*Corresponding author(s). E-mail(s): igor.matematico@gmail.com; Contributing authors: ander.porto@ict.ufvjm.edu.br; millena_andrade10@hotmail.com; †These authors contributed equally to this work.
Introduction
A subgroup H H H G G G subnormal in G G G
H = G m ⊴ G m − 1 ⊴ ⋯ ⊴ G 1 ⊴ G 0 = G ( ⋆ ) H = G_m \unlhd G_{m-1} \unlhd\cdots\unlhd G_1 \unlhd G_0 = G \quad(\star) H = G m ⊴ G m − 1 ⊴ ⋯ ⊴ G 1 ⊴ G 0 = G ( ⋆ ) for some integer m ≥ 0 m \ge0 m ≥ 0 H H H H H H subnormal of defect at most m m m H ⊴ m G H \unlhd^{m} G H ⊴ m G ( ⋆ ) (\star) ( ⋆ ) H H H G G G subnormal defect (or simply the defect ) of H H H G G G
The systematic study of subnormal subgroups originated with Wielandt [37 ] , who established the foundational theory for finite groups. The concept of defect was later formalized to analyze the length of subnormal chains, with significant contributions from Roseblade [29 ] and Möhrës [22 ] regarding infinite groups.
It is a classical result that if a group G G G c c c G G G c c c [7 p. 154] ). Roseblade [29 ] established a strong converse to this theorem: any group in which every subgroup is subnormal of defect at most d d d μ ( d ) \mu(d) μ ( d ) d d d [11 ] constructed a group in which every subgroup is subnormal, yet the center is trivial, which implies that the group cannot be nilpotent.
Let H H H G G G H H H subnormal in G G G ( ⋆ ) (\star) ( ⋆ ) G G G [5 ] ). We write H ⊴ c d G H \trianglelefteq_{c}^{d} G H ⊴ c d G H H H G G G d d d
Clearly, G G G [15 , 24 ] ). Moreover, any group in which every subgroup has defect at most 1 is nilpotent of class at most 2, and this bound is sharp; in Roseblade’s terminology, this is the content of μ ( 1 ) = 2 \mu(1)=2 μ ( 1 ) = 2 D 8 D_8 D 8 G G G [25 ] ). Furthermore, it is known that non-normal (closed) subgroups of (profinite) Cappitt groups have subnormal defect exactly 2 (see [2 , 23 ] ). Note that the only torsion-free (profinite) Dedekind groups are (profinite) abelian, and hence have nilpotency class equal to 1.
The structure of groups whose subgroups have subnormal defect at most 2 has been thoroughly investigated by authors such as Mahdavianary [18 , 20 ] , Cappitt [3 ] , and Heineken [9 ] . A fundamental result, valid for both abstract and profinite groups, states that if every cyclic (respectively, procyclic) subgroup is normal in a group (respectively, profinite group) G G G G G G G G G [24 , 25 ] ). However, for groups in which all subgroups are 2-subnormal, this equivalence does not hold in general. Mahdavianary proved that if a group G G G G G G G G G [20 Theorem 1(i)] ). For 2-generated groups, this equivalence holds unconditionally (see [21 ] ). Several other classes of groups are discussed in Mahdavianary’s work, including examples proving that the corresponding group classes are strictly distinct ([20 Theorem 1(ii)–(iii)] ).
It is known that a group in which every cyclic subgroup is 2-subnormal is nilpotent of class at most 3 (see Theorem 1 in [18 ] ). Consequently, the same bound holds for any group in which every subgroup is 2-subnormal, since in this case every cyclic subgroup is in particular 2-subnormal. In Roseblade’s terminology, this is the content of μ ( 2 ) = 3 \mu(2) = 3 μ ( 2 ) = 3 [36 ] or [32 ] ). However, this condition fails for higher defects. The requirement that every cyclic subgroup be 3-subnormal is insufficient to guarantee nilpotency, in contrast to the case where every subgroup is 3-subnormal, which by Roseblade’s theorem implies nilpotency of class at most μ ( 3 ) \mu(3) μ ( 3 ) [35 ] established that, for groups which are either torsion-free or of prime exponent different from 7, the property that every subgroup is 3-subnormal is equivalent to nilpotency of class at most 3. In contrast, for groups of exponent 7, the same condition still implies nilpotency, but with class at most 4, and this bound is sharp. More recently, Smith and Traustason [31 ] showed that torsion-free groups with all subgroups subnormal of defect at most 4 are nilpotent of class at most 4. Taken together, these results reveal a sharp contrast between the behaviour of periodic and non-periodic groups with respect to subnormality restrictions.
The following problem, due to Traustason (see Problem 16.97 in the Kourovka Notebook [14 ] ), asks whether a torsion-free group all of whose subgroups are subnormal of defect at most n n n n n n n ≤ 4 n \le4 n ≤ 4 [1 ] provides a 2-generated torsion-free nilpotent group of nilpotency class six in which every subgroup is subnormal of defect at most five, thus answering the case n = 5 n = 5 n = 5
Question 1 (Traustason – Problem 16.97, revised). Does every torsion-free group all of whose subgroups are subnormal of defect at most n n n n n n n ≥ 6 n \ge6 n ≥ 6
An analogous question may be asked for profinite groups, although only for n ≥ 2 n \ge2 n ≥ 2 n ≥ 6 n \ge6 n ≥ 6 n = 1 , 2 , 3 , 4 n = 1, 2, 3, 4 n = 1 , 2 , 3 , 4 n = 5 n = 5 n = 5 n ≥ 2 n \ge2 n ≥ 2 Question 1 ). In this paper, we take a step in this direction by settling the case n = 2 n = 2 n = 2 Theorem 4.9 ). More precisely, taking a non-periodic profinite group such that all its infinite procyclic subgroups are torsion-free, we prove a profinite version of Heineken’s theorem on 2-Engel groups [9 Theorem 1(I)] , which is used in the proof of Theorem A, where we establish a profinite analogue of Mahdavianary’s theorem [20 Theorem 1] . We also show that if G G G π \pi π G G G G G G π \pi π
Notation, commutator identities and auxiliary results
Let G G G x ∈ G x \in G x ∈ G ∣ x ∣ |x| ∣ x ∣ x x x ∣ x ∣ = ∞ |x| = \infty ∣ x ∣ = ∞ ∣ x ∣ < ∞ |x| < \infty ∣ x ∣ < ∞ G G G tor ( G ) \operatorname{tor}(G) tor ( G ) g , h ∈ G g,h \in G g , h ∈ G h h h g g g h g = g − 1 h g h^g = g^{-1}hg h g = g − 1 h g h h h g g g [ h , g ] = h − 1 g − 1 h g = h − 1 h g [h,g] = h^{-1}g^{-1}hg = h^{-1}h^g [ h , g ] = h − 1 g − 1 h g = h − 1 h g x 1 , x 2 , … , x n ∈ G x_1,x_2,\ldots,x_n \in G x 1 , x 2 , … , x n ∈ G n ≥ 2 n \ge2 n ≥ 2 left-normed commutator of weight n n n [ x 1 , x 2 , … , x n ] = [ [ x 1 , … , x n − 1 ] , x n ] [x_1,x_2,\ldots,x_n] = [[x_1,\ldots,x_{n-1}],x_n] [ x 1 , x 2 , … , x n ] = [[ x 1 , … , x n − 1 ] , x n ] x 1 = x x_1=x x 1 = x x 2 = ⋯ = x n = y x_2=\cdots=x_n=y x 2 = ⋯ = x n = y [ x , y , … , y ] [x,y,\ldots,y] [ x , y , … , y ] [ x , m y ] [x,{}_m y] [ x , m y ] m = n − 1 m=n-1 m = n − 1 y y y g ∈ G g \in G g ∈ G τ g \tau_g τ g G G G τ g ( x ) = g − 1 x g \tau_g(x)=g^{-1}xg τ g ( x ) = g − 1 xg x ∈ G x \in G x ∈ G
For a subgroup H H H G G G C G ( H ) C_G(H) C G ( H ) N G ( H ) N_G(H) N G ( H ) H G = ⟨ h g ∣ h ∈ H , g ∈ G ⟩ H^G=\langle h^g \mid h \in H,\ g \in G\rangle H G = ⟨ h g ∣ h ∈ H , g ∈ G ⟩ H H H G G G
Let G G G X ‾ \overline{X} X X ⊆ G X \subseteq G X ⊆ G H ≤ c G H \leq_c G H ≤ c G H ⊴ c G H \unlhd_c G H ⊴ c G H ≤ o G H \leq_o G H ≤ o G H ⊴ o G H \unlhd_o G H ⊴ o G H H H G G G H H H C G ( H ) C_G(H) C G ( H ) N G ( H ) N_G(H) N G ( H )
Now let H H H G G G G ^ \widehat{G} G ( H ‾ ) G ^ (\overline{H})^{\widehat{G}} ( H ) G H G ‾ \overline{H^G} H G H ‾ \overline{H} H G ^ \widehat{G} G G ^ \widehat{G} G H G H^G H G H H H G G G H G ‾ \overline{H^G} H G G ^ \widehat{G} G H ‾ \overline{H} H
Let G G G ( H i ) i = 0 ∞ (H_i)_{i=0}^{\infty} ( H i ) i = 0 ∞
H 0 = G , H i + 1 = H H i ‾ ( i ≥ 0 ) , H_0=G,\qquad H_{i+1}=\overline{H^{H_i}}\quad(i\geq0), H 0 = G , H i + 1 = H H i ( i ≥ 0 ) , where the closure is taken in G G G
H ≤ c ⋯ ⊴ c H i + 1 ⊴ c H i ⊴ c ⋯ ⊴ c H 1 ⊴ c H 0 = G . H\leq_c\cdots\unlhd_c H_{i+1}\unlhd_c H_i\unlhd_c\cdots\unlhd_c H_1\unlhd_c H_0=G. H ≤ c ⋯ ⊴ c H i + 1 ⊴ c H i ⊴ c ⋯ ⊴ c H 1 ⊴ c H 0 = G . In fact, this series descends as rapidly as possible among all normal series of closed subgroups of G G G H H H ( K i ) i = 0 ∞ (K_i)_{i=0}^{\infty} ( K i ) i = 0 ∞ K 0 = G K_0=G K 0 = G K i + 1 ⊴ c K i K_{i+1}\unlhd_c K_i K i + 1 ⊴ c K i H ⊆ K i H\subseteq K_i H ⊆ K i i i i i i i H i ⊆ K i H_i\subseteq K_i H i ⊆ K i G G G H H H [15 p. 1–2] , adapted to the profinite setting by taking topological closures. Consequently, H H H G G G H H H H n = H H_n=H H n = H n ≥ 0 n\geq0 n ≥ 0 n n n H H H G G G H ≤ c G H\leq_c G H ≤ c G 2 2 2 H = H 2 = H H G ‾ ‾ H=H_2=\overline{H^{\overline{H^G}}} H = H 2 = H H G H ≤ c H G ‾ ≤ c G H\leq_c\overline{H^G}\leq_c G H ≤ c H G ≤ c G [15 ] .
A (profinite) group G G G n n n [ x , n y ] = 1 [x,{}_n y]=1 [ x , n y ] = 1 x , y ∈ G x,y\in G x , y ∈ G x ∈ G x\in G x ∈ G n n n G G G [ x , n g ] = 1 [x,{}_n g]=1 [ x , n g ] = 1 g ∈ G g\in G g ∈ G x ∈ G x\in G x ∈ G n n n [ g , n x ] = 1 [g,{}_n x]=1 [ g , n x ] = 1 g ∈ G g\in G g ∈ G
For further background on profinite groups, we refer the reader to Ribes and Zalesskii [26 ] and Wilson [38 ] . For the abstract theory of groups, see Robinson [27 ] and Rotman [30 ] .
The following commutator identities will be used throughout the remainder of this paper without explicit reference. Identities (1) and (2) are immediate; identity (3) follows from (2). Identities (4) through (7) can be found in Heineken [9 ] and Mahdavianary [18 ] . The Hall-Witt identity is a standard result (see [27 5.1.5] ).
Lemma 2.1 (Commutator identities). Let a , b , c a,b,c a , b , c G G G
[ a , b ] = 1 ⟺ [ a , b − 1 ] = 1 [a,b]=1\Longleftrightarrow[a,b^{-1}]=1 [ a , b ] = 1 ⟺ [ a , b − 1 ] = 1
b a = b [ b , a ] b^a=b[b,a] b a = b [ b , a ] a b = b a [ a , b ] ab=ba[a,b] ab = ba [ a , b ] [ a , b ] c = [ a c , b c ] [a,b]^c=[a^c,b^c] [ a , b ] c = [ a c , b c ]
[ a c − 1 , b ] = [ a , b [ b , c ] ] c − 1 [a^{c^{-1}},b]=[a,b[b,c]]^{c^{-1}} [ a c − 1 , b ] = [ a , b [ b , c ] ] c − 1
[ a b , c ] = [ a , c ] b [ b , c ] = [ a , c ] [ a , c , b ] [ b , c ] [ab,c]=[a,c]^b[b,c]=[a,c][a,c,b][b,c] [ ab , c ] = [ a , c ] b [ b , c ] = [ a , c ] [ a , c , b ] [ b , c ]
[ a , b c ] = [ a , c ] [ a , b ] c = [ a , c ] [ a , b ] [ a , b , c ] [a,bc]=[a,c][a,b]^c=[a,c][a,b][a,b,c] [ a , b c ] = [ a , c ] [ a , b ] c = [ a , c ] [ a , b ] [ a , b , c ]
[ a − 1 , b ] = [ b , a ] a − 1 [a^{-1},b]=[b,a]^{a^{-1}} [ a − 1 , b ] = [ b , a ] a − 1 [ a , b − 1 ] = [ b , a ] b − 1 [a,b^{-1}]=[b,a]^{b^{-1}} [ a , b − 1 ] = [ b , a ] b − 1
b a b = b [ b , [ b , a ] ] b^{ab}=b[b,[b,a]] b ab = b [ b , [ b , a ]]
(The Hall-Witt identity) [ a , b − 1 , c ] b ⋅ [ b , c − 1 , a ] c ⋅ [ c , a − 1 , b ] a = 1 [a,b^{-1},c]^b\cdot[b,c^{-1},a]^c\cdot[c,a^{-1},b]^a=1 [ a , b − 1 , c ] b ⋅ [ b , c − 1 , a ] c ⋅ [ c , a − 1 , b ] a = 1
Remark 2.2. Let H ≤ G H\leq G H ≤ G H G = H ⋅ [ H , G ] H^G=H\cdot[H,G] H G = H ⋅ [ H , G ] h g = h ⋅ [ h , g ] h^g=h\cdot[h,g] h g = h ⋅ [ h , g ] h ∈ H h\in H h ∈ H g ∈ G g\in G g ∈ G H H G = H H [ G , H ] = H H [ G , H ] H^{H^G}=H^{H[G,H]}=H^{H[G,H]} H H G = H H [ G , H ] = H H [ G , H ] H G H^G H G H H H [ G , H ] [G,H] [ G , H ] H H H G G G H ≤ c G H\leq_c G H ≤ c G H H H N G ( H ) N_G(H) N G ( H ) G G G [ G , H ‾ ] ⊆ N G ( H ) [G,\overline H]\subseteq N_G(H) [ G , H ] ⊆ N G ( H ) [ G ‾ , H ‾ ] ⊆ N G ( H ) [\overline G,\overline H]\subseteq N_G(H) [ G , H ] ⊆ N G ( H ) H H H [ G , H ] [G,H] [ G , H ] [ G ‾ , H ‾ ] [\overline G,\overline H] [ G , H ] H G = H ⋅ [ H , G ] = H ⋅ [ G , H ] H^G=H\cdot[H,G]=H\cdot[G,H] H G = H ⋅ [ H , G ] = H ⋅ [ G , H ] H G H^G H G H H H [ G , H ] [G,H] [ G , H ] H H H
We end this section with some basic results used throughout the article, with proofs included for convenience.
Let G G G G G G G ^ \widehat G G G G G G ^ \widehat G G H ≤ G H\leq G H ≤ G G G G H = H ‾ ∩ G H=\overline H\cap G H = H ∩ G H ‾ \overline H H H H H G ^ \widehat G G H H H G G G H H H G G G G G G G G G G G G G G G [27 Theorem 5.4.16] ), every polycyclic group is ERF, and consequently LERF.
