Finite normal subgroups of strongly verbally closed groups

Does there exist a finitely generated nilpotent nonabelian strongly verbally closed group?

Let 𝐺 be a finitely generated nilpotent group with nonabelian torsion subgroup. Then 𝐺 is not strongly verbally closed.

What is an arbitrary finite strong retract?

Let 𝐺 be a nilpotent strong retract. Then 𝐺 is abelian. Moreover, 𝐺 is either divisible, or it has bounded period and its decomposition as a direct sum of primary cyclic factors has the property that the orders of any distinct summands are either equal or coprime.

Varieties of abelian groups are precisely the following classes of groups: (1) the class of all abelian groups; (2) the class of all abelian groups with period dividing some fixed positive integer 𝑛.

Proof

Let 𝒦 be a variety of abelian groups. Since any set 𝑋 of identities follows from the set of identities X ′ , consisting of relation x n = 1 , where n ⩾ 0 , and of commutator relations, i.e. relations of the form w ⁢ ( x 1 , … , x k ) = 1 , where w ⁢ ( x 1 , … , x k ) ∈ F k and k ∈ N (see [15, Theorem 12.12]), 𝒦 is the class of all groups satisfying the identities x n = 1 and [ x , y ] = 1 , where n ⩾ 0 . This completes the proof. ∎

If 𝐺 is a divisible abelian group and 𝐻 is an abelian group such that G ⩽ H , then 𝐺 is a direct summand of 𝐻.

An abelian group 𝐺 of unbounded period is a strong retract if and only if it is divisible.

Proof

Sufficiency follows from Lemma 2.2. Let 𝐺 be an abelian group of unbounded period. From Lemma 2.1, it follows that var ⁡ G is the class of all abelian groups. In particular, var ⁡ G contains a divisible group 𝐻 containing 𝐺 (see [10, §23]). Though, if 𝐺 is not divisible itself, it is not a direct summand of 𝐻 (as direct summands of a divisible group are divisible themselves; see [10, §23]), so 𝐺 is not a strong retract. ∎

An abelian group 𝐺 of bounded period 𝑑 is a direct sum of primary cyclic groups, i.e. G ≅ ⨁ i ∈ I Z p i k i , where p i are prime numbers and k i are natural numbers such that p i k i ∣ d , i ∈ I (𝐼 is an index set).

An abelian group 𝐺 of bounded period is a strong retract if and only if, in its decomposition into the direct sum of primary cyclic groups, the orders of any distinct direct summands are either equal or coprime. That is,

where C p i k i ⁢ ( n i ) is equal to the direct sum of n i copies of the group Z p i k i , p i are distinct prime numbers, 𝑚 and k i are positive integers, and n i are some cardinal numbers.

Proof

Suppose that 𝐺 cannot be decomposed into such a direct sum. We may assume that

where m ∈ N , | I i | = n i , and among k j , j ∈ I i , there are only finitely many different numbers (since 𝐺 is a group of bounded period), but there exists i ∈ { 1 , … , m } such that, for some j 1 , j 2 ∈ I i , k j 1 ≠ k j 2 .

Consider the group H = ⨁ i = 1 m C p i s i ⁢ ( n i ) , where

Since 𝐺 and 𝐻 have the same period ∏ i = 1 m s i , it follows from Lemma 2.1 that H ∈ var ⁡ G .

Consider the injection f : G → H , which is defined on each direct summand in (2.1) as follows: let i ∈ { 1 , … , m } , j ∈ I i , f : Z p i k j ↪ Z p i s i , where Z p i s i is the 𝑗th summand from the decomposition of C p i s i ⁢ ( n i ) into the direct sum. Every direct summand from (2.1) is mapped into the corresponding direct summand of the decomposition of 𝐻 so that the restriction of 𝑓 to Z p i k j is a natural injection: if k j = s i , then it is the identity map; otherwise it is a mapping to the subgroup of Z p i s i of order p i k j . From the uniqueness of the decomposition of an abelian group of bounded period into a direct sum of primary cyclic groups [3, Theorem 17.4], it follows that f ⁢ ( G ) is not a direct summand of 𝐻. Thus, 𝐺 is not a strong retract.

