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Strong Lie derived length of group algebras vs. derived length of their group of units

Tibor Juhász

Institute of Mathematics and Informatics Eszterházy Károly University Eger, Hungary

Groups, Rings and the Yang-Baxter equation Spa, June 18–24, 2017 This research was supported by the European Union and the State of Hungary, co-financed by the European Social Fund in the framework of TÁMOP 4.2.4. A/2-11-1-2012-0001 ’National Excellence Program’

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 1 / 14

What we want to do Let R be an associative ring with unity, and U := U(R) be its group of units, and δ0(U) = U δ1(U) = U′ = (U, U) . . . δi(U) = (δi−1(U), δi−1(U)) U is said to be solvable if δn(U) = 1 for some n; the smallest such n is denoted by dl(U) and called the derived length of U. For x, y ∈ R set [x, y] = xy − yx. R is called strongly Lie solvable if δ(n)(R) = 0 for some n; the smallest such n is denoted by dlL(R) and called the strong Lie derived length of R. If x, y are units, then (x, y) = 1 + x−1y−1[x, y]. If R is strongly Lie solvable, then U is solvable with dl(U) ≤ dlL(R).

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 2 / 14

What we want to do Let R be an associative ring with unity, and U := U(R) be its group of units, and δ0(U) = U δ1(U) = U′ = (U, U) . . . δi(U) = (δi−1(U), δi−1(U)) U is said to be solvable if δn(U) = 1 for some n; the smallest such n is denoted by dl(U) and called the derived length of U. For x, y ∈ R set [x, y] = xy − yx. R is called strongly Lie solvable if δ(n)(R) = 0 for some n; the smallest such n is denoted by dlL(R) and called the strong Lie derived length of R. If x, y are units, then (x, y) = 1 + x−1y−1[x, y]. If R is strongly Lie solvable, then U is solvable with dl(U) ≤ dlL(R).

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 2 / 14

What we want to do Let R be an associative ring with unity, and U := U(R) be its group of units, and δ0(U) = U δ1(U) = U′ = (U, U) . . . δi(U) = (δi−1(U), δi−1(U)) U is said to be solvable if δn(U) = 1 for some n; the smallest such n is denoted by dl(U) and called the derived length of U. For x, y ∈ R set [x, y] = xy − yx. R is called strongly Lie solvable if δ(n)(R) = 0 for some n; the smallest such n is denoted by dlL(R) and called the strong Lie derived length of R. If x, y are units, then (x, y) = 1 + x−1y−1[x, y]. If R is strongly Lie solvable, then U is solvable with dl(U) ≤ dlL(R).

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 2 / 14

What we want to do Let R be an associative ring with unity, and U := U(R) be its group of units, and δ0(U) = U δ(0)(R) = R δ1(U) = U′ = (U, U) δ(1)(R) = [R, R]R . . . . . . δi(U) = (δi−1(U), δi−1(U)) δ(i)(R) = [δ(i−1)(R), δ(i−1)(R)]R U is said to be solvable if δn(U) = 1 for some n; the smallest such n is denoted by dl(U) and called the derived length of U. For x, y ∈ R set [x, y] = xy − yx. R is called strongly Lie solvable if δ(n)(R) = 0 for some n; the smallest such n is denoted by dlL(R) and called the strong Lie derived length of R. If x, y are units, then (x, y) = 1 + x−1y−1[x, y]. If R is strongly Lie solvable, then U is solvable with dl(U) ≤ dlL(R).

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 2 / 14

What we want to do Let R be an associative ring with unity, and U := U(R) be its group of units, and δ0(U) = U δ(0)(R) = R δ1(U) = U′ = (U, U) δ(1)(R) = [R, R]R . . . . . . δi(U) = (δi−1(U), δi−1(U)) δ(i)(R) = [δ(i−1)(R), δ(i−1)(R)]R U is said to be solvable if δn(U) = 1 for some n; the smallest such n is denoted by dl(U) and called the derived length of U. For x, y ∈ R set [x, y] = xy − yx. R is called strongly Lie solvable if δ(n)(R) = 0 for some n; the smallest such n is denoted by dlL(R) and called the strong Lie derived length of R. If x, y are units, then (x, y) = 1 + x−1y−1[x, y]. If R is strongly Lie solvable, then U is solvable with dl(U) ≤ dlL(R).

