On semiconcise words

Theorem A.

Let w be a semiconcise word, and let G be an FC ⁢ ( w ) -group. Then [ w ⁢ ( G ) , G ] is FC -embedded in G.

Theorem B.

Let w be a semiconcise word, and let G be a BFC ⁢ ( w ) -group. Then [ w ⁢ ( G ) , G ] is BFC -embedded in G.

Lemma 2.1.

Let w be a group-word. If G is an FC ⁢ ( w ) -group, then the conjugacy class x G w * is finite for all x ∈ G .

Proof.

By the first statement in [2, Proposition 2.9], a group G is an FC ⁢ ( w ) -group if and only if it is an FC ⁢ ( w - 1 ) -group. For all g ∈ G , we have g ∈ G w if and only if g - 1 ∈ G w - 1 . Thus G w - 1 = ( G w ) - 1 , and the result follows. ∎

Lemma 2.2.

Let w be a group-word, and let G be an FC ⁢ ( w ) -group. Choose x ∈ G , and denote by A a finite subset of G w * such that x G w * = x A . Then, for any j ≥ 1 and y 1 , … , y j ∈ G w * , there exist a 1 , … , a j ∈ A such that x y 1 ⁢ … ⁢ y j = x a 1 ⁢ … ⁢ a j .

Proof.

We argue by induction on j. The case j = 1 is clear. Let j > 1 , and assume that x y 1 ⁢ … ⁢ y j - 1 = x a 1 ⁢ … ⁢ a j - 1 with a 1 , … , a j - 1 ∈ A . Then

where b = ( a 1 ⁢ … ⁢ a j - 1 ) - 1 . Since y j b ∈ G w * , we have x y j b = x a j for some a j ∈ A , and so x y 1 ⁢ … ⁢ y j = x a j ⁢ a 1 ⁢ … ⁢ a j - 1 . After renumbering the elements a i ∈ A , we obtain the required result. ∎

Lemma 2.3.

Let w = w ⁢ ( x 1 , … , x n ) be a group-word, and set

If G is an FC ⁢ ( w ) -group, then it is an FC ⁢ ( v ) -group.

Proof.

Let y ∈ G v . Then there exist g 1 , … , g n , g n + 1 ∈ G such that y = z ⁢ t , with z = w ⁢ ( g 1 , … , g n ) - 1 ∈ ( G w ) - 1 = G w - 1 and t = w ⁢ ( g 1 , … , g n ) g n + 1 ∈ G w . By Lemma 2.2, for any x ∈ G , the conjugate x y can only take finitely many values. This proves the result. ∎

Lemma 2.4.

Let w = w ⁢ ( x 1 , … , x n ) be a semiconcise word, and set

Let G be an FC ⁢ ( w ) -group and B a finite subset of G v * . Then, for any x ∈ G , there exists a positive integer e such that b e ∈ Z ⁢ ( 〈 x , B 〉 ) for all b ∈ B .

Proof.

Write B = { b 1 , … , b r } , and let x be an arbitrary element of G. For any b i ∈ B , there exist elements g i 1 , … , g i n , g i n + 1 ∈ G such that either

Put

By [2, Lemma 2.7 (i)], the set ( J / Z ⁢ ( J ) ) w is finite. As w is semiconcise, the subgroup [ w ⁢ ( J / Z ⁢ ( J ) ) , J / Z ⁢ ( J ) ] is finite. Thus v ⁢ ( J ) has finite order modulo Z ⁢ ( J ) , say e. Since B ⊆ v ⁢ ( J ) , it follows that b i e ∈ Z ⁢ ( J ) for all i. As 〈 x , B 〉 ≤ J , the result follows. ∎

Proof of Theorem A.