Now, suppose that 𝐺 has the decomposition from the statement of the proposition. Let H ∈ var ⁡ G and let f : G ↪ H be a monomorphism. Since any monomorphism preserves the order of an element, the p i th component of 𝐺 is mapped into the p i th component of 𝐻 under 𝑓, so it suffices to prove the proposition only for the case G = C p k ⁢ ( n ) , where 𝑝 is prime, k ∈ N , and 𝑛 is some cardinal number.

Let us show that there exists a subgroup X ⩽ H such that H = f ⁢ ( G ) ⊕ X . Let

be the set of all subgroups of 𝐻 having trivial intersection with f ⁢ ( G ) . Note that { 0 } ∈ M . An order on 𝑀 is introduced as follows: for X , Y ∈ M , X ⪯ Y if 𝑋 is a subgroup of 𝑌. It can be verified directly that this is an order on 𝑀. Any chain { Y α } ⊆ M of subgroups having trivial intersection with f ⁢ ( G ) is bounded by an element Y ∈ M , where Y = ⋃ α Y α . So, by Zorn’s lemma, 𝑀 contains a maximal element 𝑋: X ⩽ H , X ∩ f ⁢ ( G ) = { 0 } , and 𝑋 is not a subgroup of any bigger (with respect to the given ordering) subgroup satisfying this property.

From X ∩ f ⁢ ( G ) = { 0 } , it follows that f ⁢ ( G ) + X = f ⁢ ( G ) ⊕ X . It remains to prove that H = f ⁢ ( G ) + X . Let h ∈ H . Then there exists such k ∈ N that k ⁢ h ∈ f ⁢ ( G ) + X . Indeed, otherwise ⟨ h ⟩ ∩ ( f ⁢ ( G ) + X ) = { 0 } , which means that ( ⟨ h ⟩ + X ) ∩ f ⁢ ( G ) = { 0 } , leading to a contradiction with the maximality of 𝑋.

Let 𝑠 be the smallest such number. Replacing ℎ by a suitable power, if necessary, we may assume that s = 1 or 𝑠 is a prime. Two cases are possible:

(1) s = p . Then p ⁢ h = f ⁢ ( g ) + x for some g ∈ G , x ∈ X . If g = p ⁢ g 1 , g 1 ∈ G ( g 1 may be equal to zero), then p ⁢ h − f ⁢ ( p ⁢ g 1 ) = x . However, from the fact that h − f ⁢ ( g 1 ) ∉ X (as h ∉ f ⁢ ( G ) + X ), it can be obtained that

which leads to a contradiction with the maximality of 𝑋. Consequently, g ≠ p ⁢ g 1 for any g 1 ∈ G . As g ≠ 0 , ord ⁡ ( g ) = p k . Though, ord ⁡ ( p ⁢ h ) = p r < p k , so

As the sum f ⁢ ( G ) + X is direct, p r ⁢ f ⁢ ( g ) = p r ⁢ x = 0 , which means that p r ⁢ g = 0 , which is impossible.

(2) s ≠ p . For abelian groups of period 𝑝, the mapping g ↦ s ⁢ g is an automorphism, so, as s ⁢ h = f ⁢ ( g ) + x for some g ∈ G , x ∈ X , there exist such g 1 ∈ G , x 1 ∈ X that g = s ⁢ g 1 , x = s ⁢ x 1 . Thus, s ⁢ ( h − f ⁢ ( g 1 ) − x 1 ) = 0 . No nontrivial element of 𝐻 has order 𝑠, so h = f ⁢ ( g 1 ) + x 1 .

As a result, H = f ⁢ ( G ) ⊕ X , and 𝐺 is a strong retract. ∎

The center of a strong retract is its direct factor.