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 2 / 14

What we want to do Let R be an associative ring with unity, and U := U(R) be its group of units, and δ0(U) = U δ(0)(R) = R δ1(U) = U′ = (U, U) δ(1)(R) = [R, R]R . . . . . . δi(U) = (δi−1(U), δi−1(U)) δ(i)(R) = [δ(i−1)(R), δ(i−1)(R)]R U is said to be solvable if δn(U) = 1 for some n; the smallest such n is denoted by dl(U) and called the derived length of U. For x, y ∈ R set [x, y] = xy − yx. R is called strongly Lie solvable if δ(n)(R) = 0 for some n; the smallest such n is denoted by dlL(R) and called the strong Lie derived length of R. If x, y are units, then (x, y) = 1 + x−1y−1[x, y]. If R is strongly Lie solvable, then U is solvable with dl(U) ≤ dlL(R).

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 2 / 14

What we want to do Let R be an associative ring with unity, and U := U(R) be its group of units, and δ0(U) = U δ1(U) = U′ = (U, U) ⊆ 1 + δ(1)(R) . . . . . . δi(U) = (δi−1(U), δi−1(U)) ⊆ 1 + δ(i)(R) U is said to be solvable if δn(U) = 1 for some n; the smallest such n is denoted by dl(U) and called the derived length of U. For x, y ∈ R set [x, y] = xy − yx. R is called strongly Lie solvable if δ(n)(R) = 0 for some n; the smallest such n is denoted by dlL(R) and called the strong Lie derived length of R. If x, y are units, then (x, y) = 1 + x−1y−1[x, y]. If R is strongly Lie solvable, then U is solvable with dl(U) ≤ dlL(R).

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 2 / 14

What we want to do Let R be an associative ring with unity, and U := U(R) be its group of units, and δ0(U) = U δ1(U) = U′ = (U, U) ⊆ 1 + δ(1)(R) . . . . . . δi(U) = (δi−1(U), δi−1(U)) ⊆ 1 + δ(i)(R) U is said to be solvable if δn(U) = 1 for some n; the smallest such n is denoted by dl(U) and called the derived length of U. For x, y ∈ R set [x, y] = xy − yx. R is called strongly Lie solvable if δ(n)(R) = 0 for some n; the smallest such n is denoted by dlL(R) and called the strong Lie derived length of R. If x, y are units, then (x, y) = 1 + x−1y−1[x, y]. If R is strongly Lie solvable, then U is solvable with dl(U) ≤ dlL(R).

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 2 / 14

A third series Let δ[0](R) = R, and for i ≥ 1, let δ[i](R) be the additive subgroup of R generated by all Lie commutators [x, y] with x, y ∈ δ[i−1](R). R is called Lie solvable if δ[n](R) = 0 for some n; the smallest such n is denoted by dlL(R) and called the Lie derived length of R. If R is strongly Lie solvable, then R is Lie solvable with dlL(R) ≤ dlL(R). Whether is there any relation between dl(U) and dlL(R)?

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 3 / 14

A third series Let δ[0](R) = R, and for i ≥ 1, let δ[i](R) be the additive subgroup of R generated by all Lie commutators [x, y] with x, y ∈ δ[i−1](R). R is called Lie solvable if δ[n](R) = 0 for some n; the smallest such n is denoted by dlL(R) and called the Lie derived length of R. If R is strongly Lie solvable, then R is Lie solvable with dlL(R) ≤ dlL(R). Whether is there any relation between dl(U) and dlL(R)?

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 3 / 14

A third series Let δ[0](R) = R, and for i ≥ 1, let δ[i](R) be the additive subgroup of R generated by all Lie commutators [x, y] with x, y ∈ δ[i−1](R). R is called Lie solvable if δ[n](R) = 0 for some n; the smallest such n is denoted by dlL(R) and called the Lie derived length of R. If R is strongly Lie solvable, then R is Lie solvable with dlL(R) ≤ dlL(R). Whether is there any relation between dl(U) and dlL(R)?

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 3 / 14

A third series Let δ[0](R) = R, and for i ≥ 1, let δ[i](R) be the additive subgroup of R generated by all Lie commutators [x, y] with x, y ∈ δ[i−1](R). R is called Lie solvable if δ[n](R) = 0 for some n; the smallest such n is denoted by dlL(R) and called the Lie derived length of R. If R is strongly Lie solvable, then R is Lie solvable with dlL(R) ≤ dlL(R). Whether is there any relation between dl(U) and dlL(R)?