Lemma 2.3. Let H H H G G G G ^ \widehat{G} G
H ^ ‾ = H G ‾ . \overline{\widehat{H}}=\overline{H^{G}}. H = H G . Proof. Since H ≤ H G H\leq H^{G} H ≤ H G H ‾ ≤ H G ‾ \overline{H}\leq\overline{H^{G}} H ≤ H G G ^ \widehat{G} G H G ⊴ G H^{G}\unlhd G H G ⊴ G G G G H G H^{G} H G G G G G ^ \widehat{G} G G G G H G ‾ \overline{H^{G}} H G G G G G ^ \widehat{G} G H G ‾ ⊴ c G ^ \overline{H^{G}}\unlhd_{c}\widehat{G} H G ⊴ c G y ∈ H G ‾ y\in\overline{H^{G}} y ∈ H G φ y ( x ) = x − 1 y x \varphi_y(x)=x^{-1}yx φ y ( x ) = x − 1 y x x ∈ G ^ x\in\widehat{G} x ∈ G G ^ \widehat{G} G φ y ( G ) ⊆ H G ‾ \varphi_y(G)\subseteq\overline{H^{G}} φ y ( G ) ⊆ H G φ y ( G ^ ) = φ y ( G ) ‾ ⊆ φ y ( G ) ‾ ⊆ H G ‾ \varphi_y(\widehat{G})=\overline{\varphi_y(G)}\subseteq\overline{\varphi_y(G)}\subseteq\overline{H^{G}} φ y ( G ) = φ y ( G ) ⊆ φ y ( G ) ⊆ H G H G ‾ \overline{H^{G}} H G x − 1 y x ∈ H G ‾ x^{-1}yx\in\overline{H^{G}} x − 1 y x ∈ H G x ∈ G ^ x\in\widehat{G} x ∈ G H G ‾ ⊴ c G ^ \overline{H^{G}}\unlhd_c\widehat{G} H G ⊴ c G
Since H ≤ H ‾ ≤ H G ‾ H\leq\overline{H}\leq\overline{H^{G}} H ≤ H ≤ H G H G ‾ ⊴ c G ^ \overline{H^{G}}\unlhd_c\widehat{G} H G ⊴ c G H G ^ ‾ ≤ H G ‾ \overline{\widehat{H^{G}}}\leq\overline{H^{G}} H G ≤ H G
On the other hand, H G H^{G} H G g − 1 h g g^{-1}hg g − 1 h g g ∈ G g\in G g ∈ G h ∈ H h\in H h ∈ H H ^ ‾ \overline{\widehat{H}} H H G ≤ H ^ ‾ H^{G}\leq\overline{\widehat{H}} H G ≤ H H ^ ‾ \overline{\widehat{H}} H H G ‾ \overline{H^{G}} H G H G H^{G} H G H G ‾ ≤ H ^ ‾ \overline{H^{G}}\leq\overline{\widehat{H}} H G ≤ H
Lemma 2.4. Let G G G H H H G G G
H G ‾ ∩ G = H G = H G ^ ‾ ∩ G . \overline{H^{G}}\cap G=H^{G}=\overline{\widehat{H^{G}}}\cap G. H G ∩ G = H G = H G ∩ G . Proof. Since H G ≤ G H^{G}\leq G H G ≤ G G G G H G H^{G} H G G G G H G ‾ ∩ G = H G \overline{H^{G}}\cap G=H^{G} H G ∩ G = H G H G ‾ = H G ^ ‾ \overline{H^{G}}=\overline{\widehat{H^{G}}} H G = H G H G ^ ‾ ∩ G = H G \overline{\widehat{H^{G}}}\cap G=H^{G} H G ∩ G = H G H G ^ ‾ ⊆ H ^ ‾ \overline{\widehat{H^{G}}}\subseteq\overline{\widehat{H}} H G ⊆ H H G ^ ‾ ∩ G ⊆ H ^ ‾ ∩ G \overline{\widehat{H^{G}}}\cap G\subseteq\overline{\widehat{H}}\cap G H G ∩ G ⊆ H ∩ G H ^ ‾ ∩ G = H G \overline{\widehat{H}}\cap G=H^{G} H ∩ G = H G
Let G G G ( Z i ( G ) ) i ≥ 0 (Z_i(G))_{i\geq0} ( Z i ( G ) ) i ≥ 0 ( γ n ( G ) ) n ≥ 1 (\gamma_n(G))_{n\geq1} ( γ n ( G ) ) n ≥ 1 G G G
Z 0 ( G ) = { 1 } , Z i + 1 ( G ) / Z i ( G ) = Z ( G / Z i ( G ) ) ( i ≥ 0 ) , Z_0(G)=\{1\},\qquad Z_{i+1}(G)/Z_i(G)=Z(G/Z_i(G))\quad(i\geq0), Z 0 ( G ) = { 1 } , Z i + 1 ( G ) / Z i ( G ) = Z ( G / Z i ( G )) ( i ≥ 0 ) , γ 1 ( G ) = G , γ n + 1 ( G ) = [ γ n ( G ) , G ] ‾ ( n ≥ 1 ) , \gamma_1(G)=G,\qquad\gamma_{n+1}(G)=\overline{[\gamma_n(G),G]}\quad(n\geq1), γ 1 ( G ) = G , γ n + 1 ( G ) = [ γ n ( G ) , G ] ( n ≥ 1 ) , where the bar denotes the closure in G G G G G G r r r γ r + 1 ( G ) = { 1 } \gamma_{r+1}(G)=\{1\} γ r + 1 ( G ) = { 1 } Z r ( G ) = G Z_r(G)=G Z r ( G ) = G γ n a b s ( G ) \gamma_n^{\mathrm{abs}}(G) γ n abs ( G ) γ 1 a b s ( G ) = G \gamma_1^{\mathrm{abs}}(G)=G γ 1 abs ( G ) = G γ n + 1 a b s ( G ) = [ γ n a b s ( G ) , G ] \gamma_{n+1}^{\mathrm{abs}}(G)=[\gamma_n^{\mathrm{abs}}(G),G] γ n + 1 abs ( G ) = [ γ n abs ( G ) , G ] γ n a b s ( G ) ≤ γ n ( G ) \gamma_n^{\mathrm{abs}}(G)\leq\gamma_n(G) γ n abs ( G ) ≤ γ n ( G ) N r \mathfrak{N}_r N r r r r N r \mathcal{N}_r N r
The next proposition is a well-known result about subnormality, the abstract version of which can be found in Hall [7 p. 154] . Although the proof for the profinite case is analogous, we include it here for completeness, as a direct reference seems unavailable.
Proposition 2.5. If G G G k k k G G G
Proof. Let H ≤ c G H \leq_c G H ≤ c G G G G
{ 1 } = Z 0 ( G ) ⊴ c Z 1 ( G ) = Z ( G ) ⊴ c ⋯ ⊴ c Z k − 1 ( G ) ⊴ c Z k ( G ) = G , \{1\}=Z_0(G)\trianglelefteq_c Z_1(G)=Z(G)\trianglelefteq_c\cdots\trianglelefteq_c Z_{k-1}(G)\trianglelefteq_c Z_k(G)=G, { 1 } = Z 0 ( G ) ⊴ c Z 1 ( G ) = Z ( G ) ⊴ c ⋯ ⊴ c Z k − 1 ( G ) ⊴ c Z k ( G ) = G , where Z i ( G ) / Z i − 1 ( G ) = Z ( G / Z i − 1 ( G ) ) Z_i(G)/Z_{i-1}(G)=Z(G/Z_{i-1}(G)) Z i ( G ) / Z i − 1 ( G ) = Z ( G / Z i − 1 ( G )) i = 1 , 2 , … , k i=1,2,\ldots,k i = 1 , 2 , … , k
For any j ∈ { 0 , 1 , … , k − 1 } j\in\{0,1,\ldots,k-1\} j ∈ { 0 , 1 , … , k − 1 } [ Z j + 1 ( G ) , G ] ≤ Z j ( G ) [Z_{j+1}(G),G]\leq Z_j(G) [ Z j + 1 ( G ) , G ] ≤ Z j ( G ) Z j ( G ) Z_j(G) Z j ( G ) [ Z j + 1 ( G ) , G ] ≤ c Z j ( G ) [Z_{j+1}(G),G]\leq_c Z_j(G) [ Z j + 1 ( G ) , G ] ≤ c Z j ( G ) Z j + 1 ( G ) Z_{j+1}(G) Z j + 1 ( G ) H Z j ( G ) HZ_j(G) H Z j ( G ) H H H H Z j ( G ) HZ_j(G) H Z j ( G ) H Z j + 1 ( G ) HZ_{j+1}(G) H Z j + 1 ( G ) H Z j ( G ) HZ_j(G) H Z j ( G ) H Z j ( G ) HZ_j(G) H Z j ( G )
H = H Z 0 ( G ) ⊴ c H Z 1 ( G ) ⊴ c ⋯ ⊴ c H Z k − 1 ( G ) ⊴ c H Z k ( G ) = G . H=HZ_0(G)\trianglelefteq_c HZ_1(G)\trianglelefteq_c\cdots\trianglelefteq_c HZ_{k-1}(G)\trianglelefteq_c HZ_k(G)=G. H = H Z 0 ( G ) ⊴ c H Z 1 ( G ) ⊴ c ⋯ ⊴ c H Z k − 1 ( G ) ⊴ c H Z k ( G ) = G . As a consequence, H H H G G G k k k H H H G G G k k k
Some classes of profinite groups
In this section, we introduce specific classes of profinite groups that generalize the abstract group classes investigated by Mahdavianary [20 ] . Thereafter, we prove that non-periodic profinite groups without even-order elements in which all procyclic subgroups are subnormal of defect at most 2 are precisely profinite 2-Engel groups. This establishes a profinite generalization of Heineken’s result [9 Theorem 1(I)] .
For a profinite group G G G
(i) G ∈ T ^ G\in\widehat{\mathcal{T}} G ∈ T G G G
(ii) G ∈ A ^ G\in\widehat{\mathcal{A}} G ∈ A G G G
(iii) G ∈ S ^ G\in\widehat{\mathcal{S}} G ∈ S G G G
These definitions naturally yield the inclusions S ^ ⊆ A ^ ⊆ T ^ \widehat{\mathcal{S}}\subseteq\widehat{\mathcal{A}}\subseteq\widehat{\mathcal{T}} S ⊆ A ⊆ T [20 ] defined the corresponding classes S \mathcal{S} S A \mathcal{A} A T \mathcal{T} T G G G G ∈ S ^ G\in\widehat{\mathcal{S}} G ∈ S G ∈ S G\in\mathcal{S} G ∈ S A \mathcal{A} A T \mathcal{T} T
The abstract case of the following definition was introduced by Cappitt [2 ] .
Definition 3.1. A profinite Cappitt group is a profinite group G G G S ~ ( G ) = ⟨ L ≤ c G ∣ L ⋬ G ⟩ \widetilde{S}(G)=\langle L\leq_c G\mid L\ntrianglelefteq G\rangle S ( G ) = ⟨ L ≤ c G ∣ L ⋬ G ⟩ G G G
A direct application of Proposition 2.5 yields the following corollary. It shows that (profinite) Cappitt groups belong to S \mathcal{S} S S ^ \widehat{\mathcal{S}} S
Corollary 3.2. If G G G G G G
Proof. If G G G G G G [23 ] and [2 ] ). Therefore, by Proposition 2.5 , every (closed) subgroup of G G G
Profinite groups in T ^ \widehat{\mathcal T} T A ^ \widehat{\mathcal A} A
We begin with a series of auxiliary lemmas that govern the structural behavior of these classes.
Lemma 3.3. If G ∈ T ^ G \in\widehat{\mathcal T} G ∈ T G / U ∈ T G/U \in\mathcal T G / U ∈ T U ⊴ o G U \trianglelefteq_{\mathrm{o}} G U ⊴ o G
Proof. Suppose U ⊴ o G U \trianglelefteq_{\mathrm{o}} G U ⊴ o G L U = ⟨ g U ⟩ ≤ G / U L_U = \langle gU\rangle\leq G/U L U = ⟨ g U ⟩ ≤ G / U g ∈ G g \in G g ∈ G G ∈ T ^ G \in\widehat{\mathcal T} G ∈ T ⟨ g ⟩ ‾ ⊴ c ⟨ g ⟩ ‾ \overline{\langle g\rangle} \trianglelefteq_{\mathrm{c}} \overline{\langle g\rangle} ⟨ g ⟩ ⊴ c ⟨ g ⟩ ⟨ g ⟩ ‾ U ⊴ c ⟨ g ⟩ ‾ G U \overline{\langle g\rangle}U \trianglelefteq_{\mathrm{c}} \overline{\langle g\rangle}^{G}U ⟨ g ⟩ U ⊴ c ⟨ g ⟩ G U π U : G → G / U \pi_U:G \to G/U π U : G → G / U G / U G/U G / U G / U G/U G / U π U \pi_U π U
L U = ⟨ g ⟩ ‾ U U = ⟨ g U ⟩ ⊴ ⟨ g ⟩ ‾ G U U = π U ( ⟨ g ⟩ ‾ ) = π U ( ⟨ g ⟩ ‾ G ) G / U ⊴ ⟨ g U ⟩ G / U = L U G / U . L_U = \frac{\overline{\langle g\rangle}U}{U} = \langle gU\rangle\trianglelefteq\frac{\overline{\langle g\rangle}^{G}U}{U} = \pi_U\left(\overline{\langle g\rangle}\right) = \pi_U\left(\overline{\langle g\rangle}^{G}\right)^{G/U} \trianglelefteq\langle gU\rangle^{G/U} = L_U^{G/U}. L U = U ⟨ g ⟩ U = ⟨ g U ⟩ ⊴ U ⟨ g ⟩ G U = π U ( ⟨ g ⟩ ) = π U ( ⟨ g ⟩ G ) G / U ⊴ ⟨ g U ⟩ G / U = L U G / U . This proves that every cyclic subgroup of G / U G/U G / U G / U ∈ T G/U \in\mathcal T G / U ∈ T
Lemma 3.4. If G ∈ T ^ G \in\widehat{\mathcal T} G ∈ T G ∈ N ^ 3 G \in\widehat{\mathcal N}_3 G ∈ N 3 T ^ \widehat{\mathcal T} T
Proof. Let N = { U ∣ U ⊴ o G } \mathcal N = \{U \mid U \trianglelefteq_{\mathrm{o}} G\} N = { U ∣ U ⊴ o G } G G G [26 Theorem 2.1.3] that G = l i m ← U ∈ N G / U G = \varprojlim_{U \in\mathcal N} G/U G = lim U ∈ N G / U [38 Proposition 1.1.5] , G G G ∏ U ∈ N G / U \prod_{U \in\mathcal N} G/U ∏ U ∈ N G / U G ∈ T ^ G \in\widehat{\mathcal T} G ∈ T Lemma 3.3 ensures that G / U ∈ T G/U \in\mathcal T G / U ∈ T U ∈ N U \in\mathcal N U ∈ N [6 Theorem 1.3] (or Mahdavianary [18 Theorem 1] ), any abstract group in T \mathcal T T N 3 \mathcal N_3 N 3 G / U ∈ N 3 G/U \in\mathcal N_3 G / U ∈ N 3 U ∈ N U \in\mathcal N U ∈ N [17 ] , Proposition 4.2 of [17 ] ), we conclude that G ∈ N ^ 3 G \in\widehat{\mathcal N}_3 G ∈ N 3 G ′ ′ = [ γ 2 ( G ) , γ 2 ( G ) ] ≤ γ 4 ( G ) = { 1 } G^{\prime\prime} = [\gamma_2(G),\gamma_2(G)] \leq\gamma_4(G) = \{1\} G ′′ = [ γ 2 ( G ) , γ 2 ( G )] ≤ γ 4 ( G ) = { 1 } G G G
Corollary 3.5. The following inclusions hold: S ^ ⊆ A ^ ⊆ N ^ 3 \widehat{\mathcal S} \subseteq\widehat{\mathcal A} \subseteq\widehat{\mathcal N}_3 S ⊆ A ⊆ N 3
Proof. By definition, S ^ ⊆ A ^ ⊆ T ^ \widehat{\mathcal S} \subseteq\widehat{\mathcal A} \subseteq\widehat{\mathcal T} S ⊆ A ⊆ T Lemma 3.4 , T ^ ⊆ N ^ 3 \widehat{\mathcal T} \subseteq\widehat{\mathcal N}_3 T ⊆ N 3
Corollary 3.6. If G ∈ T ^ G \in\widehat{\mathcal T} G ∈ T G G G
Proof. By Lemma 3.4 , γ 4 ( G ) = 1 \gamma_4(G) = 1 γ 4 ( G ) = 1 [ v , 3 u ] = [ v , u , u , u ] = 1 [v,{}_3u] = [v,u,u,u] = 1 [ v , 3 u ] = [ v , u , u , u ] = 1 u , v ∈ G u,v \in G u , v ∈ G
The identity in the following lemma corresponds to Heineken [10 Lemma 1] , adapted to standard modern notation.