Set v = [ w ⁢ ( x 1 , … , x n ) , x n + 1 ] . Then G is an FC ⁢ ( v ) -group by Lemma 2.3. Let x be an arbitrary element of G. By Lemma 2.1, we can choose b 1 , … , b r ∈ G v * such that x G v * = { x b 1 , … , x b r } . Write B = { b 1 , … , b r } . Define the order < on the set of all (formal) products of the form b i 1 ⁢ … ⁢ b i j , with 1 ≤ i k ≤ r and j ≥ 1 , as follows. Put

if and only if one of the following conditions is satisfied: j < j ′ , or j = j ′ and there is a positive integer l ≤ j such that i l < i l ′ and i k = i k ′ for all k > l .

Let y be an arbitrary element of v ⁢ ( G ) . Then y = y 1 ⁢ … ⁢ y j , where each y i ∈ G v * . By Lemma 2.2, for all k ∈ { 1 , … , j } , there exists an integer i k ∈ { 1 , … , r } such that x y = x b i 1 ⁢ … ⁢ b i j . Clearly, we can choose b i 1 ⁢ … ⁢ b i j to be the smallest (in the sense of the order < ) product of elements from B such that x y = x b i 1 ⁢ … ⁢ b i j . Let us now show that i 1 ≥ i 2 ≥ ⋯ ≥ i j . Suppose to the contrary that i k < i k + 1 for some k. Then

where c = b i k ⁢ b i k + 1 ⁢ b i k - 1 ∈ G v * . In view of Lemma 2.2, we have

for some 1 ≤ i 1 ′ , … , i k - 1 ′ , i k + 1 ′ ≤ r so that

This contradicts the choice of the product b i 1 ⁢ … ⁢ b i j because

Thus x y = x b i 1 ⁢ … ⁢ b i j with i 1 ≥ i 2 ≥ … ≥ i j or, equivalently,

for some non-negative integers e r , … , e 1 .

By Lemma 2.4, there exists a positive integer e such that b i e ∈ Z ⁢ ( 〈 x , B 〉 ) for all i. Hence we may assume that e i < e for all i, and so | x v ⁢ ( G ) | ≤ e r . Thus x v ⁢ ( G ) is finite for all x ∈ G . We conclude therefore that v ⁢ ( G ) = [ w ⁢ ( G ) , G ] is FC -embedded in G, as required. ∎

Lemma 3.1.

Let { G i } i ∈ I be a family of groups, and let U be an ultrafilter over I. Then a sentence in the first-order language of groups holds in the ultraproduct modulo U if and only if the set of all i ∈ I for which the sentence holds in G i is a member of U .

Proposition 3.2.

Let m ≥ 1 . Suppose that w is a semiconcise word and G is a group in which w takes precisely m values. Then the order of [ w ⁢ ( G ) , G ] is m-bounded.

Proof.

Assuming that w involves n variables, write w = w ⁢ ( x 1 , … , x n ) , and set v ⁢ ( x 1 , … , x n , x n + 1 ) = [ w ⁢ ( x 1 , … , x n ) , x n + 1 ] . Then [ w ⁢ ( G ) , G ] = v ⁢ ( G ) .

By way of contradiction, suppose there exists a family of groups 𝒢 = { G i } i ∈ ℕ with the property that | ( G i ) w | ≤ m for all i ∈ ℕ but

Consider a non-principal ultrafilter 𝒰 over ℕ , and let Q be the ultraproduct modulo 𝒰 of 𝒢 . Then, by (a) and Lemma 3.1, we have | Q w | ≤ m . As w is semiconcise, it follows that v ⁢ ( Q ) = [ w ⁢ ( Q ) , Q ] is finite. In particular, v has finite width, say k, in Q. Now (b) and Lemma 3.1 yield that there exists X ∈ 𝒰 such that v has finite width at most k in G i for all i ∈ X . Hence every element of v ⁢ ( G i ) = [ w ⁢ ( G i ) , G i ] can be written as a product of at most k elements of ( G i ) v * . Moreover, as

from | ( G i ) w | ≤ m , we get | ( G i ) v | ≤ m 2 for all i ∈ ℕ . Therefore, | v ⁢ ( G i ) | ≤ m 2 ⁢ k for all i ∈ X . As noted above, 𝒰 contains the cofinite filter over ℕ , so X ∩ Y ∈ 𝒰 for every cofinite subset Y of ℕ . In particular, X ∩ Y is non-empty. Therefore, every cofinite subset of ℕ contains some element i for which | v ⁢ ( G i ) | ≤ m 2 ⁢ k . This is incompatible with the assumption that | v ⁢ ( G i ) | goes to infinity. ∎

Lemma 3.3.