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 3 / 14

Strongly Lie solvable group algebras Write FG for the group algebra (or group ring) of a group G over a field F. Passi-Passman-Sehgal, 1973: FG is strongly Lie solvable iff either G is abelian, or char F = p and G′ is a finite p-group. Motose-Tominaga, Bateman, Bovdi-Khripta, Passman, Taylor, Sehgal, Bovdi (1968–2005): Classification of group algebras with solvable group of units Bovdi, 2005: Let char(F) = p > 3 and G be a group with a nontrivial p-Sylow subgroup P such that if G is non-torsion, then P is infinite. Then U(FG) is a solvable iff FG is strongly Lie solvable. In the sequel we suppose that FG is strongly Lie solvable. Then dlL(FG) ≤ dlL(FG) and dl(U(FG)) ≤ dlL(FG).

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 4 / 14

Strongly Lie solvable group algebras Write FG for the group algebra (or group ring) of a group G over a field F. Passi-Passman-Sehgal, 1973: FG is strongly Lie solvable iff either G is abelian, or char F = p and G′ is a finite p-group. Motose-Tominaga, Bateman, Bovdi-Khripta, Passman, Taylor, Sehgal, Bovdi (1968–2005): Classification of group algebras with solvable group of units Bovdi, 2005: Let char(F) = p > 3 and G be a group with a nontrivial p-Sylow subgroup P such that if G is non-torsion, then P is infinite. Then U(FG) is a solvable iff FG is strongly Lie solvable. In the sequel we suppose that FG is strongly Lie solvable. Then dlL(FG) ≤ dlL(FG) and dl(U(FG)) ≤ dlL(FG).

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 4 / 14

Strongly Lie solvable group algebras Write FG for the group algebra (or group ring) of a group G over a field F. Passi-Passman-Sehgal, 1973: FG is strongly Lie solvable iff either G is abelian, or char F = p and G′ is a finite p-group. Motose-Tominaga, Bateman, Bovdi-Khripta, Passman, Taylor, Sehgal, Bovdi (1968–2005): Classification of group algebras with solvable group of units Bovdi, 2005: Let char(F) = p > 3 and G be a group with a nontrivial p-Sylow subgroup P such that if G is non-torsion, then P is infinite. Then U(FG) is a solvable iff FG is strongly Lie solvable. In the sequel we suppose that FG is strongly Lie solvable. Then dlL(FG) ≤ dlL(FG) and dl(U(FG)) ≤ dlL(FG).

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 4 / 14

Strongly Lie solvable group algebras Write FG for the group algebra (or group ring) of a group G over a field F. Passi-Passman-Sehgal, 1973: FG is strongly Lie solvable iff either G is abelian, or char F = p and G′ is a finite p-group. Motose-Tominaga, Bateman, Bovdi-Khripta, Passman, Taylor, Sehgal, Bovdi (1968–2005): Classification of group algebras with solvable group of units Bovdi, 2005: Let char(F) = p > 3 and G be a group with a nontrivial p-Sylow subgroup P such that if G is non-torsion, then P is infinite. Then U(FG) is a solvable iff FG is strongly Lie solvable. In the sequel we suppose that FG is strongly Lie solvable. Then dlL(FG) ≤ dlL(FG) and dl(U(FG)) ≤ dlL(FG).

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 4 / 14

When the derived length is at most 2 Levin-Rosenberger, 1986: FG is (strongly) Lie metabelian iff one of the following conditions holds: G is abelian; char(F) = 3, and G′ is central of order 3; char(F) = 2, and G′ is central elementary abelian of order dividing 4. Shalev, 1991: Let char(F) > 2 and G be a group. U(FG) is metabelian iff FG is (strongly) Lie metabelian. Kurdics, 1996; Coleman-Sandling, 1998: Let char(F) = 2 and G be a and nilpotent group. U(FG) is metabelian iff FG is (strongly) Lie metabelian.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 5 / 14

When the derived length is at most 2 Levin-Rosenberger, 1986: FG is (strongly) Lie metabelian iff one of the following conditions holds: G is abelian; char(F) = 3, and G′ is central of order 3; char(F) = 2, and G′ is central elementary abelian of order dividing 4. Shalev, 1991: Let char(F) > 2 and G be a finite group. U(FG) is metabelian iff FG is (strongly) Lie metabelian. Kurdics, 1996; Coleman-Sandling, 1998: Let char(F) = 2 and G be a and nilpotent group. U(FG) is metabelian iff FG is (strongly) Lie metabelian.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 5 / 14