Lemma 3.7. If G G G [ a , [ b , a ] , [ b , a ] ] = 1 [a,[b,a],[b,a]]=1 [ a , [ b , a ] , [ b , a ]] = 1 a , b ∈ G a,b\in G a , b ∈ G
Proof. First, we expand the commutator as follows:
[ a , [ b , a ] , [ b , a ] ] = [ a , [ b , a ] , ( a − 1 ) b a ] = [ a , [ b , a ] , a ] ⋅ [ a , [ b , a ] , ( a − 1 ) b a ] . (1) [a,[b,a],[b,a]]=[a,[b,a],(a^{-1})^ba]=[a,[b,a],a]\cdot[a,[b,a],(a^{-1})^{b}a]. \tag*{(1)} [ a , [ b , a ] , [ b , a ]] = [ a , [ b , a ] , ( a − 1 ) b a ] = [ a , [ b , a ] , a ] ⋅ [ a , [ b , a ] , ( a − 1 ) b a ] . (1) Since G G G [ b , 3 a ] = [ [ [ b , a ] , a ] , a ] = 1 [b,{}_3a]=[[[b,a],a],a]=1 [ b , 3 a ] = [[[ b , a ] , a ] , a ] = 1 Lemma 2.1 (1), we find:
[ b , a , a , a ] = 1 ⟺ [ [ a , [ b , a ] ] , a ] = 1 ⟺ [ [ a , [ b , a ] ] , a − 1 ] = 1. (2) [b,a,a,a]=1\Longleftrightarrow[[a,[b,a]],a]=1\Longleftrightarrow[[a,[b,a]],a^{-1}]=1. \tag*{(2)} [ b , a , a , a ] = 1 ⟺ [[ a , [ b , a ]] , a ] = 1 ⟺ [[ a , [ b , a ]] , a − 1 ] = 1. (2) Therefore, the first term on the right-hand side of (1) vanishes, and it suffices to prove that [ a , [ b , a ] , ( a − 1 ) b a ] = 1 [a,[b,a],(a^{-1})^ba]=1 [ a , [ b , a ] , ( a − 1 ) b a ] = 1
Let x , y ∈ G x,y\in G x , y ∈ G
1 = [ [ [ y − 1 , x ] , x ] , x ] = [ [ ( x − 1 ) y − 1 x , x ] , x ] = [ [ ( x − 1 ) y − 1 , x ] x [ x , x ] , x ] = [ [ ( x − 1 ) y − 1 , x ] x , x ] . 1=[[[y^{-1},x],x],x]=[[(x^{-1})^{y^{-1}}x,x],x]=[[(x^{-1})^{y^{-1}},x]^x[x,x],x]=[[(x^{-1})^{y^{-1}},x]^x,x]. 1 = [[[ y − 1 , x ] , x ] , x ] = [[( x − 1 ) y − 1 x , x ] , x ] = [[( x − 1 ) y − 1 , x ] x [ x , x ] , x ] = [[( x − 1 ) y − 1 , x ] x , x ] . This simplifies to:
[ [ [ ( x − 1 ) y − 1 , x ] , x ] ] x = 1 ⟺ [ [ [ ( x − 1 ) y − 1 , x ] , x ] ] = 1. (3) [[[(x^{-1})^{y^{-1}},x],x]]^x=1\Longleftrightarrow[[[(x^{-1})^{y^{-1}},x],x]]=1. \tag*{(3)} [[[( x − 1 ) y − 1 , x ] , x ] ] x = 1 ⟺ [[[( x − 1 ) y − 1 , x ] , x ]] = 1. (3) By substituting x = ( a − 1 ) b x=(a^{-1})^b x = ( a − 1 ) b y = b y=b y = b (3) (so that ( x − 1 ) y − 1 = a (x^{-1})^{y^{-1}}=a ( x − 1 ) y − 1 = a
1 = [ a , ( a − 1 ) b , ( a − 1 ) b ] . 1=[a,(a^{-1})^b,(a^{-1})^b]. 1 = [ a , ( a − 1 ) b , ( a − 1 ) b ] . Expanding this applying Lemma 2.1 (2) and Lemma 2.1 (5), we arrive at:
1 = [ a , ( a − 1 ) b , ( a − 1 ) b ] = [ [ a , [ b , a ] a − 1 ] , ( a − 1 ) b ] = [ [ a , a − 1 ] ⋅ [ a , [ b , a ] ] ⋅ [ a , [ b , a ] , a − 1 ] , ( a − 1 ) b ] = [ [ a , [ b , a ] ] ⋅ [ a , [ b , a ] , a − 1 ] , ( a − 1 ) b ] . (4) \begin{aligned}
1&=[a,(a^{-1})^b,(a^{-1})^b]=[[a,[b,a]a^{-1}],(a^{-1})^b]\\
&=[[a,a^{-1}]\cdot[a,[b,a]]\cdot[a,[b,a],a^{-1}],(a^{-1})^b]\\
&=[[a,[b,a]]\cdot[a,[b,a],a^{-1}],(a^{-1})^b]. \tag*{(4)}
\end{aligned} 1 = [ a , ( a − 1 ) b , ( a − 1 ) b ] = [[ a , [ b , a ] a − 1 ] , ( a − 1 ) b ] = [[ a , a − 1 ] ⋅ [ a , [ b , a ]] ⋅ [ a , [ b , a ] , a − 1 ] , ( a − 1 ) b ] = [[ a , [ b , a ]] ⋅ [ a , [ b , a ] , a − 1 ] , ( a − 1 ) b ] . (4) Substituting the result of (2) into (4) yields
1 = [ a , [ b , a ] , ( a − 1 ) b ] , 1=[a,[b,a],(a^{-1})^b], 1 = [ a , [ b , a ] , ( a − 1 ) b ] , which completes the proof. □
The following result is the profinite adaptation of Lemma 1 in [9 ] .
Lemma 3.8. Let G G G G ∈ T ^ G\in\widehat{\mathcal T} G ∈ T [ v , u , u ] ∈ ⟨ u ⟩ ‾ [v,u,u]\in\overline{\langle u\rangle} [ v , u , u ] ∈ ⟨ u ⟩ u , v ∈ G u,v\in G u , v ∈ G
Proof. For any u , v ∈ G u,v\in G u , v ∈ G
u v = u [ u , [ u , v ] ] ⟺ ( u v ) − 1 u = [ u , [ u , v ] ] − 1 = [ u , v , u ] . (5) u^v=u[u,[u,v]]\Longleftrightarrow(u^v)^{-1}u=[u,[u,v]]^{-1}=[u,v,u]. \tag*{(5)} u v = u [ u , [ u , v ]] ⟺ ( u v ) − 1 u = [ u , [ u , v ] ] − 1 = [ u , v , u ] . (5) Suppose G ∈ T ^ G\in\widehat{\mathcal T} G ∈ T ⟨ u ⟩ ‾ ⊴ c ⟨ u ⟩ G ‾ \overline{\langle u\rangle}\trianglelefteq_c\overline{\langle u\rangle^G} ⟨ u ⟩ ⊴ c ⟨ u ⟩ G u ∈ G u\in G u ∈ G (5) and Lemma 2.1 (6), we obtain
( [ v , u , u ] − 1 ) [ u , v ] = [ u , [ v , u ] ] [ u , v ] = [ u , [ v , u ] ] [ v , u ] − 1 = [ u , v , u ] = ( u u v ) − 1 u . ([v,u,u]^{-1})^{[u,v]}=[u,[v,u]]^{[u,v]}=[u,[v,u]]^{[v,u]^{-1}}=[u,v,u]=(u^{uv})^{-1}u. ([ v , u , u ] − 1 ) [ u , v ] = [ u , [ v , u ] ] [ u , v ] = [ u , [ v , u ] ] [ v , u ] − 1 = [ u , v , u ] = ( u uv ) − 1 u . Since u u g ∈ ⟨ u ⟩ ‾ u^{ug}\in\overline{\langle u\rangle} u ug ∈ ⟨ u ⟩ g ∈ G g\in G g ∈ G
( [ v , u , u ] − 1 ) [ u , v ] = [ u , v , u ] = ( u u v ) − 1 u ∈ ⟨ u ⟩ ‾ . ([v,u,u]^{-1})^{[u,v]}=[u,v,u]=(u^{uv})^{-1}u\in\overline{\langle u\rangle}. ([ v , u , u ] − 1 ) [ u , v ] = [ u , v , u ] = ( u uv ) − 1 u ∈ ⟨ u ⟩ . Every conjugate u g u^g u g u u u N G ( ⟨ u ⟩ ‾ ) N_G(\overline{\langle u\rangle}) N G ( ⟨ u ⟩ ) ⟨ u ⟩ G \langle u\rangle^G ⟨ u ⟩ G N G ( ⟨ u ⟩ ‾ ) N_G(\overline{\langle u\rangle}) N G ( ⟨ u ⟩ ) ⟨ u ⟩ G ⊆ N G ( ⟨ u ⟩ ‾ ) \langle u\rangle^G \subseteq N_G(\overline{\langle u\rangle}) ⟨ u ⟩ G ⊆ N G ( ⟨ u ⟩ ) N G ( ⟨ u ⟩ ‾ ) N_G(\overline{\langle u\rangle}) N G ( ⟨ u ⟩ ) ⟨ u ⟩ G ‾ \overline{\langle u\rangle^G} ⟨ u ⟩ G ⟨ u ⟩ ‾ ⊆ ⟨ u ⟩ G ‾ \overline{\langle u\rangle} \subseteq\overline{\langle u\rangle^G} ⟨ u ⟩ ⊆ ⟨ u ⟩ G ⟨ u ⟩ ‾ ⊆ N G ( ⟨ u ⟩ ‾ ) \overline{\langle u\rangle} \subseteq N_G(\overline{\langle u\rangle}) ⟨ u ⟩ ⊆ N G ( ⟨ u ⟩ ) [ v , u ] = ( u − 1 ) v u ∈ ⟨ u ⟩ G [v,u]=(u^{-1})^v u\in\langle u\rangle^G [ v , u ] = ( u − 1 ) v u ∈ ⟨ u ⟩ G [ v , u ] [v,u] [ v , u ] ⟨ u ⟩ ‾ \overline{\langle u\rangle} ⟨ u ⟩ ( [ v , u , u ] − 1 ) ∈ ⟨ u ⟩ ‾ ([v,u,u]^{-1})\in\overline{\langle u\rangle} ([ v , u , u ] − 1 ) ∈ ⟨ u ⟩ [ v , u ] = [ u , v ] − 1 [v,u]=[u,v]^{-1} [ v , u ] = [ u , v ] − 1 [ v , u , u ] − 1 ∈ ⟨ u ⟩ ‾ [v,u,u]^{-1}\in\overline{\langle u\rangle} [ v , u , u ] − 1 ∈ ⟨ u ⟩ [ v , u , u ] ∈ ⟨ u ⟩ ‾ [v,u,u]\in\overline{\langle u\rangle} [ v , u , u ] ∈ ⟨ u ⟩
Conversely, assume that [ v , u , u ] ∈ ⟨ u ⟩ ‾ [v,u,u]\in\overline{\langle u\rangle} [ v , u , u ] ∈ ⟨ u ⟩ u , v ∈ G u,v\in G u , v ∈ G Lemma 2.1 ), we derive:
( [ u , v , u ] − 1 ) u − 2 = [ u , [ u , v ] ] u − 2 = ( [ u , [ u , v ] ] u − 1 ) u − 1 = [ [ u , v ] , u − 1 ] u − 1 = [ [ u , v ] u − 1 , ( u − 1 ) u − 1 ] = [ [ v , u − 1 ] , u − 1 ] = [ v , u − 1 , u − 1 ] ∈ ⟨ u − 1 ⟩ = ⟨ u ⟩ . (6) \begin{aligned}
([u,v,u]^{-1})^{u^{-2}} &= [u,[u,v]]^{u^{-2}} = \left([u,[u,v]]^{u^{-1}}\right)^{u^{-1}} = [[u,v],u^{-1}]^{u^{-1}} \\
&= [[u,v]^{u^{-1}},(u^{-1})^{u^{-1}}] = [[v,u^{-1}],u^{-1}] \\
&= [v,u^{-1},u^{-1}] \in\langle u^{-1}\rangle=\langle u\rangle. \tag*{(6)}
\end{aligned} ([ u , v , u ] − 1 ) u − 2 = [ u , [ u , v ] ] u − 2 = ( [ u , [ u , v ] ] u − 1 ) u − 1 = [[ u , v ] , u − 1 ] u − 1 = [[ u , v ] u − 1 , ( u − 1 ) u − 1 ] = [[ v , u − 1 ] , u − 1 ] = [ v , u − 1 , u − 1 ] ∈ ⟨ u − 1 ⟩ = ⟨ u ⟩ . (6) Conjugating (6) by u 2 u^2 u 2
[ u , [ u , v ] ] = [ u , v , u ] − 1 = [ v , u − 1 , u − 1 ] u 2 ∈ ⟨ u ⟩ ‾ . [u,[u,v]]=[u,v,u]^{-1}=[v,u^{-1},u^{-1}]^{u^2}\in\overline{\langle u\rangle}. [ u , [ u , v ]] = [ u , v , u ] − 1 = [ v , u − 1 , u − 1 ] u 2 ∈ ⟨ u ⟩ . Finally, observing that u ⋅ [ u , [ u , v ] ] = u u v u\cdot[u,[u,v]]=u^{u^v} u ⋅ [ u , [ u , v ]] = u u v v ∈ G v\in G v ∈ G u u v ∈ ⟨ u ⟩ ‾ u^{u^v}\in\overline{\langle u\rangle} u u v ∈ ⟨ u ⟩ u u u N G ( ⟨ u ⟩ ‾ ) N_G(\overline{\langle u\rangle}) N G ( ⟨ u ⟩ ) ⟨ u ⟩ G \langle u\rangle^G ⟨ u ⟩ G ⟨ u ⟩ ‾ ⊴ ⟨ u ⟩ G \overline{\langle u\rangle}\trianglelefteq\langle u\rangle^G ⟨ u ⟩ ⊴ ⟨ u ⟩ G N G ( ⟨ u ⟩ ‾ ) N_G(\overline{\langle u\rangle}) N G ( ⟨ u ⟩ ) ⟨ u ⟩ G ‾ ⊆ N G ( ⟨ u ⟩ ‾ ) \overline{\langle u\rangle^G}\subseteq N_G(\overline{\langle u\rangle}) ⟨ u ⟩ G ⊆ N G ( ⟨ u ⟩ ) ⟨ u ⟩ ‾ ⊴ ⟨ u ⟩ G ‾ \overline{\langle u\rangle}\trianglelefteq\overline{\langle u\rangle^G} ⟨ u ⟩ ⊴ ⟨ u ⟩ G ⟨ u ⟩ G ‾ = ⟨ u ⟩ ‾ G \overline{\langle u\rangle^G}=\overline{\langle u\rangle}^G ⟨ u ⟩ G = ⟨ u ⟩ G ⟨ u ⟩ ‾ ⊴ c ⟨ u ⟩ ‾ G \overline{\langle u\rangle}\trianglelefteq_c\overline{\langle u\rangle}^G ⟨ u ⟩ ⊴ c ⟨ u ⟩ G G ∈ T ^ G\in\widehat{\mathcal T} G ∈ T
Remark 3.9. Let G G G a ∈ G a\in G a ∈ G Z ^ \widehat{\mathbb Z} Z { 1 } \{1\} { 1 } ζ : Z ^ → ⟨ a ⟩ ‾ \zeta:\widehat{\mathbb Z}\to\overline{\langle a\rangle} ζ : Z → ⟨ a ⟩ ζ ( 1 ) = a \zeta(1)=a ζ ( 1 ) = a [26 Section 4.1] ). Since ζ \zeta ζ ⟨ a ⟩ ‾ \overline{\langle a\rangle} ⟨ a ⟩ a λ = ζ ( λ ) a^\lambda=\zeta(\lambda) a λ = ζ ( λ ) λ ∈ Z ^ \lambda\in\widehat{\mathbb Z} λ ∈ Z
The following lemma adapts Heineken [10 Lemma 2] to the profinite framework.
Lemma 3.10. If G ∈ T ^ G\in\widehat{\mathcal T} G ∈ T [ b , a , a λ ] = [ b , a , a ] λ [b,a,a^\lambda]=[b,a,a]^\lambda [ b , a , a λ ] = [ b , a , a ] λ a , b ∈ G a,b\in G a , b ∈ G λ ∈ Z ^ \lambda\in\widehat{\mathbb Z} λ ∈ Z
Proof. By Corollary 3.6, G G G [ b , a , a ] = 1 [b,a,a]=1 [ b , a , a ] = 1 [ b , a ] ∈ C G ( ⟨ a ⟩ ‾ ) [b,a]\in C_G(\overline{\langle a\rangle}) [ b , a ] ∈ C G ( ⟨ a ⟩ ) [ b , a , a ] ≠ 1 [b,a,a]\ne1 [ b , a , a ] = 1 [ b , a , a ] ∈ ⟨ a ⟩ ‾ [b,a,a]\in\overline{\langle a\rangle} [ b , a , a ] ∈ ⟨ a ⟩ Lemma 3.8 , we may write [ b , a , a ] = a μ [b,a,a]=a^\mu [ b , a , a ] = a μ μ ∈ Z ^ \mu\in\widehat{\mathbb Z} μ ∈ Z ker ( ζ ) \ker(\zeta) ker ( ζ )
By induction on n ∈ N n\in\mathbb N n ∈ N [ b , a , a n ] = [ b , a , a ] n [b,a,a^n]=[b,a,a]^n [ b , a , a n ] = [ b , a , a ] n a , b ∈ G a,b\in G a , b ∈ G n = − m n=-m n = − m m > 0 m>0 m > 0 [ b , a , a ] [b,a,a] [ b , a , a ] C G ( ⟨ a ⟩ ‾ ) C_G(\overline{\langle a\rangle}) C G ( ⟨ a ⟩ )
[ b , a , a − m ] = a m ⋅ a − m ⋅ ( [ b , a ] − 1 a m [ b , a ] a − m ) = a m ( [ b , a m ] ) − 1 a − m = a m [ b , a , a ] − m a − m = ( [ b , a , a ] a − 1 ) − m = [ b , a , a ] − m . \begin{aligned}
[b,a,a^{-m}] &= a^m\cdot a^{-m}\cdot\left([b,a]^{-1}a^m[b,a]a^{-m}\right) \\
&=a^m([b,a^m])^{-1}a^{-m} \\
&=a^m[b,a,a]^{-m}a^{-m}=([b,a,a]a^{-1})^{-m}=[b,a,a]^{-m}.