Proof

Let 𝐺 be a strong retract. The center of any group is a normal subgroup, so it suffices to prove that Z ⁢ ( G ) is a retract of 𝐺. Consider the central product of 𝐺 with its copy G ̃ with joined center

The group G ̃ is isomorphic to the group 𝐺, so it is a strong retract too. Let 𝜌 be a retraction from 𝐾 to its subgroup G ̃ . From the fact that, in the group 𝐾, the subgroup 𝐺 commutes with the subgroup G ̃ , we obtain ρ ⁢ ( G ) ⩽ Z ⁢ ( G ) . By definition of the retraction, ρ ⁢ ( g ) = g for any element g ∈ Z ⁢ ( G ) . Thus, the restriction of 𝜌 to the subgroup 𝐺 of the group 𝐾 is the desired retraction to Z ⁢ ( G ) . ∎

Let 𝐺 be a nilpotent strong retract. Then 𝐺 is abelian.

Proof

Any nontrivial normal subgroup of a nilpotent group intersects the center of this group nontrivially (see [5, Theorem 16.2.3]). By combining this with Proposition 2.6, we deduce that any nilpotent strong retract is equal to its center. ∎

Suppose that a group presentation ⟨ X ∣ R ⟩ is finitely presented over a group presentation ⟨ Y ∣ S ⟩ and ⟨ Y ∣ S ⟩ ≅ ⟨ Y ′ ∣ S ′ ⟩ . Then ⟨ X ∣ R ⟩ is finitely presented over ⟨ Y ′ ∣ S ′ ⟩ .

Proof

We may assume that X = Y ∪ A and R = S ∪ B for some finite sets 𝐴 and 𝐵. It can be easily shown (for similar fact, refer to [11, Chapter II, Proposition 2.1]) that groups defined by group presentations ⟨ Y ∣ S ⟩ and ⟨ Y ′ ∣ S ′ ⟩ are isomorphic if and only if the presentation ⟨ Y ′ ∣ S ′ ⟩ can be obtained from ⟨ Y ∣ S ⟩ by applying a finite number of Tietze transformations (and their inverses):

• adding to the set 𝑆 an arbitrary set T ⊆ ⟨ ⟨ S ⟩ ⟩ ⊴ F ⁢ ( Y ) ,

• adding to the set 𝑌 an arbitrary set Y ̃ while adding to the set 𝑆 a set

  { y ̃ = w y ̃ ∣ y ̃ ∈ Y ̃ , w y ̃ ∈ F ⁢ ( Y ) } .

It is sufficient to prove the lemma only for the case when ⟨ Y ′ ∣ S ′ ⟩ is obtained from ⟨ Y ∣ S ⟩ by applying one Tietze transformation. One can easily verify that, in case of the first transformation, X ′ = X and R ′ = R ∪ T , while in case of the second transformation, X ′ = X ∪ Y ̃ and R ′ = R ∪ { y ̃ = w y ̃ ∣ y ̃ ∈ Y ̃ , w y ̃ ∈ F ⁢ ( Y ) } provide the desired group presentation. ∎

Suppose that 𝐺 contains a subgroup 𝐻 and a finite normal subgroup 𝑁 such that G / N is finitely presented over H / ( H ∩ N ) . Then 𝐺 is finitely presented over 𝐻.

Proof

This is similar to the proof of Hall’s theorem [4] on the preservation of finite presentability of a group under extensions (see also [16, Theorem 2.2.4]).

Write H = ⟨ X ∣ R ⟩ ⩽ G , and N = ⟨ Y ∣ S ⟩ ⊴ G , where 𝑌 and 𝑆 are finite. We are assuming that G / N is finitely presented over H / ( H ∩ N ) ≅ ⟨ X ∣ R ∪ C ⟩ , where ⟨ ⟨ C ⟩ ⟩ = H ∩ N and the set 𝐶 is finite. Consequently,

where sets 𝐴 and 𝐵 are finite.