When the derived length is at most 2 Levin-Rosenberger, 1986: FG is (strongly) Lie metabelian iff one of the following conditions holds: G is abelian; char(F) = 3, and G′ is central of order 3; char(F) = 2, and G′ is central elementary abelian of order dividing 4. Shalev, 1991: Let char(F) > 2 and G be a finite group. U(FG) is metabelian iff FG is (strongly) Lie metabelian. Kurdics, 1996; Coleman-Sandling, 1998: Let char(F) = 2 and G be a finite and nilpotent group. U(FG) is metabelian iff FG is (strongly) Lie metabelian.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 5 / 14

When the derived length is at most 2 Levin-Rosenberger, 1986: FG is (strongly) Lie metabelian iff one of the following conditions holds: G is abelian; char(F) = 3, and G′ is central of order 3; char(F) = 2, and G′ is central elementary abelian of order dividing 4. Shalev, 1991: Let char(F) > 2 and G be a finite group. U(FG) is metabelian iff FG is (strongly) Lie metabelian. Kurdics, 1996; Coleman-Sandling, 1998: Let char(F) = 2 and G be a finite and nilpotent group. U(FG) is metabelian iff FG is (strongly) Lie metabelian. (Exception for non-nilpotent G.)

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 5 / 14

When the derived length is at most 2 Levin-Rosenberger, 1986: FG is (strongly) Lie metabelian iff one of the following conditions holds: G is abelian; char(F) = 3, and G′ is central of order 3; char(F) = 2, and G′ is central elementary abelian of order dividing 4. Catino-Spinelli, 2010: Let char(F) > 2 and G be a torsion group. U(FG) is metabelian iff FG is (strongly) Lie metabelian. Catino-Spinelli, 2010: Let char(F) = 2 and G be a torsion and nilpotent group. U(FG) is metabelian iff FG is (strongly) Lie metabelian.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 5 / 14

The derived length of the group of units Baginski, 2002: If char(F) = p > 2, and G is a finite p-group with cyclic commutator subgroup, then dl(U(FG)) = ⌈log2(|G′| + 1)⌉. Questions: What happens if G is not a finite p-group? What happens for p = 2? How much is dlL(FG) in these cases?

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 6 / 14

The derived length of the group of units Baginski, 2002: If char(F) = p > 2, and G is a finite p-group with cyclic commutator subgroup, then dl(U(FG)) = ⌈log2(|G′| + 1)⌉. Questions: What happens if G is not a finite p-group? What happens for p = 2? How much is dlL(FG) in these cases?

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 6 / 14

Lie derived lengths Theorem (Balogh-J, 2008) Let FG be a (strongly) Lie solvable group algebra. If G′ is cyclic of order pn, where p > 2, and G/CG(G′) has order 2mprs, where (2p, s) = 1, then dlL(FG) = dlL(FG) = ⌈log2 2pnνm⌉, where νm = 1 if s > 1, otherwise νm = 1 −

1 2m+1 .

G is nilpotent ⇒ G/CG(G′) is a p-group ⇒ m = 0, s = 1 ⇒ ν0 = 1/2 ⇒ dlL(FG) = dlL(FG) = ⌈log2 pn⌉ = ⌈log2(pn + 1)⌉ = ⌈log2(|G′| + 1)⌉

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 7 / 14

Lie derived lengths Theorem (Balogh-J, 2008) Let FG be a (strongly) Lie solvable group algebra. If G′ is cyclic of order pn, where p > 2, and G/CG(G′) has order 2mprs, where (2p, s) = 1, then dlL(FG) = dlL(FG) = ⌈log2 2pnνm⌉, where νm = 1 if s > 1, otherwise νm = 1 −

1 2m+1 .