\end{aligned} [ b , a , a − m ] = a m ⋅ a − m ⋅ ( [ b , a ] − 1 a m [ b , a ] a − m ) = a m ([ b , a m ] ) − 1 a − m = a m [ b , a , a ] − m a − m = ([ b , a , a ] a − 1 ) − m = [ b , a , a ] − m . Thus, the property holds for all n ∈ Z n\in\mathbb Z n ∈ Z
Take λ ∈ Z ^ \lambda\in\widehat{\mathbb Z} λ ∈ Z Z ^ = Z \widehat{\mathbb Z}=\mathbb Z Z = Z ( n i ) i ∈ N (n_i)_{i\in\mathbb N} ( n i ) i ∈ N λ \lambda λ Z ^ \widehat{\mathbb Z} Z ζ \zeta ζ a n i → a λ a^{n_i}\to a^\lambda a n i → a λ [26 Lemma 4.1.1] ) in ⟨ a ⟩ ‾ \overline{\langle a\rangle} ⟨ a ⟩ i i i
[ b , a , a ] n i = ( a μ ) n i = a μ n i . (7) [b,a,a]^{n_i}=(a^\mu)^{n_i}=a^{\mu n_i}. \tag*{(7)} [ b , a , a ] n i = ( a μ ) n i = a μ n i . (7) Since n i → λ n_i \to\lambda n i → λ Z ^ \widehat{\mathbb{Z}} Z Z ^ \widehat{\mathbb{Z}} Z μ n i → μ λ \mu n_i \to\mu\lambda μ n i → μ λ Z ^ \widehat{\mathbb{Z}} Z a a a
[ b , a , a ] n i = a μ n i → a μ λ = ( a μ ) λ = [ b , a , a ] λ in ⟨ a ⟩ ‾ . (8) [b,a,a]^{n_i}=a^{\mu n_i}\to a^{\mu\lambda}=(a^\mu)^\lambda=[b,a,a]^\lambda\quad\text{in }\overline{\langle a\rangle}. \tag*{(8)} [ b , a , a ] n i = a μ n i → a μ λ = ( a μ ) λ = [ b , a , a ] λ in ⟨ a ⟩ . (8) On the other hand, from [ b , a , a ] = a μ [b,a,a]=a^\mu [ b , a , a ] = a μ ( a − 1 ) [ b , a ] = a μ − 1 ∈ ⟨ a ⟩ ‾ (a^{-1})^{[b,a]}=a^{\mu-1}\in\overline{\langle a\rangle} ( a − 1 ) [ b , a ] = a μ − 1 ∈ ⟨ a ⟩ [ b , a ] [b,a] [ b , a ] ⟨ a ⟩ ‾ \overline{\langle a\rangle} ⟨ a ⟩ τ [ b , a ] \tau_{[b,a]} τ [ b , a ] G G G
τ [ b , a ] ( x ) = [ b , a ] − 1 x [ b , a ] , x ∈ G , \tau_{[b,a]}(x)=[b,a]^{-1}x[b,a],\quad x\in G, τ [ b , a ] ( x ) = [ b , a ] − 1 x [ b , a ] , x ∈ G , restricts to a continuous automorphism of ⟨ a ⟩ ‾ \overline{\langle a\rangle} ⟨ a ⟩ τ [ b , a ] \tau_{[b,a]} τ [ b , a ] ⟨ a ⟩ ‾ \overline{\langle a\rangle} ⟨ a ⟩ a n i → a λ a^{n_i}\to a^\lambda a n i → a λ
τ [ b , a ] ( a n i ) → τ [ b , a ] ( a λ ) in ⟨ a ⟩ ‾ . \tau_{[b,a]}(a^{n_i})\to\tau_{[b,a]}(a^\lambda)\quad\text{in }\overline{\langle a\rangle}. τ [ b , a ] ( a n i ) → τ [ b , a ] ( a λ ) in ⟨ a ⟩ . Using the continuity of inversion and multiplication in ⟨ a ⟩ ‾ \overline{\langle a\rangle} ⟨ a ⟩
[ b , a , a n i ] = [ b , a ] − 1 ( a n i ) − 1 [ b , a ] a n i → [ b , a ] − 1 ( a λ ) − 1 [ b , a ] a λ = [ b , a , a λ ] in ⟨ a ⟩ ‾ . (9) [b,a,a^{n_i}]=[b,a]^{-1}(a^{n_i})^{-1}[b,a]a^{n_i}\to[b,a]^{-1}(a^\lambda)^{-1}[b,a]a^\lambda=[b,a,a^\lambda]\quad\text{in }\overline{\langle a\rangle}. \tag*{(9)} [ b , a , a n i ] = [ b , a ] − 1 ( a n i ) − 1 [ b , a ] a n i → [ b , a ] − 1 ( a λ ) − 1 [ b , a ] a λ = [ b , a , a λ ] in ⟨ a ⟩ . (9) Since [ b , a , a ] n i = [ b , a , a n i ] [b,a,a]^{n_i}=[b,a,a^{n_i}] [ b , a , a ] n i = [ b , a , a n i ] i i i ⟨ a ⟩ ‾ \overline{\langle a\rangle} ⟨ a ⟩ (8) and (9) must coincide:
[ b , a , a ] λ = [ b , a , a λ ] . [b,a,a]^\lambda=[b,a,a^\lambda]. [ b , a , a ] λ = [ b , a , a λ ] . Lemma 3.11. Let w w w x , y , z x,y,z x , y , z G G G
(i) w w w
(ii) [ w , x , y ] = [ w , y , x ] − 1 [w,x,y]=[w,y,x]^{-1} [ w , x , y ] = [ w , y , x ] − 1
(iii) [ w , [ x , y ] ] = [ w , x , y ] 2 [w,[x,y]]=[w,x,y]^2 [ w , [ x , y ]] = [ w , x , y ] 2
(iv) [ w , [ x , y ] , z ] = 1 [w,[x,y],z]=1 [ w , [ x , y ] , z ] = 1
The next lemma outlines essential identities for right 2-Engel elements. A detailed proof is available in Robinson [28, Theorem 7.13].
We shall use the following identity, due to Kappe [12, Corollary 2.2], to prove that the right-hand 2-Engel element set R ( G ) R(G) R ( G ) G G G
Lemma 3.12. If x , y x,y x , y G G G [ y x , g , g ] = 1 [yx,g,g]=1 [ y x , g , g ] = 1 [ g , [ g , y x ] ] = 1 [g,[g,yx]]=1 [ g , [ g , y x ]] = 1 g ∈ G g\in G g ∈ G
Proof. Expanding the commutator with the identity in Lemma 2.1 (4) and applying the property that x x x
[ y x , g , g ] = [ [ y x , g ] , g ] = [ [ y , g ] x [ x , g ] , g ] = [ [ y , g ] x , g ] [ x , g ] ⋅ [ [ x , g ] , g ] = [ [ y , g ] x , g ] [ x , g ] , ∀ g ∈ G . [yx,g,g]=[[yx,g],g]=[[y,g]^x[x,g],g]=[[y,g]^x,g]^{[x,g]}\cdot[[x,g],g]=[[y,g]^x,g]^{[x,g]},\quad\forall g\in G. [ y x , g , g ] = [[ y x , g ] , g ] = [[ y , g ] x [ x , g ] , g ] = [[ y , g ] x , g ] [ x , g ] ⋅ [[ x , g ] , g ] = [[ y , g ] x , g ] [ x , g ] , ∀ g ∈ G . Further expansion of the right-hand term provides:
[ [ y , g ] x , g ] [ x , g ] = [ [ y , g ] , g x − 1 ] x [ x , g ] = [ [ y , g ] , g [ g , x − 1 ] ] x [ x , g ] = [ [ y , g ] , [ g , x − 1 ] ] x [ x , g ] , ∀ g ∈ G . [[y,g]^x,g]^{[x,g]}=[[y,g],g^{x^{-1}}]^{x[x,g]}=[[y,g],g[g,x^{-1}]]^{x[x,g]}=[[y,g],[g,x^{-1}]]^{x[x,g]},\quad\forall g\in G. [[ y , g ] x , g ] [ x , g ] = [[ y , g ] , g x − 1 ] x [ x , g ] = [[ y , g ] , g [ g , x − 1 ] ] x [ x , g ] = [[ y , g ] , [ g , x − 1 ] ] x [ x , g ] , ∀ g ∈ G . Applying Lemma 3.11 (ii) and (iv), along with the fact that y y y
[ y x , g , g ] = [ [ y x , g ] , g ] = [ [ y , g ] , [ g , x − 1 ] ] x [ x , g ] = ( [ [ y , [ g , x − 1 ] , g ] − 1 ) x [ x , g ] = 1 , ∀ g ∈ G . [yx,g,g]=[[yx,g],g]=[[y,g],[g,x^{-1}]]^{x[x,g]}=\left([[y,[g,x^{-1}],g]^{-1}\right)^{x[x,g]}=1,\quad\forall g\in G. [ y x , g , g ] = [[ y x , g ] , g ] = [[ y , g ] , [ g , x − 1 ] ] x [ x , g ] = ( [[ y , [ g , x − 1 ] , g ] − 1 ) x [ x , g ] = 1 , ∀ g ∈ G . Since [ y x , g ] [yx,g] [ y x , g ] g g g [ g , [ g , y x ] ] = 1 [g,[g,yx]]=1 [ g , [ g , y x ]] = 1
Proposition 3.13. The set R ( G ) R(G) R ( G ) G G G
Proof. We first prove that R ( G ) R(G) R ( G ) g ∈ G g\in G g ∈ G ρ g : G → G \rho_g:G\to G ρ g : G → G ρ g ( x ) = [ x , g , g ] \rho_g(x)=[x,g,g] ρ g ( x ) = [ x , g , g ] R ( G ) = ⋂ g ∈ G ρ g − 1 ( { 1 } ) R(G)=\bigcap_{g\in G}\rho_g^{-1}(\{1\}) R ( G ) = ⋂ g ∈ G ρ g − 1 ({ 1 }) R ( G ) R(G) R ( G ) G G G 1 ∈ R ( G ) ≠ ∅ 1\in R(G)\ne\varnothing 1 ∈ R ( G ) = ∅ R ( G ) R(G) R ( G ) a ∈ R ( G ) a\in R(G) a ∈ R ( G ) a − 1 ∈ R ( G ) a^{-1}\in R(G) a − 1 ∈ R ( G ) a ∈ R ( G ) a\in R(G) a ∈ R ( G ) a a a x ∈ G x\in G x ∈ G C G ( ⟨ [ x , a ] ⟩ ) C_G(\langle[x,a]\rangle) C G (⟨[ x , a ]⟩) a a a a − 1 a^{-1} a − 1 x x x Lemma 2.1 (6) that
[ a − 1 , x , x ] = [ [ x , a ] a − 1 , x ] = [ x , a , x ] = 1. [a^{-1},x,x]=[[x,a]^{a^{-1}},x]=[x,a,x]=1. [ a − 1 , x , x ] = [[ x , a ] a − 1 , x ] = [ x , a , x ] = 1. Finally, let φ \varphi φ G G G a ∈ R ( G ) a\in R(G) a ∈ R ( G ) φ ( [ a , g , g ] ) = [ φ ( a ) , φ ( g ) , φ ( g ) ] = 1 \varphi([a,g,g])=[\varphi(a),\varphi(g),\varphi(g)]=1 φ ([ a , g , g ]) = [ φ ( a ) , φ ( g ) , φ ( g )] = 1 φ ( a ) ∈ R ( G ) \varphi(a)\in R(G) φ ( a ) ∈ R ( G ) R ( G ) R(G) R ( G )
:::
The following lemma characterizes the groups in class A ^ \widehat{\mathcal{A}} A [20 Corollary 3(ii)] to profinite groups.
Lemma 3.14. Let G G G G ∈ A ^ G\in\widehat{\mathcal{A}} G ∈ A [ v , x , y ] ∈ ⟨ x , y ⟩ ‾ [v,x,y]\in\overline{\langle x,y\rangle} [ v , x , y ] ∈ ⟨ x , y ⟩ v , x , y ∈ G v,x,y\in G v , x , y ∈ G [ x , y ] = 1 [x,y]=1 [ x , y ] = 1
Proof. Assume G ∈ A ^ G\in\widehat{\mathcal{A}} G ∈ A x , y ∈ G x,y\in G x , y ∈ G [ x , y ] = 1 [x,y]=1 [ x , y ] = 1 ⟨ x , y ⟩ ‾ \overline{\langle x,y\rangle} ⟨ x , y ⟩ ⟨ x , y ⟩ ‾ ⊴ c ⟨ x , y ⟩ ‾ G \overline{\langle x,y\rangle}\trianglelefteq_c\overline{\langle x,y\rangle}^{G} ⟨ x , y ⟩ ⊴ c ⟨ x , y ⟩ G v ∈ G v\in G v ∈ G y [ v , x ] y^{[v,x]} y [ v , x ] y [ v , x ] = y ( x − 1 ) v x y^{[v,x]}=y^{(x^{-1})^{vx}} y [ v , x ] = y ( x − 1 ) v x ( x − 1 ) v x ∈ ⟨ x ⟩ G ≤ ⟨ x , y ⟩ ‾ G (x^{-1})^{vx}\in\langle x\rangle^{G}\leq\overline{\langle x,y\rangle}^{G} ( x − 1 ) v x ∈ ⟨ x ⟩ G ≤ ⟨ x , y ⟩ G ( x − 1 ) v x ∈ ⟨ x ⟩ ‾ G ≤ c ⟨ x , y ⟩ ‾ G (x^{-1})^{vx}\in\overline{\langle x\rangle}^{G}\leq_c\overline{\langle x,y\rangle}^{G} ( x − 1 ) v x ∈ ⟨ x ⟩ G ≤ c ⟨ x , y ⟩ G y [ v , x ] ∈ ⟨ x , y ⟩ ‾ y^{[v,x]}\in\overline{\langle x,y\rangle} y [ v , x ] ∈ ⟨ x , y ⟩
y [ v , x ] = y [ y , [ v , x ] ] = y [ [ v , x ] , y ] − 1 = y [ v , x , y ] − 1 . y^{[v,x]}=y[y,[v,x]]=y[[v,x],y]^{-1}=y[v,x,y]^{-1}. y [ v , x ] = y [ y , [ v , x ]] = y [[ v , x ] , y ] − 1 = y [ v , x , y ] − 1 . Therefore, [ v , x , y ] = ( y [ v , x ] ) − 1 y ∈ ⟨ x , y ⟩ ‾ [v,x,y]=(y^{[v,x]})^{-1}y\in\overline{\langle x,y\rangle} [ v , x , y ] = ( y [ v , x ] ) − 1 y ∈ ⟨ x , y ⟩
Conversely, assume [ v , x , y ] ∈ ⟨ x , y ⟩ ‾ [v,x,y]\in\overline{\langle x,y\rangle} [ v , x , y ] ∈ ⟨ x , y ⟩ x , y x,y x , y G G G H H H G G G H = H 2 = H H G ‾ H=H_2=H^{\overline{H^G}} H = H 2 = H H G H ⊴ c H G ‾ ⊴ c G H\trianglelefteq_c\overline{H^G}\trianglelefteq_c G H ⊴ c H G ⊴ c G H H H x , y ∈ H x,y\in H x , y ∈ H v ∈ G v\in G v ∈ G y [ v , x ] ∈ H y^{[v,x]}\in H y [ v , x ] ∈ H y [ x , v ] ∈ H y^{[x,v]}\in H y [ x , v ] ∈ H H H H [ G , H ] [G,H] [ G , H ] H G ‾ \overline{H^G} H G x , y ∈ H x,y\in H x , y ∈ H v ∈ G v\in G v ∈ G y [ v , x ] = y [ v , x , y ] − 1 ∈ ⟨ x , y ⟩ ≤ H y^{[v,x]}=y[v,x,y]^{-1}\in\langle x,y\rangle\leq H y [ v , x ] = y [ v , x , y ] − 1 ∈ ⟨ x , y ⟩ ≤ H H H H x x x x − 1 x^{-1} x − 1 y [ v , x − 1 ] = y [ v , x − 1 , y ] − 1 ∈ ⟨ x − 1 , y ⟩ = ⟨ x , y ⟩ ≤ H y^{[v,x^{-1}]}=y[v,x^{-1},y]^{-1}\in\langle x^{-1},y\rangle=\langle x,y\rangle\leq H y [ v , x − 1 ] = y [ v , x − 1 , y ] − 1 ∈ ⟨ x − 1 , y ⟩ = ⟨ x , y ⟩ ≤ H y [ x , v ] = y [ v , x − 1 ] x = y x − 1 [ v , x − 1 ] x = ( y [ v , x − 1 ] ) x ∈ ⟨ x , y ⟩ x = ⟨ x , y ⟩ ≤ H y^{[x,v]}=y^{[v,x^{-1}]^x}=y^{x^{-1}[v,x^{-1}]x}=(y^{[v,x^{-1}]})^x\in\langle x,y\rangle^x=\langle x,y\rangle\leq H y [ x , v ] = y [ v , x − 1 ] x = y x − 1 [ v , x − 1 ] x = ( y [ v , x − 1 ] ) x ∈ ⟨ x , y ⟩ x = ⟨ x , y ⟩ ≤ H Lemma 2.1 (6). Therefore H = H 2 H=H_2 H = H 2 G ∈ A ^ G\in\widehat{\mathcal{A}} G ∈ A
Profinite groups with procyclic subgroups whose subnormal defect is at most 2
The first result below (Proposition 4.1) is the profinite analogue of Theorem 1(1) of [9 ] ; it is the key ingredient for the rest of the article. In the abstract case the torsion-freeness of an infinite cyclic subgroup is automatic, since it is isomorphic to Z \mathbb{Z} Z
Proposition 4.1 (Profinite version of Theorem 1(1) in [9 ] ). Let G ∈ T G \in\mathcal{T} G ∈ T G G G
Proof. The proof will be divided into three distinct steps.