Let us construct a presentation of 𝐺. For the generators, take X ̄ ∪ A ̄ ∪ Y ̄ , where the sets X ̄ , A ̄ , Y ̄ are in one-to-one correspondence with the sets 𝑋, 𝐴, 𝑌 respectively. The sets 𝑅, 𝑆, 𝐶, and 𝐵 are in correspondence with the sets R ̄ , S ̄ , C ̄ , and B ̄ respectively. For the defining relations, take the union of the following sets: R ̄ , S ̄ , C ̄ 1 , B ̄ 1 , and T ̄ , where

( c ∈ C ̄ and b ∈ B ̄ are considered as words from F ⁢ ( X ̄ ) and from F ⁢ ( X ̄ ∪ A ̄ ) respectively). Denote the group with the above mentioned generators and defining relations as G ̃ ,

Consider a surjective homomorphism θ : G ̃ → G , defined by the following bijections X ̄ → X , A ̄ → A , Y ̄ → Y on the generators (defining relations are mapped into true identities under such a map on generators, so such a homomorphism exists). The restriction θ | K : K → N on the subgroup K = ⟨ Y ̄ ⟩ ⩽ G ̃ is an isomorphism as all the relations in the alphabet Y ̄ in G ̃ are consequences of the defining relations S ̄ . Moreover, K ⊴ G ̃ .

The homomorphism θ ̃ : G ̃ / K → G / N generated by 𝜃, is an isomorphism too. Now, let g ∈ ker ⁡ θ . Then g ⁢ K ∈ ker ⁡ θ ̃ , but θ ̃ is an isomorphism, so g ∈ K . Finally, θ | K is an isomorphism, so g = 1 . ∎

Suppose that H = ⟨ X ∣ R ⟩ is a subgroup of 𝐺, and 𝐺 is finitely presented over 𝐻. The subgroup 𝐻 is algebraically closed in 𝐺 if and only if 𝐻 is a retract of 𝐺.

Proof

Suppose 𝐻 is algebraically closed in 𝐺 and

are the sets from the definition of finite presentability of 𝐺 over 𝐻. The relations s i ⁢ ( a 1 , … , a m , X ) = 1 , i = 1 , … , n , correspond to a system of equations with coefficients from 𝐻,

which, by condition, has a solution t 1 = a 1 , … , t m = a m . By virtue of the algebraic closedness of 𝐻 in 𝐺, this system has a solution t 1 = h 1 , … , t m = h m in 𝐻. Mapping X ⊔ { a 1 , … , a m } → H , X ∋ x ↦ x , a i ↦ h i extends to a surjective homomorphism φ : G → H , as defining relations of 𝐺 are mapped into true identities under such a mapping of generators (note that 𝑅 is the set of words in the alphabet 𝑋).

This homomorphism is the desired retraction: let h ∈ H , h = v ⁢ ( x 1 , … , x r ) , x i ∈ X . Applying to this word the homomorphism 𝜑, we get

Algebraic closedness of a subgroup 𝐻 of a group 𝐺 follows from retractness of 𝐻 in 𝐺 for any group 𝐺 (see [14, Proposition 2.2 (1)]). ∎

Let 𝐶 be a finite elementary abelian 𝑝-group (where 𝑝 is a prime number). For any k ∈ N , there exists t ⩾ k such that the direct product

of copies C i of 𝐶 contains a subgroup 𝑅 invariant with respect to the diagonal action on 𝑃 of the endomorphism algebra End ⁡ C with the following properties:

1. R ⊆ ⋃ ker ⁡ ρ j , where ρ j : P → C j , j = 1 , … , t , are the natural projections,

2. R ⋅ × j ∉ J C j = P for any subset J ⊆ { 1 , … , t } of cardinality | J | = k ,

3. moreover, each such 𝐽 is contained in a set J ′ ⊇ J such that

   P = R × ( × j ∉ J ′ C j ) ,

   and there exist integers n i ⁢ j ∈ Z such that the projection π : P → × j ∉ J ′ C j with the kernel 𝑅 acts as C i ∋ c i ↦ ∏ j ∉ J ′ c j n i ⁢ j , where c j ∈ C j is the element corresponding to c i under the isomorphism C i ≅ C ≅ C j .