Let w = w ⁢ ( x 1 , … , x n ) be a group-word. If G is a BFC ⁢ ( w ) -group such that | x G w | ≤ m for all x ∈ G , then the conjugacy class x G w * has { m , n } -bounded order for all x ∈ G .

Proof.

By the second statement in [2, Proposition 2.9], a group G is a BFC ⁢ ( w ) -group if and only if it is a BFC ⁢ ( w - 1 ) -group. More precisely, if G is a BFC ⁢ ( w ) -group such that | x G w | ≤ m for all x ∈ G , then x G w - 1 has { m , n } -bounded order. Since G w * = G w ∪ G w - 1 , the result follows. ∎

Lemma 3.4.

Let w = w ⁢ ( x 1 , … , x n ) be a group-word, and set

If G is a BFC ⁢ ( w ) -group such that | x G w | ≤ m for all x ∈ G , then G is a BFC ⁢ ( v ) -group and x G v has { m , n } -bounded order for all x ∈ G .

Proof.

This is similar to the proof of Lemma 2.3 taking into account that x y can only take { m , n } -boundedly many values by Lemma 3.3. ∎

Lemma 3.5.

Let w = w ⁢ ( x 1 , … , x n ) be a group-word, and set

Let G be a BFC ⁢ ( w ) -group such that | x G w | ≤ m for all x ∈ G , and let B be a finite subset of G v * . Then, for any x ∈ G , there exists an { m , n , | B | } -bounded positive integer e such that b e ∈ Z ⁢ ( 〈 x , B 〉 ) for all b ∈ B .

Proof.

It follows as in the proof of Lemma 2.4: by [2, Lemma 2.7 (ii)], the set ( J / Z ⁢ ( J ) ) w is finite of { m , n , | B | } -bounded order and, by Proposition 3.2, the number e is { m , n , | B | } -bounded. ∎

Proof of Theorem B.

Set v = [ w ⁢ ( x 1 , … , x n ) , x n + 1 ] . Then Lemma 3.4 tells us that G is a BFC ⁢ ( v ) -group and x G v has { m , n } -bounded order for all x ∈ G . Let x be an arbitrary element of G, and choose

Write B = { b 1 , … , b r } . Define the order < on the set of all (formal) products of the form b i 1 ⁢ … ⁢ b i j , with 1 ≤ i k ≤ r and j ≥ 1 , as in (2.1) in the proof of Theorem A.

Let y be an arbitrary element of v ⁢ ( G ) . Arguing as in the proof of Theorem A, write

for some non-negative integers e r , … , e 1 . Since the number r is { m , n } -bounded by Lemma 3.4, it follows from Lemma 3.5 that there exists an { m , n } -bounded positive integer e such that b i e ∈ Z ⁢ ( 〈 x , B 〉 ) for all i. Hence we may assume that e i < e for all i, and so | x v ⁢ ( G ) | ≤ e r . Thus x v ⁢ ( G ) is finite of { m , n } -bounded order for all x ∈ G , and so v ⁢ ( G ) is BFC -embedded in G. ∎

Lemma 4.1.

Let G = 〈 g 1 , … , g m 〉 be a group, and suppose that, for a group-word w, the set { [ x , g i ] ∣ x ∈ G w , i = 1 , … , m } is finite. Then G w is contained in finitely many right cosets of w ⁢ ( G ) ∩ Z ⁢ ( G ) .

Proof.