Step 1: We first show that [ g , u , u ] = 1 [g,u,u]=1 [ g , u , u ] = 1 u ∈ G u \in G u ∈ G g ∈ G g \in G g ∈ G
Assume for a contradiction that [ g , u , u ] ≠ 1 [g,u,u] \ne1 [ g , u , u ] = 1 ζ : Z ^ → ⟨ u ⟩ ‾ \zeta:\widehat{\mathbb{Z}}\to\overline{\langle u\rangle} ζ : Z → ⟨ u ⟩ α ↦ u α \alpha\mapsto u^\alpha α ↦ u α Lemma 3.8 , [ g , u , u ] ∈ ⟨ u ⟩ ‾ [g,u,u]\in\overline{\langle u\rangle} [ g , u , u ] ∈ ⟨ u ⟩
[ g , u , u ] = u λ [g,u,u]=u^\lambda [ g , u , u ] = u λ for some λ ∈ Z ^ \lambda\in\widehat{\mathbb{Z}} λ ∈ Z λ ∉ ker ( ζ ) \lambda\notin\ker(\zeta) λ ∈ / ker ( ζ ) u u u
⟨ u ⟩ ‾ ≅ ∏ p ∈ π Z p , \overline{\langle u\rangle}\cong\prod_{p\in\pi}\mathbb{Z}_p, ⟨ u ⟩ ≅ p ∈ π ∏ Z p , with π \pi π G G G Lemma 3.7 , we have
1 = [ u , [ g , u ] , [ g , u ] ] . 1=[u,[g,u],[g,u]]. 1 = [ u , [ g , u ] , [ g , u ]] . Applying the commutator identity (6) from Lemma 2.1 , we obtain:
1 = [ g , u , u ] − 1 [ g , u ] = [ ( u λ ) − 1 , [ g , u ] ] = [ g , u , u λ ] ( u λ ) − 1 . 1=[g,u,u]^{-1}[g,u]=[(u^\lambda)^{-1},[g,u]]=[g,u,u^\lambda]^{(u^\lambda)^{-1}}. 1 = [ g , u , u ] − 1 [ g , u ] = [( u λ ) − 1 , [ g , u ]] = [ g , u , u λ ] ( u λ ) − 1 . Conjugating this equation by u λ u^\lambda u λ [ g , u , u λ ] = 1 [g,u,u^\lambda]=1 [ g , u , u λ ] = 1 Lemma 3.10 , we know that [ g , u , u λ ] = [ g , u , u ] λ [g,u,u^\lambda]=[g,u,u]^\lambda [ g , u , u λ ] = [ g , u , u ] λ
1 = [ g , u , u λ ] = [ g , u , u ] λ = ( u λ ) λ = u λ 2 ⟹ λ 2 ∈ ker ( ζ ) . 1=[g,u,u^\lambda]=[g,u,u]^\lambda=(u^\lambda)^\lambda=u^{\lambda^2}\Longrightarrow\lambda^2\in\ker(\zeta). 1 = [ g , u , u λ ] = [ g , u , u ] λ = ( u λ ) λ = u λ 2 ⟹ λ 2 ∈ ker ( ζ ) . Since λ ∉ ker ( ζ ) \lambda\notin\ker(\zeta) λ ∈ / ker ( ζ ) λ + ker ( ζ ) \lambda+\ker(\zeta) λ + ker ( ζ ) ( λ + ker ( ζ ) ) 2 = 0 (\lambda+\ker(\zeta))^2=0 ( λ + ker ( ζ ) ) 2 = 0
Z ^ / ker ( ζ ) ≅ ∏ p ∈ π ′ Z p \widehat{\mathbb{Z}}/\ker(\zeta)\cong\prod_{p\in\pi'}\mathbb{Z}_p Z / ker ( ζ ) ≅ p ∈ π ′ ∏ Z p has no nonzero nilpotent elements. Hence λ ∈ ker ( ζ ) \lambda\in\ker(\zeta) λ ∈ ker ( ζ ) u λ = 1 u^\lambda=1 u λ = 1 [ g , u , u ] = u λ [g,u,u]=u^\lambda [ g , u , u ] = u λ
[ g , u , u ] = 1 , [g,u,u]=1, [ g , u , u ] = 1 , for any u ∈ G u\in G u ∈ G g ∈ G g\in G g ∈ G
Step 2: We now prove that [ u , g , g ] = 1 [u,g,g]=1 [ u , g , g ] = 1 u ∈ G u\in G u ∈ G g ∈ G g\in G g ∈ G
If g g g u u u g g g g g g W = ⟨ u , g ⟩ W=\langle u,g\rangle W = ⟨ u , g ⟩ G G G u u u g g g G G G Lemma 3.4 , we have γ 4 ( G ) = { 1 } \gamma_4(G)=\{1\} γ 4 ( G ) = { 1 } W ≤ G W\leq G W ≤ G γ 4 a b s ( W ) ≤ γ 4 ( G ) = { 1 } \gamma_4^{\mathrm{abs}}(W)\leq\gamma_4(G)=\{1\} γ 4 abs ( W ) ≤ γ 4 ( G ) = { 1 } W W W W W W tor ( W ) \operatorname{tor}(W) tor ( W ) W W W [13 ] , Lemma 16.2.1, Theorem 16.2.7). Since u u u g g g g u gu g u u = g − 1 ( g u ) u=g^{-1}(gu) u = g − 1 ( g u )
Applying the result of Step 1 to the elements u u u g u gu g u Lemma 2.1 , we find:
1 = [ g , g u , g u ] = [ [ g , u ] , g u ] = [ [ g , u ] , u ] ⋅ [ [ g , u ] , g ] u = 1 ⋅ [ g , u , g ] u . 1=[g,gu,gu]=[[g,u],gu]=[[g,u],u]\cdot[[g,u],g]^u=1\cdot[g,u,g]^u. 1 = [ g , g u , g u ] = [[ g , u ] , g u ] = [[ g , u ] , u ] ⋅ [[ g , u ] , g ] u = 1 ⋅ [ g , u , g ] u . This forces [ g , u , g ] = 1 [g,u,g]=1 [ g , u , g ] = 1 [1 ] , [ g , u , g ] = 1 [g,u,g]=1 [ g , u , g ] = 1 [ u , g , g ] = 1 [u,g,g]=1 [ u , g , g ] = 1
Step 3: Finally, we prove that G G G R ( G ) = { x ∈ G ∣ [ x , g , g ] = 1 for all g ∈ G } R(G)=\{x\in G\mid[x,g,g]=1\text{ for all }g\in G\} R ( G ) = { x ∈ G ∣ [ x , g , g ] = 1 for all g ∈ G } G G G
From Step 2, we know that all infinite-order elements of G G G R ( G ) R(G) R ( G )
Let v ∈ G v\in G v ∈ G u ∈ G u\in G u ∈ G G G G u v uv uv u u u u − 1 u^{-1} u − 1 u v uv uv ( u v ) − 1 (uv)^{-1} ( uv ) − 1 R ( G ) R(G) R ( G ) R ( G ) R(G) R ( G ) v = u − 1 ( u v ) ∈ R ( G ) v=u^{-1}(uv)\in R(G) v = u − 1 ( uv ) ∈ R ( G ) [ v , g , g ] = 1 [v,g,g]=1 [ v , g , g ] = 1 g ∈ G g\in G g ∈ G G G G R ( G ) R(G) R ( G ) G G G
Corollary 4.2. Let G ∈ T ^ G\in\widehat{\mathcal{T}} G ∈ T p p p G G G
Proof First, G G G G G G [26 ] , G G G p p p G = ∏ p G p G=\prod_pG_p G = ∏ p G p G p G_p G p p p p G G G G = { 1 } G=\{1\} G = { 1 } G ≠ { 1 } G\ne\{1\} G = { 1 } G p ≠ { 1 } G_p\ne\{1\} G p = { 1 } u u u G G G
Let w ∈ G w\in G w ∈ G ⟨ w ⟩ \langle w\rangle ⟨ w ⟩ [17 ] ), it is pronilpotent. Hence ⟨ w ⟩ \langle w\rangle ⟨ w ⟩ p p p ⟨ w ⟩ = ∏ p ∈ π ′ W p \langle w\rangle=\prod_{p\in\pi'}W_p ⟨ w ⟩ = ∏ p ∈ π ′ W p π ′ \pi' π ′ W p W_p W p p p p ⟨ w ⟩ \langle w\rangle ⟨ w ⟩ [26 ] , W p ≤ G p W_p\leq G_p W p ≤ G p p p p G G G W p W_p W p ⟨ w ⟩ \langle w\rangle ⟨ w ⟩ ⟨ w ⟩ ≅ ∏ p ∈ π ′ Z p \langle w\rangle\cong\prod_{p\in\pi'}\mathbb{Z}_p ⟨ w ⟩ ≅ ∏ p ∈ π ′ Z p
Remark 4.3. The hypothesis that all infinite procyclic subgroups are torsion-free was crucial in the proof of Proposition 4.1: without it, the argument that a non-periodic profinite group all of whose infinite procyclic subgroups are 2-subnormal is 2-Engel would not go through. This contrasts with the abstract case, where Theorem 1 [1 ] of [9 ] does not require any restriction on the torsion of cyclic subgroups. It is not known, however, whether this hypothesis can be removed in the profinite case: there is neither a counterexample nor a proof that it is unnecessary.
The following result gives a profinite version of a theorem of Heineken on torsion-free 2-Baer groups (see [9 ] and [36 ] ).
Theorem 4.4. A torsion-free profinite group G G G G ∈ T ^ G\in\widehat{\mathcal{T}} G ∈ T
Proof This follows immediately from Lemma 3.8 and Proposition 4.1, since [ g , h , h ] = 1 [g,h,h]=1 [ g , h , h ] = 1 [ g , h , h ] ∈ ⟨ h ⟩ [g,h,h]\in\langle h\rangle [ g , h , h ] ∈ ⟨ h ⟩ g , h ∈ G g,h\in G g , h ∈ G
Corollary 4.5. A non-periodic pro-p p p G G G infinite procyclic subgroups are torsion-free is 2-Engel if and only if G ∈ T ^ G \in\widehat{\mathcal{T}} G ∈ T
Proof. The proof uses the same arguments as that of Theorem 4.4 . □
Example 4.6. Consider the generalized quaternion group of order 16 (identified as SmallGroup(16,9) in [34 ] ), defined by
Q 16 = ⟨ a , b ∣ a 8 = 1 , b 2 = a 4 , a b = a − 1 ⟩ . Q_{16}=\langle a,b\mid a^{8}=1,\ b^{2}=a^{4},\ a^{b}=a^{-1}\rangle. Q 16 = ⟨ a , b ∣ a 8 = 1 , b 2 = a 4 , a b = a − 1 ⟩ . It is nilpotent of class 3, with center ⟨ a 4 ⟩ ≅ C 2 \langle a^{4}\rangle\cong C_{2} ⟨ a 4 ⟩ ≅ C 2 ⟨ a 2 ⟩ ≅ C 4 \langle a^{2}\rangle\cong C_{4} ⟨ a 2 ⟩ ≅ C 4 b b b b a ba ba b a 2 ba^{2} b a 2 b a 3 ba^{3} b a 3 Q 16 Q_{16} Q 16 Q 16 Q_{16} Q 16 T \mathcal{T} T T ^ \widehat{\mathcal{T}} T a b = a − 1 a^{b}=a^{-1} a b = a − 1 [ a , b ] = a − 1 a b = a − 2 [a,b]=a^{-1}a^{b}=a^{-2} [ a , b ] = a − 1 a b = a − 2
[ a , b , b ] = [ a − 2 , b ] = ( a 2 ) ( a − 2 ) b = a 2 a 2 = a 4 ≠ 1. [a,b,b]=[a^{-2},b]=(a^{2})(a^{-2})^{b}=a^{2}a^{2}=a^{4}\ne1. [ a , b , b ] = [ a − 2 , b ] = ( a 2 ) ( a − 2 ) b = a 2 a 2 = a 4 = 1. Therefore Q 16 Q_{16} Q 16
Similarly, by Theorem 1 of [19 ] , the group H ( 3 , 1 ) H(3,1) H ( 3 , 1 )
H ( 3 , 1 ) = ⟨ x , y ∣ [ y , x , x ] = x 3 = [ y , x , y ] = y 3 , x 9 = 1 , γ 4 a b s ( H ( 3 , 1 ) ) = { 1 } ⟩ H(3,1)=\langle x,y\mid[y,x,x]=x^{3}=[y,x,y]=y^{3},\ x^{9}=1,\ \gamma_{4}^{\mathrm{abs}}(H(3,1))=\{1\}\rangle H ( 3 , 1 ) = ⟨ x , y ∣ [ y , x , x ] = x 3 = [ y , x , y ] = y 3 , x 9 = 1 , γ 4 abs ( H ( 3 , 1 )) = { 1 }⟩ is a finite 3-group of order 81. Moreover, H ( 3 , 1 ) H(3,1) H ( 3 , 1 ) T \mathcal{T} T T ^ \widehat{\mathcal{T}} T [ y , x , x ] = x 3 ≠ 1 [y,x,x]=x^{3}\ne1 [ y , x , x ] = x 3 = 1 H ( 3 , 1 ) H(3,1) H ( 3 , 1 )
Thus the non-periodicity hypothesis in Proposition 4.1 cannot be removed: the examples Q 16 Q_{16} Q 16 H ( 3 , 1 ) H(3,1) H ( 3 , 1 ) T \mathcal{T} T Q 16 Q_{16} Q 16 a 4 a^{4} a 4 H ( 3 , 1 ) H(3,1) H ( 3 , 1 ) Theorem 4.4 cannot be relaxed either: for instance, H ( 3 , 1 ) H(3,1) H ( 3 , 1 ) T ^ \widehat{\mathcal{T}} T ■
Let G G G n ≥ 1 n\geq1 n ≥ 1 φ n : G → G \varphi_n:G\to G φ n : G → G φ n ( x ) = x n \varphi_n(x)=x^n φ n ( x ) = x n x ∈ G x\in G x ∈ G T n = φ n − 1 ( { 1 } ) = { x ∈ G : x n = 1 } T_n=\varphi_n^{-1}(\{1\})=\{x\in G:x^n=1\} T n = φ n − 1 ({ 1 }) = { x ∈ G : x n = 1 } n n n G G G T n = φ n − 1 ( { 1 } ) T_n=\varphi_n^{-1}(\{1\}) T n = φ n − 1 ({ 1 }) G G G
The following result is well-known for abstract groups (see Robinson [28 Theorem 7.14] ). The proof relies on the fact that the group satisfies the 2-Engel condition, which we established for profinite groups in Proposition 4.1.