Let 𝐺 be a group and N ⊴ G . If x ⁢ C ⁢ ( N ) = y ⁢ C ⁢ ( N ) for some x , y ∈ G , then 𝑥 and 𝑦 act on 𝑁 (by conjugations) identically.

Proof

From x ⁢ C ⁢ ( N ) = y ⁢ C ⁢ ( N ) , it follows that, for some c ∈ C ⁢ ( N ) , x = y ⁢ c . Then, for n ∈ N , we have

The last identity is true, as (due to normality) y − 1 ⁢ n ⁢ y ∈ N and c ∈ C ⁢ ( N ) . ∎

Let 𝐻 be a strongly verbally closed group. For any finite normal subgroup 𝑇 of 𝐻, for any abelian subgroup 𝐴 of 𝑇, normal in 𝐻, it is true that Z ⁢ ( C T ⁢ ( A ) ) is a direct factor of C T ⁢ ( A ) , and some complement is normal in 𝐻.

Proof

Let 𝐻 be such a group, and let L = C T ⁢ ( A ) . It suffices, for each prime 𝑝, to find a homomorphism ψ p : L → Z ⁢ ( L ) commuting with the conjugation action of 𝐻 on 𝐿 (this action is well-defined as L ⊴ H ) and injective on the 𝑝-component Z p ⁢ ( L ) of the center of 𝐿. Then the homomorphism

where π p : Z ⁢ ( L ) → Z p ⁢ ( L ) is the projection on the 𝑝-component, is injective on Z ⁢ ( L ) , so its kernel is the desired complement 𝐷 (normality of 𝐷 in 𝐻 follows from the fact that 𝜓 commutes with the action of 𝐻 on 𝐿).

Suppose that there are no such homomorphisms for some prime number 𝑝, i.e. every homomorphism f : L → Z ⁢ ( L ) commuting with the action of 𝐻 on 𝐿 is not injective on Z p ⁢ ( L ) . Then it is not injective on the maximal elementary abelian 𝑝-subgroup C ⩽ Z p ⁢ ( L ) (it is finite as 𝐿 is finite). Indeed, if x ∈ Z p ⁢ ( L ) , x ≠ 1 is an element such that f ⁢ ( x ) = 1 , then, raising it to the appropriate power 𝑑, we get f ⁢ ( x d ) = 1 and x d ∈ C , x d ≠ 1 .

Use Lemma 3.4 (with respect to 𝐶) to choose a positive integer 𝑡 as in the lemma (for some 𝑘 to be specified later). Then consider the fibered product of 𝑡 copies of the group 𝐻,

First of all, let us show that the subgroup R ⩽ C t ⩽ Q from Lemma 3.4 is normal in 𝑄. Note that the subgroup 𝑅 is invariant under the diagonal action of automorphisms Aut ⁡ C ⩽ End ⁡ C . It remains to show that 𝑄 acts diagonally on P = C t by conjugation. This follows from Lemma 3.5.

Indeed, let

As q 1 ⁢ L = q 2 ⁢ L = ⋯ = q t ⁢ L , according to Lemma 3.5, q − 1 ⁢ p ⁢ q = q ̃ − 1 ⁢ p ⁢ q ̃ , where q ̃ = ( q 1 , … , q 1 ) . It means that the conjugation action of 𝑄 on 𝑃 is diagonal. On the other hand, the diagonal action by conjugation induces an endomorphism of C t (due to normality of C ⊴ H ), and 𝑅 is invariant with respect to the diagonal action of such endomorphisms, leading to normality of 𝑅 in 𝑄.