G is nilpotent ⇒ G/CG(G′) is a p-group ⇒ m = 0, s = 1 ⇒ ν0 = 1/2 ⇒ dlL(FG) = dlL(FG) = ⌈log2 pn⌉ = ⌈log2(pn + 1)⌉ = ⌈log2(|G′| + 1)⌉

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 7 / 14

The derived length of the group of units Corollary If char(F) = p > 2, and G is a finite p-group with cyclic commutator subgroup, then dl(U(FG)) = dlL(FG) = dlL(FG) = ⌈log2(|G′| + 1)⌉ Balogh-Li, 2007: If G is torsion, or non-nilpotent, and G′ is a cyclic p-group, then dl(U(FG)) = dlL(FG) = dlL(FG). Theorem (J, 2016) dl(U(FG)) does not always equal to dlL(FG) for nilpotent and non-torsion G.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 8 / 14

The derived length of the group of units Corollary If char(F) = p > 2, and G is a finite p-group with cyclic commutator subgroup, then dl(U(FG)) = dlL(FG) = dlL(FG) = ⌈log2(|G′| + 1)⌉ Balogh-Li, 2007: If G is torsion, or non-nilpotent, and G′ is a cyclic p-group, then dl(U(FG)) = dlL(FG) = dlL(FG). Theorem (J, 2016) dl(U(FG)) does not always equal to dlL(FG) for nilpotent and non-torsion G.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 8 / 14

The derived length of the group of units Corollary If char(F) = p > 2, and G is a finite p-group with cyclic commutator subgroup, then dl(U(FG)) = dlL(FG) = dlL(FG) = ⌈log2(|G′| + 1)⌉ Balogh-Li, 2007: If G is torsion, or non-nilpotent, and G′ is a cyclic p-group, then dl(U(FG)) = dlL(FG) = dlL(FG). Theorem (J, 2016) dl(U(FG)) does not always equal to dlL(FG) for nilpotent and non-torsion G.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 8 / 14

When the derived length may be smaller Theorem (J, 2016) Let G be a nilpotent group whose commutator subgroup is finite abelian, and let char(F) = p > 2. If G′ = Sylp(G), and γ3(G) ⊆ (G′)p, then dl(U(FG)) ≤

• log2

2 3

• t(G′) + 1
• .

Furthermore, if G′ is cyclic, then the equality holds. Theorem (J, 2006) Let G be a nilpotent group with G′ a finite p-group, and let char(F) = p. If γ3(G) ⊆ (G′)p, then dlL(FG) = ⌈log2(t(G′) + 1)⌉.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 9 / 14

When the derived length may be smaller Theorem (J, 2016) Let G be a nilpotent group whose commutator subgroup is finite abelian, and let char(F) = p > 2. If G′ = Sylp(G), and γ3(G) ⊆ (G′)p, then dl(U(FG)) ≤

• log2

2 3

• t(G′) + 1
• .

Furthermore, if G′ is cyclic, then the equality holds. Theorem (J, 2006) Let G be a nilpotent group with G′ a finite p-group, and let char(F) = p. If γ3(G) ⊆ (G′)p, then dlL(FG) = ⌈log2(t(G′) + 1)⌉.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 9 / 14

When the derived length may be smaller Theorem (J, 2016) Let G be a nilpotent group whose commutator subgroup is finite abelian, and let char(F) = p > 2. If G′ = Sylp(G), and γ3(G) ⊆ (G′)p, then dl(U(FG)) ≤

• log2

2 3

• t(G′) + 1
• .

Furthermore, if G′ is cyclic, then the equality holds. Theorem (J, 2006) Let G be a nilpotent group with G′ a finite p-group, and let char(F) = p. If γ3(G) ⊆ (G′)p, then dlL(FG) = ⌈log2(t(G′) + 1)⌉.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 9 / 14

Examples Example Let G = a, b, c | c5 = 1, b−1ab = ac, ac = ca, bc = cb, and char(F) = 5. Then dl(U(FG)) = 2, but dlL(FG) = dlL(FG) = 3. Example Let G′ = Syl3(G) ∼ = C3 × C3, G′ is central, and let char(F) = 3. Then dl(U(FG)) = 2, but dlL(FG) = dlL(FG) = 3. Corollary U(FG) can be metabelian, even if FG is strongly Lie solvable, but not (strongly) Lie metabelian.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 10 / 14

Open question We still don’t know dl(U(FG)) when G is nilpotent, non-torsion, and G′ is cyclic p-group (p = char(F) is odd), but G′ = Sylp(G). All we can say about it at present is

• log2

2 3(|G′| + 1)

• ≤ dl(U(FG)) ≤ ⌈log2(|G′| + 1)⌉.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 11 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 25 23 26 24 27 25 28 26