Corollary 4.7 (Profinite version of Theorem 1 (2) in [9 ] ). Let G ∈ T ^ G\in\widehat{\mathcal{T}} G ∈ T ( γ 3 ( G ) ) 3 = { 1 } (\gamma_{3}(G))^{3}=\{1\} ( γ 3 ( G ) ) 3 = { 1 } G ∈ T ^ G\in\widehat{\mathcal{T}} G ∈ T
Proof. We invoke the Hall-Witt identity (Lemma 2.1 (8)):
1 = [ x , y − 1 , z ] y ⋅ [ y , z − 1 , x ] z ⋅ [ z , x − 1 , y ] x (10) 1=[x,y^{-1},z]^y\cdot[y,z^{-1},x]^z\cdot[z,x^{-1},y]^x \tag*{(10)} 1 = [ x , y − 1 , z ] y ⋅ [ y , z − 1 , x ] z ⋅ [ z , x − 1 , y ] x (10) for all x , y , z ∈ G x,y,z\in G x , y , z ∈ G
By Proposition 4.1, G G G (iii) along with Lemma 2.1 (5), we simplify the individual terms of the Hall-Witt identity:
[ x , y − 1 , z ] y = ( [ x , z , y − 1 ] y ) − 1 = ( [ y , [ x , z ] ] ) − 1 = [ x , z , y ] = [ x , y , z ] − 1 , [x,y^{-1},z]^y=([x,z,y^{-1}]^y)^{-1}=([y,[x,z]])^{-1}=[x,z,y]=[x,y,z]^{-1}, [ x , y − 1 , z ] y = ([ x , z , y − 1 ] y ) − 1 = ([ y , [ x , z ]] ) − 1 = [ x , z , y ] = [ x , y , z ] − 1 , [ y , z − 1 , x ] z = ( [ y , x , z − 1 ] z ) − 1 = [ z , [ y , x ] ] − 1 = [ y , x , z ] = [ y , z , x ] − 1 = [ x , [ y , z ] ] = [ x , y , z ] 2 , [y,z^{-1},x]^z=([y,x,z^{-1}]^z)^{-1}=[z,[y,x]]^{-1}=[y,x,z]=[y,z,x]^{-1}=[x,[y,z]]=[x,y,z]^2, [ y , z − 1 , x ] z = ([ y , x , z − 1 ] z ) − 1 = [ z , [ y , x ] ] − 1 = [ y , x , z ] = [ y , z , x ] − 1 = [ x , [ y , z ]] = [ x , y , z ] 2 , [ z , x − 1 , y ] x = ( [ z , y , x − 1 ] x ) − 1 = [ x , [ z , y ] ] − 1 = [ x , z , y ] − 2 = [ x , y , z ] 2 . [z,x^{-1},y]^x=([z,y,x^{-1}]^x)^{-1}=[x,[z,y]]^{-1}=[x,z,y]^{-2}=[x,y,z]^2. [ z , x − 1 , y ] x = ([ z , y , x − 1 ] x ) − 1 = [ x , [ z , y ] ] − 1 = [ x , z , y ] − 2 = [ x , y , z ] 2 . Substituting these simplified terms back into equation (10) yields:
1 = [ x , y , z ] − 1 ⋅ [ x , y , z ] 2 ⋅ [ x , y , z ] 2 = [ x , y , z ] 3 1=[x,y,z]^{-1}\cdot[x,y,z]^2\cdot[x,y,z]^2=[x,y,z]^3 1 = [ x , y , z ] − 1 ⋅ [ x , y , z ] 2 ⋅ [ x , y , z ] 2 = [ x , y , z ] 3 for all x , y , z ∈ G x,y,z\in G x , y , z ∈ G
Since G G G Lemma 3.4 , we have γ 4 ( G ) = { 1 } \gamma_4(G)=\{1\} γ 4 ( G ) = { 1 } [ γ 3 ( G ) , G ] = { 1 } [\gamma_3(G),G]=\{1\} [ γ 3 ( G ) , G ] = { 1 } γ 3 ( G ) \gamma_3(G) γ 3 ( G ) γ 3 ( G ) = γ 3 a b s ( G ) \gamma_3(G)=\gamma_3^{\mathrm{abs}}(G) γ 3 ( G ) = γ 3 abs ( G ) γ 3 a b s ( G ) ≤ γ 3 ( G ) \gamma_3^{\mathrm{abs}}(G)\leq\gamma_3(G) γ 3 abs ( G ) ≤ γ 3 ( G ) N ⊴ o G N\trianglelefteq_o G N ⊴ o G G / N G/N G / N γ 3 ( G ) N / N = γ 3 a b s ( G ) N / N \gamma_3(G)N/N=\gamma_3^{\mathrm{abs}}(G)N/N γ 3 ( G ) N / N = γ 3 abs ( G ) N / N γ 3 ( G ) ≤ γ 3 a b s ( G ) N \gamma_3(G)\leq\gamma_3^{\mathrm{abs}}(G)N γ 3 ( G ) ≤ γ 3 abs ( G ) N N N N γ 3 a b s ( G ) = ⋂ N ⊴ o G γ 3 a b s ( G ) N \gamma_3^{\mathrm{abs}}(G)=\bigcap_{N\trianglelefteq_o G}\gamma_3^{\mathrm{abs}}(G)N γ 3 abs ( G ) = ⋂ N ⊴ o G γ 3 abs ( G ) N [38 Proposition 0.3.3] ), we obtain γ 3 ( G ) = γ 3 a b s ( G ) \gamma_3(G)=\gamma_3^{\mathrm{abs}}(G) γ 3 ( G ) = γ 3 abs ( G ) γ 3 ( G ) \gamma_3(G) γ 3 ( G ) [ x , y , z ] [x,y,z] [ x , y , z ] ≥ 4 \geq4 ≥ 4 T 3 = { x ∈ G : x 3 = 1 } T_3=\{x\in G:x^3=1\} T 3 = { x ∈ G : x 3 = 1 } G G G [ x , y , z ] 3 = 1 [x,y,z]^3=1 [ x , y , z ] 3 = 1 x , y , z ∈ G x,y,z\in G x , y , z ∈ G { [ x , y , z ] ∣ x , y , z ∈ G } ⊆ T 3 \{[x,y,z]\mid x,y,z\in G\}\subseteq T_3 {[ x , y , z ] ∣ x , y , z ∈ G } ⊆ T 3 γ 3 ( G ) \gamma_3(G) γ 3 ( G ) w ↦ w 3 w\mapsto w^3 w ↦ w 3 γ 3 ( G ) \gamma_3(G) γ 3 ( G ) T 3 ∩ γ 3 ( G ) T_3\cap\gamma_3(G) T 3 ∩ γ 3 ( G ) γ 3 ( G ) \gamma_3(G) γ 3 ( G ) [ x , y , z ] [x,y,z] [ x , y , z ] γ 3 ( G ) = ⟨ { [ x , y , z ] ∣ x , y , z ∈ G } ⟩ ‾ ⊆ T 3 ∩ γ 3 ( G ) \gamma_3(G)=\overline{\langle\{[x,y,z]\mid x,y,z\in G\}\rangle}\subseteq T_3\cap\gamma_3(G) γ 3 ( G ) = ⟨{[ x , y , z ] ∣ x , y , z ∈ G }⟩ ⊆ T 3 ∩ γ 3 ( G ) ( γ 3 ( G ) ) 3 = { 1 } (\gamma_3(G))^3=\{1\} ( γ 3 ( G ) ) 3 = { 1 }
Note that, in the proof of Corollary 4.7 , the hypothesis on infinite procyclic subgroups serves only to deduce that G G G G G G
Proposition 4.8. Let G ∈ T G\in\mathcal{T} G ∈ T ( γ 3 ( G ) ) 3 = { 1 } (\gamma_3(G))^3=\{1\} ( γ 3 ( G ) ) 3 = { 1 }
Proof. Since G G G [ x , y , y ] = 1 [x,y,y]=1 [ x , y , y ] = 1 x , y ∈ G x,y\in G x , y ∈ G [ x , y , z ] 3 = 1 [x,y,z]^3=1 [ x , y , z ] 3 = 1 Corollary 4.7 , mutatis mutandis). As γ 3 ( G ) \gamma_3(G) γ 3 ( G ) ( γ 3 ( G ) ) 3 = { 1 } (\gamma_3(G))^3=\{1\} ( γ 3 ( G ) ) 3 = { 1 }
The following result answers Question 1 for torsion-free profinite groups with n = 2 n=2 n = 2 n = 1 n=1 n = 1
Theorem 4.9. Let G G G π \pi π G G G G G G
Proof. As every closed subgroup is 2-subnormal, so is every cyclic subgroup, and therefore G ∈ T ^ G \in\widehat{\mathcal{T}} G ∈ T Corollary 4.7 , we have ( γ 3 ( G ) ) 3 = { 1 } (\gamma_3(G))^3 = \{1\} ( γ 3 ( G ) ) 3 = { 1 } G G G γ 3 ( G ) = { 1 } \gamma_3(G) = \{1\} γ 3 ( G ) = { 1 } G G G
The following result is a profinite version of a theorem of Stadelmann (see [33 ] ).
Corollary 4.10. Let G G G G G G G G G
Proof. It follows directly from Proposition 2.5 and Theorem 4.9 . □
Remark 4.11. The group H ( 3 , 1 ) H(3,1) H ( 3 , 1 ) Example 4.6 shows that the torsion-free hypothesis in Corollary 4.10 cannot be relaxed. By Corollary 1 of [19 ] , for a 2-generator p p p p ≥ 3 p \ge3 p ≥ 3 S \mathcal{S} S T \mathcal{T} T T ^ \widehat{\mathcal{T}} T S ^ \widehat{\mathcal{S}} S H ( 3 , 1 ) ∈ S ^ H(3,1) \in\widehat{\mathcal{S}} H ( 3 , 1 ) ∈ S H ( 3 , 1 ) H(3,1) H ( 3 , 1 ) H ( 3 , 1 ) H(3,1) H ( 3 , 1 )
Proposition 4.12. Let G ∈ T ^ G \in\widehat{\mathcal{T}} G ∈ T G ∈ A ^ G \in\widehat{\mathcal{A}} G ∈ A
Proof. By [26 Theorem 2.1.3] , G G G G = l i m ← U ∈ N G / U G = \varprojlim_{U\in\mathcal{N}} G/U G = lim U ∈ N G / U N \mathcal{N} N 1 ∈ G 1 \in G 1 ∈ G G G G U ∈ N U \in\mathcal{N} U ∈ N π U : G ↠ G / U \pi_U:G \twoheadrightarrow G/U π U : G ↠ G / U
Let H H H G G G G ∈ A ^ G \in\widehat{\mathcal{A}} G ∈ A H H H H H H π U ( H ) = H U / U \pi_U(H)=HU/U π U ( H ) = H U / U U ∈ N U \in\mathcal{N} U ∈ N G G G 2 ′ 2' 2 ′ G / U G/U G / U 2 ′ 2' 2 ′ Lemma 3.3 , G / U ∈ T G/U \in\mathcal{T} G / U ∈ T U ∈ N U \in\mathcal{N} U ∈ N [20 Theorem 1(i)] , any finite 2 ′ 2' 2 ′ T \mathcal{T} T A \mathcal{A} A G / U ∈ A G/U \in\mathcal{A} G / U ∈ A U ∈ N U \in\mathcal{N} U ∈ N H U / U HU/U H U / U G / U G/U G / U H U / U ⊴ 2 G / U HU/U \trianglelefteq^{2} G/U H U / U ⊴ 2 G / U U ∈ N U \in\mathcal{N} U ∈ N G ∈ T ^ G \in\widehat{\mathcal{T}} G ∈ T Lemma 3.4 ensures that G ∈ N ^ 3 G \in\widehat{\mathcal{N}}_3 G ∈ N 3 Proposition 2.5 , every closed subgroup of G G G H H H H H H
H = H 3 = H H 2 ‾ ⊴ c H 2 = H H G ‾ ⊴ c H 1 = H G ‾ ⊴ c H 0 = G . H = H_3 = \overline{H^{H_2}} \trianglelefteq_c H_2 = \overline{H^{H_G}} \trianglelefteq_c H_1 = \overline{H^G} \trianglelefteq_c H_0 = G. H = H 3 = H H 2 ⊴ c H 2 = H H G ⊴ c H 1 = H G ⊴ c H 0 = G . On the other hand, since H U / U HU/U H U / U G / U G/U G / U
π U ( H ) = H U U ⊴ π U ( H G ‾ ) = π U ( H G ) = H G U U = ( H U U ) G / U . \pi_U(H) = \frac{HU}{U} \trianglelefteq\pi_U(\overline{H^G}) = \pi_U(H^G) = \frac{H^G U}{U} = \left(\frac{HU}{U}\right)^{G/U}. π U ( H ) = U H U ⊴ π U ( H G ) = π U ( H G ) = U H G U = ( U H U ) G / U . Hence, by the Correspondence Theorem, pulling back through π U \pi_U π U H U ⊴ c H G ‾ U HU \trianglelefteq_c \overline{H^G}U H U ⊴ c H G U U ∈ N U \in\mathcal{N} U ∈ N H ⊴ c H G ‾ = H 1 H \trianglelefteq_c \overline{H^G} = H_1 H ⊴ c H G = H 1 H ⊴ c H 2 ⊴ c H 1 H \trianglelefteq_c H_2 \trianglelefteq_c H_1 H ⊴ c H 2 ⊴ c H 1 H H H G G G G ∈ A ^ G \in\widehat{\mathcal{A}} G ∈ A
:::
Corollary 4.13. Let G G G π \pi π π \pi π G G G G G G π \pi π G G G S \mathcal{S} S G G G
Proof. By hypothesis, G ∈ T ^ G \in\widehat{\mathcal{T}} G ∈ T G G G G G G p p p [26 Proposition 2.3.8] ). According to Zel’manov’s celebrated solution to the Restricted Burnside Problem [39 ] , any finitely generated periodic pro-p p p p p p G G G π \pi π G G G π \pi π G G G A \mathcal{A} A Proposition 4.12 (or [20 Theorem 1(i)] ). The final statement follows from the established result that for 2-generated groups, the classes collapse to S = A = T \mathcal{S} = \mathcal{A} = \mathcal{T} S = A = T [21 ] ). □
Proposition 4.14. Let G ∈ T ^ G \in\widehat{\mathcal{T}} G ∈ T G ∈ A ^ G \in\widehat{\mathcal{A}} G ∈ A
Proof. By Lemma 3.4 , G G G G G G γ 3 ( G ) ≤ Z ( G ) \gamma_{3}(G) \le Z(G) γ 3 ( G ) ≤ Z ( G ) Lemma 3.14 , to prove that G ∈ A ^ G \in\widehat{\mathcal{A}} G ∈ A x , y ∈ G x,y \in G x , y ∈ G z ∈ G z \in G z ∈ G [ z , x , y ] [z,x,y] [ z , x , y ] H = ⟨ x , y ⟩ H = \langle x,y\rangle H = ⟨ x , y ⟩
Because γ 3 ( G ) \gamma_{3}(G) γ 3 ( G ) H H H [ x , y ] = 1 [x,y] = 1 [ x , y ] = 1
1 = [ x , z , y ] z − 1 ⋅ [ z , y , x ] y − 1 ⋅ [ y , x , z ] x − 1 = [ x , z , y ] ⋅ [ z , y , x ] for all z ∈ G . 1 = [x,z,y]^{z^{-1}} \cdot[z,y,x]^{y^{-1}} \cdot[y,x,z]^{x^{-1}} = [x,z,y] \cdot[z,y,x] \quad\text{for all } z \in G. 1 = [ x , z , y ] z − 1 ⋅ [ z , y , x ] y − 1 ⋅ [ y , x , z ] x − 1 = [ x , z , y ] ⋅ [ z , y , x ] for all z ∈ G . This establishes that [ z , y , x ] = [ x , z , y ] − 1 [z,y,x] = [x,z,y]^{-1} [ z , y , x ] = [ x , z , y ] − 1 G G G γ 2 ( G ) = [ G , G ] \gamma_{2}(G) = [G,G] γ 2 ( G ) = [ G , G ] ( [ x , z ] − 1 ) y ([x,z]^{-1})^{y} ([ x , z ] − 1 ) y γ 2 ( G ) \gamma_{2}(G) γ 2 ( G )
[ z , y , x ] = [ x , z , y ] − 1 = [ y , [ x , z ] ] = ( [ x , z ] − 1 ) y [ x , z ] = [ x , z ] ( [ x , z ] − 1 ) y = [ [ x , z ] − 1 , y ] = [ z , x , y ] . (11) \begin{aligned}
[z,y,x] &= [x,z,y]^{-1} = [y,[x,z]] = ([x,z]^{-1})^{y}[x,z]\\
&= [x,z]([x,z]^{-1})^{y} = [[x,z]^{-1},y] = [z,x,y]. \tag*{(11)}
\end{aligned} [ z , y , x ] = [ x , z , y ] − 1 = [ y , [ x , z ]] = ([ x , z ] − 1 ) y [ x , z ] = [ x , z ] ([ x , z ] − 1 ) y = [[ x , z ] − 1 , y ] = [ z , x , y ] . (11) By Lemma 3.8 , we know that [ z , x y , x y ] ∈ ⟨ x y ⟩ ≤ H [z,xy,xy] \in\langle xy\rangle\le H [ z , x y , x y ] ∈ ⟨ x y ⟩ ≤ H γ 3 ( G ) \gamma_{3}(G) γ 3 ( G ) a , b , c , d ∈ G a,b,c,d \in G a , b , c , d ∈ G
[ [ a , b ] , c d ] = [ [ a , b ] , d ] ⋅ [ [ a , b ] , c ] d = [ [ a , b ] , d ] ⋅ [ [ a , b ] , c ] . (12) [[a,b],cd] = [[a,b],d] \cdot[[a,b],c]^{d} = [[a,b],d] \cdot[[a,b],c]. \tag*{(12)} [[ a , b ] , c d ] = [[ a , b ] , d ] ⋅ [[ a , b ] , c ] d = [[ a , b ] , d ] ⋅ [[ a , b ] , c ] . (12) As γ 4 ( G ) = { 1 } \gamma_{4}(G) = \{1\} γ 4 ( G ) = { 1 } γ 3 ( G ) \gamma_{3}(G) γ 3 ( G ) (12) and Lemma 2.1 to expand [ z , x y , x y ] [z,xy,xy] [ z , x y , x y ]
[ z , x y , x y ] = [ [ z , x y ] , y ] ⋅ [ [ z , x y ] , x ] = [ [ z , y ] [ z , x ] [ z , x , y ] , y ] ⋅ [ [ z , y ] [ z , x ] [ z , x , y ] , x ] = [ [ z , y ] [ z , x ] , y ] ⋅ [ [ z , x , y ] , y ] ⋅ [ [ z , y ] [ z , x ] , x ] = [ [ z , y ] [ z , x ] , y ] ⋅ [ [ z , x , y ] , x ] = [ [ z , y ] , y ] ⋅ [ [ z , x ] , y ] ⋅ [ [ z , y ] , x ] ⋅ [ [ z , x ] , x ] = [ z , y , y ] ⋅ [ z , x , y ] ⋅ [ z , y , x ] ⋅ [ z , x , x ] = [ z , x , x ] ⋅ [ z , y , y ] ⋅ [ z , x , y ] ⋅ [ z , y , x ] . \begin{aligned}
[z,xy,xy] &= [[z,xy],y] \cdot[[z,xy],x]\\
&= [[z,y][z,x][z,x,y],y] \cdot[[z,y][z,x][z,x,y],x]\\
&= [[z,y][z,x],y] \cdot[[z,x,y],y] \cdot[[z,y][z,x],x]\\
&= [[z,y][z,x],y] \cdot[[z,x,y],x]\\
&= [[z,y],y] \cdot[[z,x],y] \cdot[[z,y],x] \cdot[[z,x],x]\\
&= [z,y,y] \cdot[z,x,y] \cdot[z,y,x] \cdot[z,x,x]\\
&= [z,x,x] \cdot[z,y,y] \cdot[z,x,y] \cdot[z,y,x].