Put G = Q / R . First, let us show that 𝐻 embeds into 𝐺. The group 𝐻 embeds into 𝑄 diagonally: h ↦ ( h , … , h ) , h ∈ H . This homomorphism serves as an embedding into 𝐺 as well, as all projections of any nontrivial diagonal element of 𝑄 are nontrivial (and 𝑅 is contained in the union of the kernels of these projections).

Now, let us prove the verbal closedness of this diagonal subgroup (denote it as 𝐻 too) in 𝐺. Consider an equation

having a solution in 𝐺 and let x ̃ 1 , … , x ̃ n be a preimage (in 𝑄) of a solution x 1 , … , x n . Then (in 𝑄)

where ( c 1 , … , c t ) ∈ R . By the first property in Lemma 3.4, c i = 1 for some 𝑖. This means that, in 𝐻 (the group itself), w ⁢ ( x ̃ 1 i , … , x ̃ n i ) = h , where x ̃ j i is the 𝑖th coordinate of the vector x ̃ j , j = 1 , … , n .

Let us take y j = ( x ̃ j i , … , x ̃ j i ) , j = 1 , … , n . Then, in H ⩽ G , the following is true:

which proves the verbal closedness of 𝐻 in 𝐺.

Let U ⩽ L and define

(coordinate 𝑢 stands on the 𝑖th place). It remains to prove that 𝐻 is not algebraically closed in 𝐺.

From Lemma 3.3 and Claim 3.7, it follows that it suffices to show that 𝐻 is not a retract of 𝐺. Let ρ : G → H be a hypothetical retraction, and let ρ ̂ : Q → H be its composition with the natural epimorphism Q → Q / R = G . Henceforth, all centralizers and other subgroups we refer to relate to 𝑄.

Let us verify that ρ ̂ ⁢ ( L i ) ⩽ C T ⁢ ( C T ⁢ ( L ) ) ⩽ L for every 𝑖. First, we prove the left inclusion. Let h ∈ C T ⁢ ( L ) . Then ℎ commutes with every element from 𝐿; consequently, ℎ, as an element of 𝑄, commutes with L i . Applying the retraction ρ ̂ to this identity, we get that ρ ̂ ⁢ ( h ) ( = h ) commutes with the subgroup ρ ̂ ⁢ ( L i ) , which (by definition of the centralizer) proves the inclusion. The second inclusion follows from the fact that L = C T ⁢ ( A ) = C ⁢ ( A ) ∩ T , which means that

The first inclusion here is true as C ⁢ ( L ) ⩾ A , and the claim follows since A ⩽ T .

On the other hand, for i ≠ j , the mutual commutator subgroup [ L i , L j ] is trivial (for i ≠ j , L i and L j are contained in different components of the fibered product). This means that the image of this mutual commutator subgroup is trivial too: [ ρ ̂ ⁢ ( L i ) , ρ ̂ ⁢ ( L j ) ] = { 1 } . Consequently,

If ρ ̂ ⁢ ( L i ) = ρ ̂ ⁢ ( L l ) for some i ≠ l , then (by the virtue of well-known commutator identities) [ ρ ̂ ⁢ ( L i ) , ∏ j ρ ̂ ⁢ ( L j ) ] = { 1 } , which means that ρ ̂ ⁢ ( L i ) ⩽ C T ⁢ ( L ) (as L = ρ ̂ ⁢ ( L ) ⩽ ∏ j ρ ̂ ⁢ ( L j ) ).

Thereby, if, for some different 𝑖 and 𝑗, ρ ̂ ⁢ ( L i ) = ρ ̂ ⁢ ( L j ) , then ρ ̂ ⁢ ( L i ) ⩽ C T ⁢ ( L ) . From here and from the inclusion we proved earlier, we get