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 25 23 26 24 27 25 28 26 Konovalov, Rossmanith, . . . : If char(F) = 2 and G has an abelian subgroup of index 2, then dlL(FG) ≤ 3.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 3 25 23 3 26 24 3 27 25 3 28 26 3 Konovalov, Rossmanith, . . . : If char(F) = 2 and G has an abelian subgroup of index 2, then dlL(FG) ≤ 3.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 3 25 23 3 26 24 3 27 25 3 28 26 3 Theorem (J, 2006) Let G be a nilpotent group with G′ a finite p-group, and let char(F) = p. If γ3(G) ⊆ (G′)p, then dlL(FG) = ⌈log2(t(G′) + 1)⌉.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 3 3 25 23 3 4 26 24 3 5 27 25 3 6 28 26 3 7 Theorem (J, 2006) Let G be a nilpotent group with G′ a finite p-group, and let char(F) = p. If γ3(G) ⊆ (G′)p, then dlL(FG) = ⌈log2(t(G′) + 1)⌉.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 3 3 25 23 3 4 26 24 3 5 27 25 3 6 28 26 3 7 Let’s play!

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 3 3 3 25 23 3 4 26 24 3 5 27 25 3 6 28 26 3 7 Let’s play!

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 3 3 3 25 23 3 4 4 26 24 3 5 27 25 3 6 28 26 3 7 Let’s play!

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 3 3 3 25 23 3 4 4 26 24 3 5 4 27 25 3 6 28 26 3 7 Let’s play!

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 3 3 3 25 23 3 4 4 26 24 3 5 4 27 25 3 6 5 28 26 3 7 Let’s play!

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 3 3 3 25 23 3 4 4 26 24 3 5 4 27 25 3 6 5 28 26 3 7 6 Let’s play!

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 3 3 3 25 23 3 4 4 26 24 3 5 4 27 25 3 6 5 28 26 3 7 6 Theorem (J-Kurdics, 2017) Let char(F) = 2, and let G be a group with cyclic commutator subgroup of order 2n, and assume that cl(G) = n + 1. Then, for n ≥ 4, we have dl(U(FG)) < dlL(FG) = n + 1.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Write Dk for the dihedral group of order k, and F for a field of characteristic 2. k |D′

k|

dlL(FDk) dlL(FDk) dl(U(FDk)) 23 21 2 2 2 24 22 3 3 3 25 23 3 4 4 26 24 3 5 4 27 25 3 6 5 28 26 3 7 6 Theorem (J-Kurdics, 2017) Let char(F) = 2, and let G be a group with cyclic commutator subgroup of order 2n, and assume that cl(G) = n + 1. Then, for n ≥ 4, we have dl(U(FG)) < dlL(FG) = n + 1.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 12 / 14

The case p = 2 Theorem (J, 2016) Let char(F) = 2, and let G be a group with cyclic commutator subgroup of order 2n, where n > 1, and assume that cl(G) ≤ n. Then either dl(U(FG)) = dlL(FG) = n + 1,

• r

dl(U(FG)) = dlL(FG) − 1 = n. Furthermore, if G′ = Syl2(G), then dl(U(FG)) = n < dlL(FG). Corollary For non-torsion G, U(FG) can be metabelian, even if G′ is cyclic of order 4, that is when FG is strongly Lie solvable, but not (strongly) Lie metabelian.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 13 / 14

The case p = 2 Theorem (J, 2016) Let char(F) = 2, and let G be a group with cyclic commutator subgroup of order 2n, where n > 1, and assume that cl(G) ≤ n. Then either dl(U(FG)) = dlL(FG) = n + 1,

• r

dl(U(FG)) = dlL(FG) − 1 = n. Furthermore, if G′ = Syl2(G), then dl(U(FG)) = n < dlL(FG). Corollary For non-torsion G, U(FG) can be metabelian, even if G′ is cyclic of order 4, that is when FG is strongly Lie solvable, but not (strongly) Lie metabelian.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 13 / 14

The case p = 2 Theorem (J, 2016) Let char(F) = 2, and let G be a group with cyclic commutator subgroup of order 2n, where n > 1, and assume that cl(G) ≤ n. Then either dl(U(FG)) = dlL(FG) = n + 1,

• r

dl(U(FG)) = dlL(FG) − 1 = n. Furthermore, if G′ = Syl2(G), then dl(U(FG)) = n < dlL(FG). Corollary For non-torsion G, U(FG) can be metabelian, even if G′ is cyclic of order 4, that is when FG is strongly Lie solvable, but not (strongly) Lie metabelian.

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 13 / 14

Tibor Juhász (EKU - Hungary) Lie derived lengths vs. derived length Spa, 2017 14 / 14