\end{aligned} [ z , x y , x y ] = [[ z , x y ] , y ] ⋅ [[ z , x y ] , x ] = [[ z , y ] [ z , x ] [ z , x , y ] , y ] ⋅ [[ z , y ] [ z , x ] [ z , x , y ] , x ] = [[ z , y ] [ z , x ] , y ] ⋅ [[ z , x , y ] , y ] ⋅ [[ z , y ] [ z , x ] , x ] = [[ z , y ] [ z , x ] , y ] ⋅ [[ z , x , y ] , x ] = [[ z , y ] , y ] ⋅ [[ z , x ] , y ] ⋅ [[ z , y ] , x ] ⋅ [[ z , x ] , x ] = [ z , y , y ] ⋅ [ z , x , y ] ⋅ [ z , y , x ] ⋅ [ z , x , x ] = [ z , x , x ] ⋅ [ z , y , y ] ⋅ [ z , x , y ] ⋅ [ z , y , x ] . By Lemma 3.8, [ z , x , x ] ∈ ⟨ x ⟩ ≤ H [z,x,x] \in\langle x\rangle\le H [ z , x , x ] ∈ ⟨ x ⟩ ≤ H [ z , y , y ] ∈ ⟨ y ⟩ ≤ H [z,y,y] \in\langle y\rangle\le H [ z , y , y ] ∈ ⟨ y ⟩ ≤ H [ z , x , x ] [ z , y , y ] [z,x,x][z,y,y] [ z , x , x ] [ z , y , y ] H H H (11) , which states [ z , y , x ] = [ z , x , y ] [z,y,x]=[z,x,y] [ z , y , x ] = [ z , x , y ]
[ z , x , y ] 2 = [ z , y , y ] − 1 [ z , x , x ] − 1 [ z , x y , x y ] ∈ ⟨ x , y ⟩ = H . [z,x,y]^2=[z,y,y]^{-1}[z,x,x]^{-1}[z,xy,xy]\in\langle x,y\rangle=H. [ z , x , y ] 2 = [ z , y , y ] − 1 [ z , x , x ] − 1 [ z , x y , x y ] ∈ ⟨ x , y ⟩ = H . Finally, by Corollary 4.7, ( γ 3 ( G ) ) 3 = { 1 } (\gamma_3(G))^3=\{1\} ( γ 3 ( G ) ) 3 = { 1 }
1 = [ z , x , y ] 3 = [ z , x , y ] 2 [ z , x , y ] ⟹ [ z , x , y ] − 1 = [ z , x , y ] 2 ∈ H . 1=[z,x,y]^3=[z,x,y]^2[z,x,y]\Longrightarrow[z,x,y]^{-1}=[z,x,y]^2\in H. 1 = [ z , x , y ] 3 = [ z , x , y ] 2 [ z , x , y ] ⟹ [ z , x , y ] − 1 = [ z , x , y ] 2 ∈ H . Consequently, [ z , x , y ] ∈ H [z,x,y]\in H [ z , x , y ] ∈ H G ∈ A ^ G\in\widehat{\mathcal A} G ∈ A
Proposition 4.15. Let G ∈ T ^ G\in\widehat{\mathcal T} G ∈ T non-periodic 2-Engel profinite group . Then G ∈ A ^ G\in\widehat{\mathcal A} G ∈ A
Proof. The proof is identical to that of Proposition 4.14, except that the step invoking ( γ 3 ( G ) ) 3 = { 1 } (\gamma_3(G))^3=\{1\} ( γ 3 ( G ) ) 3 = { 1 } G G G ( γ 3 ( G ) ) 3 = { 1 } (\gamma_3(G))^3=\{1\} ( γ 3 ( G ) ) 3 = { 1 }
We summarize our last results in the profinite analogue of Mahdavianary’s [20 Theorem 1(i)] .
Theorem A. If G ∈ T ^ G\in\widehat{\mathcal T} G ∈ T G ∈ A ^ G\in\widehat{\mathcal A} G ∈ A
Proof. This follows directly from Propositions 4.12 and 4.14.
Corollary 4.16. Let G G G G ∈ A ^ G\in\widehat{\mathcal A} G ∈ A
Strict inclusions and Cartesian products
Although the strictness of the inclusions S ^ ⊆ A ^ ⊆ T ^ \widehat{\mathcal S}\subseteq\widehat{\mathcal A}\subseteq\widehat{\mathcal T} S ⊆ A ⊆ T [20 ] to construct explicit infinite profinite groups that separate these classes. Furthermore, we prove that the class T ^ \widehat{\mathcal T} T
We begin by recalling the properties of a specific finite group constructed by Mahdavianary [20 Theorem 1(ii)] .
Remark 5.1. Let D D D
D = ⟨ a , b , c ∣ [ x 1 , x 2 , x 2 ] = 1 for x 1 , x 2 ∈ { a , b , c } , c 2 = b 8 = 1 , [ a , c , b ] = [ a , c ] = 1 , D=\langle a,b,c\mid[x_1,x_2,x_2]=1\text{ for }x_1,x_2\in\{a,b,c\},\ c^2=b^8=1,\ [a,c,b]=[a,c]=1, D = ⟨ a , b , c ∣ [ x 1 , x 2 , x 2 ] = 1 for x 1 , x 2 ∈ { a , b , c } , c 2 = b 8 = 1 , [ a , c , b ] = [ a , c ] = 1 , [ b , c , a ] = [ a , b , c ] = b 4 , [ a , b ] 2 = [ b , c ] 2 = a 2 = 1 , γ 4 ( D ) = 1. [b,c,a]=[a,b,c]=b^{4},\ [a,b]^{2}=[b,c]^{2}=a^{2}=1,\ \gamma_{4}(D)=1. [ b , c , a ] = [ a , b , c ] = b 4 , [ a , b ] 2 = [ b , c ] 2 = a 2 = 1 , γ 4 ( D ) = 1. Using the relations of D D D SmallGroup(128, 523) in the GAP library [34 ] ), one can verify that D D D Z ( D ) = ⟨ b 2 ⟩ ≅ C 4 Z(D)=\langle b^{2}\rangle\cong C_{4} Z ( D ) = ⟨ b 2 ⟩ ≅ C 4 D ′ = ⟨ [ a , b ] , [ b , c ] , b 4 ⟩ D'=\langle[a,b],[b,c],b^{4}\rangle D ′ = ⟨[ a , b ] , [ b , c ] , b 4 ⟩ D ∈ T D\in\mathcal{T} D ∈ T D ∉ A D\notin\mathcal{A} D ∈ / A ⟨ a , c ⟩ \langle a,c\rangle ⟨ a , c ⟩ D D D D D D [ c , a b , a b ] = b 4 ≠ 1 [c,ab,ab]=b^{4}\ne1 [ c , ab , ab ] = b 4 = 1
Theorem 5.2. There exists an infinite periodic profinite group V V V V ∈ T ^ V\in\widehat{\mathcal{T}} V ∈ T V ∉ A ^ V\notin\widehat{\mathcal{A}} V ∈ / A
Proof. Let V = D × E V=D\times E V = D × E D D D E E E D D D E E E V V V u ∈ V u\in V u ∈ V ⟨ u ⟩ = ⟨ u ⟩ ‾ \langle u\rangle=\overline{\langle u\rangle} ⟨ u ⟩ = ⟨ u ⟩
First, we show that V ∈ T ^ V\in\widehat{\mathcal{T}} V ∈ T V ∈ T ^ V\in\widehat{\mathcal{T}} V ∈ T [ u , v , v ] ∈ ⟨ v ⟩ [u,v,v]\in\langle v\rangle [ u , v , v ] ∈ ⟨ v ⟩ u , v ∈ V u,v\in V u , v ∈ V u = ( d 1 , u 1 ) u=(d_{1},u_{1}) u = ( d 1 , u 1 ) v = ( d 2 , u 2 ) v=(d_{2},u_{2}) v = ( d 2 , u 2 ) D × E D\times E D × E E E E [ u , v , v ] = ( [ d 1 , d 2 , d 2 ] , 1 ) [u,v,v]=([d_{1},d_{2},d_{2}],1) [ u , v , v ] = ([ d 1 , d 2 , d 2 ] , 1 ) D ∈ T D\in\mathcal{T} D ∈ T [20 ] gives [ d 1 , d 2 , d 2 ] ∈ ⟨ d 2 ⟩ [d_{1},d_{2},d_{2}]\in\langle d_{2}\rangle [ d 1 , d 2 , d 2 ] ∈ ⟨ d 2 ⟩ d 2 = 1 d_{2}=1 d 2 = 1 [ u , v , v ] = 1 ∈ ⟨ v ⟩ [u,v,v]=1\in\langle v\rangle [ u , v , v ] = 1 ∈ ⟨ v ⟩ d 2 ≠ 1 d_{2}\ne1 d 2 = 1 γ 3 a b s ( D ) = { 1 } \gamma_{3}^{\mathrm{abs}}(D)=\{1\} γ 3 abs ( D ) = { 1 } γ 3 ( D ) ≤ Z ( D ) = ⟨ b 2 ⟩ \gamma_{3}(D)\leq Z(D)=\langle b^{2}\rangle γ 3 ( D ) ≤ Z ( D ) = ⟨ b 2 ⟩ [ d 1 , d 2 , d 2 ] ∈ Z ( D ) ∩ ⟨ d 2 ⟩ [d_{1},d_{2},d_{2}]\in Z(D)\cap\langle d_{2}\rangle [ d 1 , d 2 , d 2 ] ∈ Z ( D ) ∩ ⟨ d 2 ⟩ d 2 ∈ Z ( D ) d_{2}\in Z(D) d 2 ∈ Z ( D ) [ d 1 , d 2 , d 2 ] = 1 = d 2 0 [d_{1},d_{2},d_{2}]=1=d_{2}^{0} [ d 1 , d 2 , d 2 ] = 1 = d 2 0 Z ( D ) ∩ ⟨ d 2 ⟩ Z(D)\cap\langle d_{2}\rangle Z ( D ) ∩ ⟨ d 2 ⟩ ⟨ d 2 ⟩ \langle d_{2}\rangle ⟨ d 2 ⟩ ⟨ d 2 2 ⟩ \langle d_{2}^{2}\rangle ⟨ d 2 2 ⟩ [ d 1 , d 2 , d 2 ] = d 2 2 k [d_{1},d_{2},d_{2}]=d_{2}^{2k} [ d 1 , d 2 , d 2 ] = d 2 2 k 0 ≤ k ≤ 3 0\leq k\leq3 0 ≤ k ≤ 3
[ u , v , v ] = ( [ d 1 , d 2 , d 2 ] , 1 ) = ( d 2 2 k , 1 ) = ( d 2 2 k , u 2 2 k ) = v 2 k ∈ ⟨ v ⟩ . [u,v,v]=([d_{1},d_{2},d_{2}],1)=(d_{2}^{2k},1)=(d_{2}^{2k},u_{2}^{2k})=v^{2k}\in\langle v\rangle. [ u , v , v ] = ([ d 1 , d 2 , d 2 ] , 1 ) = ( d 2 2 k , 1 ) = ( d 2 2 k , u 2 2 k ) = v 2 k ∈ ⟨ v ⟩ . Therefore, V ∈ T ^ V\in\widehat{\mathcal{T}} V ∈ T
Second, we prove that V ∉ A ^ V\notin\widehat{\mathcal{A}} V ∈ / A H = ⟨ a , c ⟩ ≤ D H=\langle a,c\rangle\leq D H = ⟨ a , c ⟩ ≤ D [20 Theorem 1(ii)] , H H H D D D H H H V V V H × { 1 } H\times\{1\} H × { 1 } H H H V V V H ⊴ c L ⊴ c V H\trianglelefteq_{c}L\trianglelefteq_{c}V H ⊴ c L ⊴ c V D × { 1 } ≅ D D\times\{1\}\cong D D × { 1 } ≅ D H ⊴ ( L ∩ D ) ⊴ D H\trianglelefteq(L\cap D)\trianglelefteq D H ⊴ ( L ∩ D ) ⊴ D H H H D D D D D D H H H V V V V ∉ A ^ V\notin\widehat{\mathcal{A}} V ∈ / A
Remark 5.3. A well-known fact about an abstract nilpotent group G G G i i i n ∈ N n\in\mathbb{N} n ∈ N a ∈ γ i a b s ( G ) a\in\gamma_{i}^{\mathrm{abs}}(G) a ∈ γ i abs ( G ) b ∈ G b\in G b ∈ G [ a , b n ] ≡ [ a , b ] n ( m o d γ i + 2 a b s ( G ) ) [a,b^{n}]\equiv[a,b]^{n}\pmod{\gamma_{i+2}^{\mathrm{abs}}(G)} [ a , b n ] ≡ [ a , b ] n ( mod γ i + 2 abs ( G )) [8 ] ).
Theorem 5.4. There exists a non-periodic 2-Engel profinite group W W W W ∈ A ^ W\in\widehat{\mathcal{A}} W ∈ A W ∉ T ^ W\notin\widehat{\mathcal{T}} W ∈ / T
Proof. Consider the abstract group G G G [3 ] and further analyzed by Stadelmann [33 ] and Mahdavianary [20 Theorem 1(iii)] , given by the presentation:
G = ⟨ x , y , z ∣ [ u , v , u ] = 1 for all u , v ∈ G , y 3 = [ z , y ] , z 9 = [ y , x ] 3 = [ z , y , x ] , [ z , x ] 3 = 1 ⟩ . G=\langle x,y,z\mid[u,v,u]=1\text{ for all }u,v\in G,\ y^{3}=[z,y],\ z^{9}=[y,x]^{3}=[z,y,x],\ [z,x]^{3}=1\rangle. G = ⟨ x , y , z ∣ [ u , v , u ] = 1 for all u , v ∈ G , y 3 = [ z , y ] , z 9 = [ y , x ] 3 = [ z , y , x ] , [ z , x ] 3 = 1 ⟩ . By definition, G G G [16 ] , any 2-Engel group is nilpotent of class at most 3. Moreover, the relation [ z , y , x ] = z 9 ≠ 1 [z,y,x] = z^9 \ne1 [ z , y , x ] = z 9 = 1 γ 3 ( G ) ≠ { 1 } \gamma_3(G) \ne\{1\} γ 3 ( G ) = { 1 } G G G G G G G a b = G / G ′ G^{\mathrm{ab}} = G/G' G ab = G / G ′ Z × C 3 × C 9 \mathbb{Z} \times C_3 \times C_9 Z × C 3 × C 9 G G G W = G ^ W = \widehat{G} W = G [26 ] ); since G G G
W a b = ( G ^ ) a b = G ^ [ G ^ , G ^ ] ≅ ( G [ G , G ] ) ^ = G a b ^ ≅ Z ^ × C 3 × C 9 . W^{\mathrm{ab}} = (\widehat{G})^{\mathrm{ab}} = \frac{\widehat{G}}{[\widehat{G},\widehat{G}]} \cong\left(\frac{G}{[G,G]}\right)^{\widehat{\ }} = \widehat{G^{\mathrm{ab}}} \cong\widehat{\mathbb{Z}} \times C_3 \times C_9. W ab = ( G ) ab = [ G , G ] G ≅ ( [ G , G ] G ) = G ab ≅ Z × C 3 × C 9 . Therefore the profinite group W W W
We now divide the remainder of the proof into three parts.