Let us take 𝑘 in Lemma 3.4 to be the number of all subgroups of 𝑇, and let 𝐽 be the set of all exclusive numbers 𝑖, namely such that, for any l ≠ i , ρ ̂ ⁢ ( L i ) ≠ ρ ̂ ⁢ ( L l ) . Since, among ρ ̂ ⁢ ( L i ) ⩽ T , there are no more than 𝑘 different subgroups, | J | ⩽ k . Thus, from property (3) of Lemma 3.4, we have a decomposition

where I ⊆ { 1 , … , t } ∖ J is some set of non-exclusive elements. Again, according to property (3) of Lemma 3.4, the projection π : × i = 1 t C i → × i ∈ I C i onto the second factor of this decomposition is defined by an integer matrix ( n i ⁢ j ) , namely, for c i ∈ C i , π : c i ↦ ∏ j ∈ I c j n i ⁢ j , where c j are elements corresponding to c i under the isomorphism C i ≅ C ≅ C j .

This means that the restriction of 𝜋 to C = { ( c , … , c ) ∣ c ∈ C ⩽ H } is defined as follows:

Here c j are elements corresponding to 𝑐 under the isomorphism C ≅ C j .

Then (as i ∈ I are non-exclusive, we have ρ ̂ ⁢ ( L i ) ⩽ Z ⁢ ( L ) ) consider the composition

This extends to a homomorphism Φ : L → Z ⁢ ( L ) defined by the similar formula

where g ∈ L , and g j ∈ L j are elements corresponding to 𝑔. Obviously, it is an extension of Ψ and a homomorphism since, for j ∈ I , ρ ̂ ⁢ ( L j ) ⩽ Z ⁢ ( L ) and the group Z ⁢ ( L ) is abelian. This homomorphism commutes with the conjugation action of 𝐻 on 𝐿. Indeed, let g ∈ H and let 𝔤 be the action of 𝑔 on 𝐿 by conjugation, namely, for x ∈ L , g ⁢ ( x ) = g − 1 ⁢ x ⁢ g . Let us show that Φ ∘ g = g ∘ Φ . Let h ∈ L . Then

The penultimate equality holds since ρ ̂ is a retraction to 𝐻, so it acts identically on 𝐻 itself. By the assumption we made in the beginning, the kernel of this homomorphism has nontrivial intersection with 𝐶: ker ⁡ Φ ∩ C ≠ { 1 } , so the restriction Ψ = Φ | C has a nontrivial kernel too.

On the other hand, Ψ is the identity map since Ψ = ρ ̂ | C ∘ π | C = ρ ̂ | C ∘ π ̂ = ρ ̂ | C (the final equality is true since π ̂ is a projection “forgetting” the 𝑅-coordinate, and ρ ̂ ⁢ ( R ) = { 1 } is a composition of the natural homomorphism to the quotient group and of the retraction to 𝐻) and ρ ̂ | C = id , as ρ ̂ is the retraction from 𝑄 to 𝐻, so it acts trivially on 𝐶. The contradiction thus obtained completes the proof. ∎

Let 𝐺 be a finitely generated nilpotent group with a nonabelian torsion subgroup. Then 𝐺 is not strongly verbally closed.

Proof

Let 𝑇 be the torsion subgroup of 𝐺 and set A = Z ⁢ ( T ) ≠ { 1 } . Since 𝑇 is nilpotent and nonabelian, every nontrivial normal subgroup of 𝑇 has a nontrivial intersection with 𝐴 (see [5, Theorem 16.2.3]), so 𝐴 is not a direct factor of 𝑇. ∎

Let 𝐺 be a finite group such that Z ⁢ ( G ) is not a direct factor of 𝐺. Then 𝐺 is not strongly verbally closed.

Let 𝐻 be the discrete Heisenberg group with 𝑎 and 𝑏 being its free generators, and let 𝑁 be the following subgroup:

where 𝛼, 𝑛 are non-negative integers. Then the group G = H / N is strongly verbally closed if and only if gcd ⁡ ( α , n ) = 1 .