Suppose that there exists an infinite sequence x 1 , x 2 , … of elements of G w belonging to distinct right cosets of C = w ⁢ ( G ) ∩ Z ⁢ ( G ) . For any i = 1 , … , m , put S i = { [ x , g i ] ∣ x ∈ G w } . Denote by P the Cartesian product S 1 × … × S m , and let π be the map sending x ∈ G w to ( [ x , g 1 ] , … , [ x , g m ] ) ∈ P . Now we have that x i ⁢ π = x j ⁢ π implies x i ⁢ x j - 1 ∈ C , or C ⁢ x i = C ⁢ x j which is impossible. It follows that P must be infinite, a contradiction. ∎

Proposition 4.2.

Let w = w ⁢ ( x 1 , … , x n ) be a group-word, and set

If w is semiconcise, then v is semiconcise.

Proof.

Let G be a group, and assume that G v is finite. Since v ⁢ ( G ) ′ is finite (see, for instance, [6, Proposition 1]), we may assume that v ⁢ ( G ) is abelian. It follows that every subgroup of v ⁢ ( G ) is finitely generated. Let K = 〈 g 1 , … , g m 〉 be a finitely generated subgroup of G such that v ⁢ ( G ) = v ⁢ ( K ) . For an arbitrary g = g 0 ∈ G , put H = 〈 g 0 , K 〉 . Of course, v ⁢ ( G ) = v ⁢ ( H ) . Also, | H : C H ⁢ ( v ⁢ ( H ) ) | is finite because every h ∈ H v has only finitely many conjugates in H. We claim that | v ⁢ ( H ) : v ⁢ ( H ) ∩ Z ⁢ ( H ) | is also finite, from which it follows that [ v ⁢ ( H ) , H ] is finite by a result of Baer (see [10, Corollary, p. 103]). In fact, the set

is finite, and therefore, by Lemma 4.1, H w is contained in finitely many right cosets of w ⁢ ( H ) ∩ Z ⁢ ( H ) . Hence ( H / Z ⁢ ( H ) ) w is finite. Since w is semiconcise, we obtain that

is finite.

Then [ v ⁢ ( H ) , H ] = [ v ⁢ ( G ) , H ] is finite. In particular, [ v ⁢ ( G ) , 〈 g 〉 ] is finite for any g ∈ G . Thus [ v ⁢ ( G ) , G ] is a finitely generated periodic abelian group, and so [ v ⁢ ( G ) , G ] is finite. This proves that v is semiconcise. ∎

Corollary 4.3.

Let w = [ x , n y ] be the n-Engel word, and set

where the variables x , y , z 1 , … , z m are all different. Then v is semiconcise.

Proposition 4.4.

There exist a group-word w (which is not semiconcise) and a BFC ⁢ ( w ) -group G such that [ w ⁢ ( G ) , G ] is not FC -embedded in G.

Proof.

Let G and w be as in (4.1) and (4.2), respectively. Then, by [1, Proposition 4.1], G is a BFC ⁢ ( w ) -group.

Denote by K = A × A b the base group of G. For any odd integer m ≥ 1 , let N = 〈 v 0 m , ( v 0 b ) m 〉 , where v 0 ∈ A is the nontrivial value of v ⁢ ( x , y ) in A. Notice that N is central in K and closed under conjugation by b so that N is a normal subgroup of G. Also, since K / N has odd exponent p 2 ⁢ m ⁢ n and | G / K | = 2 , we have

and consequently

Hence v 0 ⁢ N ∈ ( G / N ) w , and therefore v 0 k ⁢ N ∈ w ⁢ ( G / N ) for any integer k. It follows that

Now

where v 0 2 ⁢ ( v 0 b ) - 2 has order m in G / N . Thus

and so

In particular, | b [ w ⁢ ( G ) , G ] | ≥ m . Since m is an arbitrary odd positive integer, we conclude that b [ w ( G ) , G ) ] is infinite. This proves that [ w ( G ) , G ) ] is not FC ⁢ ( w ) -embedded in G. ∎