Part 1. We prove that W ∈ A W \in\mathcal{A} W ∈ A [ u , v , v ] = 1 [u,v,v] = 1 [ u , v , v ] = 1 G G G W = G ^ W = \widehat{G} W = G W ∈ T W \in\mathcal{T} W ∈ T W W W W ∈ A W \in\mathcal{A} W ∈ A
Part 2. Next, we prove that W ∉ S W \notin\mathcal{S} W ∈ / S H = ⟨ x 3 z , x 9 z ⟩ ≤ G H = \langle x^3z,x^9z\rangle\leq G H = ⟨ x 3 z , x 9 z ⟩ ≤ G [33 ] showed that H H H G G G H ‾ \overline{H} H H H H W W W H ‾ \overline{H} H W W W H ‾ ⊲ c H ‾ W ‾ ⊲ c W = G ^ \overline{H} \lhd_c \overline{\overline{H}^{W}} \lhd_c W = \widehat{G} H ⊲ c H W ⊲ c W = G G G G
( H ‾ ∩ G ) ⊲ ( H ‾ W ‾ ∩ G ) ⊲ G . (\overline{H}\cap G)\lhd\left(\overline{\overline{H}^{W}}\cap G\right)\lhd G. ( H ∩ G ) ⊲ ( H W ∩ G ) ⊲ G . Since G G G H ‾ W ∩ G = H ‾ W ‾ ∩ G = H G \overline{H}^{W}\cap G=\overline{\overline{H}^{W}}\cap G=H^{G} H W ∩ G = H W ∩ G = H G H H H G G G H ‾ ∩ G = H \overline{H}\cap G=H H ∩ G = H H ⊲ H G ⊲ G H\lhd H^{G}\lhd G H ⊲ H G ⊲ G H H H G G G H ‾ \overline{H} H W W W W = G ^ ∉ S W=\widehat{G}\notin\mathcal{S} W = G ∈ / S
Part 3. Finally, it remains to show that W W W G G G
Claim. γ 2 a b s ( G ) = [ G , G ] \gamma_2^{\mathrm{abs}}(G)=[G,G] γ 2 abs ( G ) = [ G , G ]
Indeed, since G G G
[ γ 2 a b s ( G ) , γ 2 a b s ( G ) ] ≤ γ 4 a b s ( G ) = { 1 } [\gamma_2^{\mathrm{abs}}(G),\gamma_2^{\mathrm{abs}}(G)]\leq\gamma_4^{\mathrm{abs}}(G)=\{1\} [ γ 2 abs ( G ) , γ 2 abs ( G )] ≤ γ 4 abs ( G ) = { 1 } by [27 Proposition 5.1.11] . Thus γ 2 a b s ( G ) \gamma_2^{\mathrm{abs}}(G) γ 2 abs ( G ) γ 3 a b s ( G ) \gamma_3^{\mathrm{abs}}(G) γ 3 abs ( G ) γ 2 a b s ( G ) \gamma_2^{\mathrm{abs}}(G) γ 2 abs ( G ) G ∈ T G\in\mathcal{T} G ∈ T [20 Theorem I (iii)] , it follows from [9 Theorem 1] that ( γ 3 a b s ( G ) ) 3 = { 1 } (\gamma_3^{\mathrm{abs}}(G))^3=\{1\} ( γ 3 abs ( G ) ) 3 = { 1 } γ 3 a b s ( G ) \gamma_3^{\mathrm{abs}}(G) γ 3 abs ( G ) x , y , z x,y,z x , y , z G G G [4 Lemma 2.6] . Hence γ 3 a b s ( G ) \gamma_3^{\mathrm{abs}}(G) γ 3 abs ( G ) [4 ] , the quotient γ 2 a b s ( G ) / γ 3 a b s ( G ) \gamma_2^{\mathrm{abs}}(G)/\gamma_3^{\mathrm{abs}}(G) γ 2 abs ( G ) / γ 3 abs ( G ) [ x , y ] [x,y] [ x , y ] [ y , z ] [y,z] [ y , z ] [ z , x ] [z,x] [ z , x ] γ 3 a b s ( G ) \gamma_3^{\mathrm{abs}}(G) γ 3 abs ( G ) G G G
[ x , z ] 3 = 1 , [ y , x ] 3 = [ z , y , x ] ≡ 1 ( m o d γ 3 a b s ( G ) ) . [x,z]^3=1,\qquad[y,x]^3=[z,y,x]\equiv1\pmod{\gamma_3^{\mathrm{abs}}(G)}. [ x , z ] 3 = 1 , [ y , x ] 3 = [ z , y , x ] ≡ 1 ( mod γ 3 abs ( G )) . Also, since y 3 = [ z , y ] ∈ γ 2 a b s ( G ) y^3=[z,y]\in\gamma_2^{\mathrm{abs}}(G) y 3 = [ z , y ] ∈ γ 2 abs ( G ) [ y 3 , z ] ∈ γ 3 a b s ( G ) [y^3,z]\in\gamma_3^{\mathrm{abs}}(G) [ y 3 , z ] ∈ γ 3 abs ( G ) a = y a=y a = y b = z b=z b = z i = 1 i=1 i = 1
[ y , z ] 3 ≡ [ y 3 , z ] = [ z , y , z ] ≡ 1 ( m o d γ 3 a b s ( G ) ) . [y,z]^3\equiv[y^3,z]=[z,y,z]\equiv1\pmod{\gamma_3^{\mathrm{abs}}(G)}. [ y , z ] 3 ≡ [ y 3 , z ] = [ z , y , z ] ≡ 1 ( mod γ 3 abs ( G )) . Hence every generator of γ 2 a b s ( G ) / γ 3 a b s ( G ) \gamma^{\mathrm{abs}}_2(G)/\gamma^{\mathrm{abs}}_3(G) γ 2 abs ( G ) / γ 3 abs ( G )
γ 2 a b s ( G ) / γ 3 a b s ( G ) ≅ C 3 s \gamma^{\mathrm{abs}}_2(G)/\gamma^{\mathrm{abs}}_3(G) \cong C_3^s γ 2 abs ( G ) / γ 3 abs ( G ) ≅ C 3 s for some s ≥ 0 s \ge0 s ≥ 0 γ 3 a b s ( G ) \gamma^{\mathrm{abs}}_3(G) γ 3 abs ( G ) γ 2 a b s ( G ) / γ 3 a b s ( G ) \gamma^{\mathrm{abs}}_2(G)/\gamma^{\mathrm{abs}}_3(G) γ 2 abs ( G ) / γ 3 abs ( G ) γ 2 a b s ( G ) \gamma^{\mathrm{abs}}_2(G) γ 2 abs ( G )
With the claim established, we now proceed to prove that W = G ^ W = \widehat{G} W = G
[ ⋅ , ⋅ ] : W × W → W , ( a , b ) ↦ [ a , b ] , [\,\cdot,\cdot\,]\colon W \times W \to W,\qquad(a,b) \mapsto[a,b], [ ⋅ , ⋅ ] : W × W → W , ( a , b ) ↦ [ a , b ] , is continuous, and G × G G \times G G × G W × W W \times W W × W
[ G ‾ , G ‾ ] = [ W ‾ , W ‾ ] = [ W , W ] , [\overline{G},\overline{G}] = [\overline{W},\overline{W}] = [W,W], [ G , G ] = [ W , W ] = [ W , W ] , where the last equality uses that [ W , W ] [W,W] [ W , W ] W W W [26 ] ). Since γ 2 a b s ( G ) = [ G , G ] \gamma^{\mathrm{abs}}_2(G) = [G,G] γ 2 abs ( G ) = [ G , G ] W W W
[ W , W ] = [ G ‾ , G ‾ ] = [ G , G ] = γ 2 a b s ( G ) . [W,W] = [\overline{G},\overline{G}] = [G,G] = \gamma^{\mathrm{abs}}_2(G). [ W , W ] = [ G , G ] = [ G , G ] = γ 2 abs ( G ) . Therefore [ W , W ] [W,W] [ W , W ] W W W w w w W a b ≅ Z ^ × C 3 × C 9 W^{\mathrm{ab}} \cong\widehat{\mathbb{Z}} \times C_3 \times C_9 W ab ≅ Z × C 3 × C 9 W a b W^{\mathrm{ab}} W ab w ∈ γ 2 ( W ) w \in\gamma_2(W) w ∈ γ 2 ( W ) γ 2 ( W ) \gamma_2(W) γ 2 ( W ) w 2 = 1 w^2 = 1 w 2 = 1 w = 1 w = 1 w = 1 W = G ^ W = \widehat{G} W = G G G G
In order to prove that the class T ^ \widehat{\mathcal{T}} T D D D
Lemma 5.5. In the group D D D ( a b ) 4 = b 4 (ab)^4 = b^4 ( ab ) 4 = b 4 ∣ a b ∣ = 8 |ab| = 8 ∣ ab ∣ = 8
Proof Recall from Remark 5.1 that from the relations in D = ⟨ a , b , c ⟩ D = \langle a,b,c\rangle D = ⟨ a , b , c ⟩ a 2 = b 8 = [ a , b ] 2 = 1 a^2 = b^8 = [a,b]^2 = 1 a 2 = b 8 = [ a , b ] 2 = 1 Z ( D ) = ⟨ b 2 ⟩ Z(D) = \langle b^2\rangle Z ( D ) = ⟨ b 2 ⟩ [ x 1 , x 2 , x 2 ] = 1 [x_1,x_2,x_2] = 1 [ x 1 , x 2 , x 2 ] = 1 x 1 , x 2 ∈ { a , b , c } x_1,x_2 \in\{a,b,c\} x 1 , x 2 ∈ { a , b , c } [ b , a , a ] = [ a , b , a ] = [ a , b , b ] = 1 [b,a,a] = [a,b,a] = [a,b,b] = 1 [ b , a , a ] = [ a , b , a ] = [ a , b , b ] = 1 [ a , b ] [a,b] [ a , b ] a a a b b b Lemma 2.1 (2), we expand ( a b ) 2 (ab)^2 ( ab ) 2
( a b ) 2 = a b a b = a 2 ( a − 1 b a b ) = 1 ⋅ a b = b [ b , a ] = b 2 [ b , a ] . (ab)^2 = abab = a^2(a^{-1}bab) = 1 \cdot{}^a b = b[b,a] = b^2[b,a]. ( ab ) 2 = abab = a 2 ( a − 1 bab ) = 1 ⋅ a b = b [ b , a ] = b 2 [ b , a ] . Squaring this result yields:
( a b ) 4 = ( b 2 [ b , a ] ) 2 = b 4 [ b , a ] 2 = b 4 ⋅ 1 = b 4 . (ab)^4 = \left(b^2[b,a]\right)^2 = b^4[b,a]^2 = b^4 \cdot1 = b^4. ( ab ) 4 = ( b 2 [ b , a ] ) 2 = b 4 [ b , a ] 2 = b 4 ⋅ 1 = b 4 . Since b 8 = 1 b^8 = 1 b 8 = 1 b 4 ≠ 1 b^4 \ne1 b 4 = 1 ( a b ) 4 ≠ 1 (ab)^4 \ne1 ( ab ) 4 = 1 ( a b ) 8 = 1 (ab)^8 = 1 ( ab ) 8 = 1 ∣ a b ∣ = 8 |ab| = 8 ∣ ab ∣ = 8
Lemma 5.6. In the group D D D [ c , b , a b ] = b 4 [c,b,ab] = b^4 [ c , b , ab ] = b 4
Proof Applying Lemma 2.1 (5), we expand the commutator:
[ c , b , a b ] = [ [ c , b ] , a b ] = [ [ c , b ] , b ] ⋅ [ [ c , b ] , a ] b . [c,b,ab] = [[c,b],ab] = [[c,b],b]\cdot[[c,b],a]^b. [ c , b , ab ] = [[ c , b ] , ab ] = [[ c , b ] , b ] ⋅ [[ c , b ] , a ] b . Since [ x 1 , x 2 , x 2 ] = 1 [x_1,x_2,x_2] = 1 [ x 1 , x 2 , x 2 ] = 1 [ c , b , b ] = 1 [c,b,b] = 1 [ c , b , b ] = 1 [ b , c ] 2 = 1 [b,c]^2 = 1 [ b , c ] 2 = 1 [ c , b ] = [ b , c ] − 1 = [ b , c ] [c,b] = [b,c]^{-1} = [b,c] [ c , b ] = [ b , c ] − 1 = [ b , c ]
[ b , c , a ] = b 4 [b,c,a]=b^{4} [ b , c , a ] = b 4 [ D , D , D ] ≤ Z ( D ) [D,D,D]\leq Z(D) [ D , D , D ] ≤ Z ( D ) b 4 b^{4} b 4 b b b
[ c , b , a b ] = ( b 4 ) b = b 4 . [c,b,ab]=(b^{4})^{b}=b^{4}. [ c , b , ab ] = ( b 4 ) b = b 4 . Theorem 5.7. The class T ^ \widehat{\mathcal{T}} T
Proof Let U = ∏ i ∈ N D i U=\prod_{i\in\mathbb{N}}D_i U = ∏ i ∈ N D i D D D D i ≅ D D_i\cong D D i ≅ D i i i D i D_i D i T ^ \widehat{\mathcal{T}} T T \mathcal{T} T D i ∈ T ^ D_i\in\widehat{\mathcal{T}} D i ∈ T T \mathcal{T} T a i a_i a i b i b_i b i c i c_i c i D i D_i D i U U U ⟨ u ⟩ = ⟨ u ⟩ ^ \langle u\rangle=\widehat{\langle u\rangle} ⟨ u ⟩ = ⟨ u ⟩ u ∈ U u\in U u ∈ U
Define two elements h , g ∈ U h,g\in U h , g ∈ U
h = ( a 1 b 1 , b 2 , 1 , 1 , … ) , h=(a_1b_1,b_2,1,1,\ldots), h = ( a 1 b 1 , b 2 , 1 , 1 , … ) , g = ( c 1 , a 2 , 1 , 1 , … ) . g=(c_1,a_2,1,1,\ldots). g = ( c 1 , a 2 , 1 , 1 , … ) . Suppose for a contradiction that U ∈ T ^ U\in\widehat{\mathcal{T}} U ∈ T Lemma 3.8 , [ g , h , h ] ∈ ⟨ h ⟩ [g,h,h]\in\langle h\rangle [ g , h , h ] ∈ ⟨ h ⟩ [ g , h , h ] [g,h,h] [ g , h , h ]
[ g , h , h ] = ( [ c 1 , a 1 b 1 , a 1 b 1 ] , [ a 2 , b 2 , b 2 ] , 1 , 1 , … ) . [g,h,h]=([c_1,a_1b_1,a_1b_1],[a_2,b_2,b_2],1,1,\ldots). [ g , h , h ] = ([ c 1 , a 1 b 1 , a 1 b 1 ] , [ a 2 , b 2 , b 2 ] , 1 , 1 , … ) . In the second coordinate, the identity [ x 1 , x 2 , x 2 ] = 1 [x_1,x_2,x_2]=1 [ x 1 , x 2 , x 2 ] = 1 [ a 2 , b 2 , b 2 ] = 1 [a_2,b_2,b_2]=1 [ a 2 , b 2 , b 2 ] = 1 Lemma 2.1 (5):
[ c 1 , a 1 b 1 ] = [ c 1 , b 1 ] ⋅ [ c 1 , a 1 ] b 1 . [c_1,a_1b_1]=[c_1,b_1]\cdot[c_1,a_1]^{b_1}. [ c 1 , a 1 b 1 ] = [ c 1 , b 1 ] ⋅ [ c 1 , a 1 ] b 1 . Since [ c 1 , a 1 ] = 1 [c_1,a_1]=1 [ c 1 , a 1 ] = 1 [ c 1 , a 1 b 1 ] = [ c 1 , b 1 ] [c_1,a_1b_1]=[c_1,b_1] [ c 1 , a 1 b 1 ] = [ c 1 , b 1 ]
[ c 1 , a 1 b 1 , a 1 b 1 ] = [ [ c 1 , b 1 ] , a 1 b 1 ] = [ c 1 , b 1 , a 1 b 1 ] . [c_1,a_1b_1,a_1b_1]=[[c_1,b_1],a_1b_1]=[c_1,b_1,a_1b_1]. [ c 1 , a 1 b 1 , a 1 b 1 ] = [[ c 1 , b 1 ] , a 1 b 1 ] = [ c 1 , b 1 , a 1 b 1 ] . By Lemma 5.6 , [ c 1 , b 1 , a 1 b 1 ] = b 1 4 [c_1,b_1,a_1b_1]=b_1^{4} [ c 1 , b 1 , a 1 b 1 ] = b 1 4 [ g , h , h ] = ( b 1 4 , 1 , 1 , … ) [g,h,h]=(b_1^{4},1,1,\ldots) [ g , h , h ] = ( b 1 4 , 1 , 1 , … ) [ g , h , h ] ∈ ⟨ h ⟩ [g,h,h]\in\langle h\rangle [ g , h , h ] ∈ ⟨ h ⟩ k k k [ g , h , h ] = h k [g,h,h]=h^k [ g , h , h ] = h k
( b 1 4 , 1 , 1 , … ) = h k = ( ( a 1 b 1 ) k , b 2 k , 1 , … ) . (b_1^{4},1,1,\ldots)=h^k=((a_1b_1)^k,b_2^k,1,\ldots). ( b 1 4 , 1 , 1 , … ) = h k = (( a 1 b 1 ) k , b 2 k , 1 , … ) . Comparing coordinates, we obtain b 2 k = 1 b_2^k=1 b 2 k = 1 ( a 1 b 1 ) k = b 1 4 (a_1b_1)^k=b_1^4 ( a 1 b 1 ) k = b 1 4 ∣ b 2 ∣ = 8 |b_2|=8 ∣ b 2 ∣ = 8 b 2 k = 1 b_2^k=1 b 2 k = 1 k k k Lemma 5.5 , ∣ a 1 b 1 ∣ = 8 |a_1b_1|=8 ∣ a 1 b 1 ∣ = 8 k k k ( a 1 b 1 ) k = 1 (a_1b_1)^k=1 ( a 1 b 1 ) k = 1 b 1 4 = 1 b_1^4=1 b 1 4 = 1 D 1 D_1 D 1 b 1 4 ≠ 1 b_1^4\ne1 b 1 4 = 1 [ g , h , h ] ∉ ⟨ h ⟩ [g,h,h]\notin\langle h\rangle [ g , h , h ] ∈ / ⟨ h ⟩ U ∉ T ^ U\notin\widehat{\mathcal{T}} U ∈ / T
Note that the group U U U P ^ \widehat{\mathcal{P}} P A ^ \widehat{\mathcal{A}} A
Remark 5.8. The failure of closure under Cartesian products is not restricted to infinite products; it also occurs for finite direct products. Consider the generalized quaternion group of order 16, defined by
Q 16 = ⟨ a , b ∣ a 8 = 1 , b 2 = a 4 , a b = a − 1 ⟩ . Q_{16}=\langle a,b\mid a^{8}=1,\ b^{2}=a^{4},\ ab=a^{-1}\rangle. Q 16 = ⟨ a , b ∣ a 8 = 1 , b 2 = a 4 , ab = a − 1 ⟩ . By Example 4.6 , Q 16 ∈ P ⊂ A ⊂ T Q_{16}\in\mathcal{P}\subset\mathcal{A}\subset\mathcal{T} Q 16 ∈ P ⊂ A ⊂ T Q 16 Q_{16} Q 16 P ^ ⊂ A ^ ⊂ T ^ \widehat{\mathcal{P}}\subset\widehat{\mathcal{A}}\subset\widehat{\mathcal{T}} P ⊂ A ⊂ T
Yet, Q 16 × Q 16 ∉ T Q_{16} \times Q_{16} \notin\mathcal{T} Q 16 × Q 16 ∈ / T g = ( a , 1 ) g=(a,1) g = ( a , 1 ) h = ( b , b ) h=(b,b) h = ( b , b ) Q 16 × Q 16 Q_{16} \times Q_{16} Q 16 × Q 16 [20 Lemma 1] , we must have [ g , h , h ] ∈ ⟨ h ⟩ [g,h,h] \in\langle h\rangle [ g , h , h ] ∈ ⟨ h ⟩ [ g , h , h ] = ( [ a , b , b ] , 1 ) [g,h,h]=([a,b,b],1) [ g , h , h ] = ([ a , b , b ] , 1 ) Example 4.6 , [ a , b , b ] = a 4 ≠ 1 [a,b,b]=a^4 \ne1 [ a , b , b ] = a 4 = 1 [ g , h , h ] = ( a 4 , 1 ) [g,h,h]=(a^4,1) [ g , h , h ] = ( a 4 , 1 )
⟨ ( b , b ) ⟩ = { ( 1 , 1 ) , ( b , b ) , ( a 4 , a 4 ) , ( a 4 b , a 4 b ) } , \langle(b,b)\rangle=\{(1,1),(b,b),(a^4,a^4),(a^4b,a^4b)\}, ⟨( b , b )⟩ = {( 1 , 1 ) , ( b , b ) , ( a 4 , a 4 ) , ( a 4 b , a 4 b )} , which does not contain ( a 4 , 1 ) (a^4,1) ( a 4 , 1 ) [ g , h , h ] ∉ ⟨ h ⟩ = ⟨ ( b , b ) ⟩ [g,h,h]\notin\langle h\rangle=\langle(b,b)\rangle [ g , h , h ] ∈ / ⟨ h ⟩ = ⟨( b , b )⟩ Q 16 × Q 16 ∉ T Q_{16} \times Q_{16} \notin\mathcal{T} Q 16 × Q 16 ∈ / T S c \mathcal{S}_c S c A c \mathcal{A}_c A c T \mathcal{T} T
Acknowledgements
The first author is grateful for the support from UFMG in his Post-Doctoral internship at this institution and for financial support from FAPDF. The second and third authors were partially supported by FAPDF and the last author is also grateful for the support from CNPq and for partial support from CAPES during her PhD.