Proof

Let T ⁢ ( G ) be the torsion subgroup of 𝐺. The center of the group 𝐻 is equal to its commutator subgroup, and it is isomorphic to the infinite cyclic group. Suppose T ⁢ ( G ) = { 1 } , so ( α , n ) = ( 0 , 0 ) or ( α , n ) = ( 0 , 1 ) , and, respectively, G = H or 𝐺 is abelian. The non-strong-verbal-closedness of 𝐻 was proven in [9], while the strong verbal closedness of abelian groups was proven in [12].

If gcd ⁢ ( α , n ) = 1 , then, once again, 𝐺 is abelian since [ a , b ] α = [ a α , b ] ∈ N ; consequently, it is strongly verbally closed.

Consider the case gcd ⁢ ( α , n ) = d ≠ 1 . Without loss of generality, we may assume that 𝛼 and 𝑛 are minimal such that a α ∈ N , [ a , b ] n ∈ N . Consider the central product of 𝐺 with its copy G ̃ with joined commutator subgroup,

The subgroup 𝐺 is not algebraically closed in 𝐾 since 𝐺 is not a retract of 𝐾. Indeed, let 𝜌 be a hypothetical retraction. The subgroup 𝐺 commutes with G ̃ in 𝐾, so ρ ⁢ ( G ̃ ) ⩽ Z ⁢ ( G ) and ρ ⁢ ( G ̃ ′ ) = { 1 } , which leads to a contradiction with the definition of retraction.

However, 𝐺 is verbally closed in 𝐾. Indeed, let w ∈ F ⁢ ( t 1 , … , t s ) be some word and

for some h ⁢ N , h i ⁢ N ∈ G , h i ′ ⁢ N ∈ G ̃ . Then, for some c ⁢ N ∈ G ′ , the following holds:

By an automorphism of the free group, the word 𝑤 can be reduced to a normal form [9]: w ⁢ ( t 1 , … , t s ) = t 1 m ⁢ w ′ ⁢ ( t 1 , … , t s ) , where m ∈ N , w ′ ∈ F s ′ . From the first equation, we get c ⁢ N ∈ G ′ ∩ φ ⁢ ( G s ) , where

is a verbal mapping. This means that, for some w 1 , w 2 ∈ N , in 𝐻, it is true that

Let us show that, in 𝐺, the identity ( a ⁢ N ) x = [ a ⁢ N , b ⁢ N ] z does not hold for x ∉ α ⁢ Z . Suppose otherwise. Then, in 𝐻, we have

for some k , l , t , s ∈ Z . After some reductions, we get a x − α ⁢ t = [ a , b ] n ⁢ s + z + α ⁢ t ⁢ k , which means that x = α ⁢ t ∈ α ⁢ Z , a contradiction. Thus, h 1 ′ = [ a , b ] γ for some γ ∈ Z , and, consequently, c ⁢ w 1 ∈ H ′ . By [9], proof of the Heisenberg-group theorem, for any verbal mapping 𝜑 in the discrete Heisenberg group, for any g ∈ φ ⁢ ( H s ) , the coset g ⁢ ( φ ⁢ ( H s ) ∩ H ′ ) is contained in φ ⁢ ( H s ) . Thus, for some g 1 , … , g s ∈ H , w ⁢ ( g 1 , … , g s ) = w ⁢ ( h 1 , … , h s ) ⁢ c ⁢ w 1 . This means that

for some w 3 ∈ W , and in 𝐺,

which proves verbal closedness of 𝐺 in 𝐾. ∎

The group H n ⁢ ( K ) is not strongly verbally closed.

Proof

Consider the central product of H n ⁢ ( K ) with its copy H ̃ n ⁢ ( K ) with joined commutator subgroup,

Denote with the symbols 𝐻 and H ̃ the first and the second factor of this central product respectively. Let us show that 𝐻 is not algebraically closed in 𝐺. The group 𝐻 is linear, and, consequently, it is equationally noetherian (see [1, Theorem B1]), so it is algebraically closed in 𝐺 if and only if it is a retract of every finitely generated over 𝐻 subgroup of 𝐺 (see [8]). In particular, of such a subgroup of 𝐺,