⋅ {B}.

Lemma 5 .
and by Theorem A, this chain of length n + m lives in CD(G).Of course, our Theorem B gives the precise relationship between heights in CD(A), CD(B), and CD(G).If a group G = AB is a central product, then for any U ≤ G, we define π A (U ) = {a g = ab with g ∈ U and a ∈ A and b ∈ B} and π B (U ) = {b g = ab with g ∈ U and a ∈ A and b ∈ B}.Suppose a group G = AB is a central product, and U ≤ G. Then π A (U ) is a subgroup of A and, furthermore, A ∩ B ≤ π A (U ) ≤ A. A similar result is true for π B (U ).

Proof.
If A ∈ CD(A) and B ∈ CD(B), then by Theorem A, AB = G ∈ CD(G).Conversely, suppose G ∈ CD(G).Then by Theorem A, we have T A T B = G ∈ CD(G), where T A is the top element of CD(A) and T B is the top element of CD(B).Now A ∩ B ≤ Z(A) ≤ T A , and A ∩ B ≤ Z(B) ≤ T B .Since T A T B = AB, it follows from Lemma 4 that T A = A and T B = B.

Proposition 4 .Proposition 5 .
There exists a finite group G with G ∈ CD(G) so that CD(G) has height 2 and CD(G) possesses both abelian and nonabelian subgroups of height 1.See[6, Proposition 10]  for an explicit construction.There exists a finite group G that admits a proper central product, so that for C the subgroup collection of G generated by all CD(X) ⋅ CD(Y ) where G = XY is a proper central product, we have that C is not equal to CD(G).

Proof.
Let G be a group as in Proposition 4. Then there exists H ∈ CD(G) nonabelian of height 1, and so C G (H) ∈ CD(G) is nonabelian of height 1, and so the central product G = HC G (H) is proper.If G = AB is an arbitrary proper central product, then by Corollary 2, we have A ∈ CD(A) and B ∈ CD(B).By Theorem A we have that CD(A) ⋅ CD(B) ⊆ CD(G), and since CD(G) has height 2, it follows that CD(A) = {A, Z(A)} and CD(B) = {B, Z(B)} both have height 1.Thus the only abelian subgroup in CD(A) ⋅ CD(B) is Z(A)Z(B) = Z(G).And so if C is the subgroup collection of G generated by all CD(X) ⋅ CD(Y ) where G = XY is a proper central product, then the only abelian subgroup in C is Z(G).Thus C is not equal to CD(G).

[ 15 ,
Corollary 4].Tȃrnȃuceanu's result occurs as a corollary to their classification of groupsG satisfying CD(G) = [Z(G) ∶ G] L(G) .Once again, we are able to present a central product based proof.

Corollary 4 .
If G is a finite group satisfying CD(G) = [Z(G) ∶ G] L(G) and H is a subgroup of G, then CD(H) = [Z(H) ∶ H] L(G) .