Sidkis Conjecture; showing finiteness of group presentations using - PowerPoint PPT Presentation

Introduction Geometries and amalgams Sidki’s problem Sidki’s Conjecture; showing finiteness of group presentations using amalgams Justin M c Inroy University of Leicester 4 th August 2013, Groups St Andrews Joint work with Sergey Shpectorov (University of Birmingham) 1 / 21

Introduction Geometries and amalgams Sidki’s problem Some presentations The following is a well-known presentation for the Alternating group A m +2 : i = 1 , ∀ i , ( a i a j ) 2 = 1 , i � = j � � a 1 , . . . , a m : a 3 2 / 21

Introduction Geometries and amalgams Sidki’s problem Some presentations The following is a well-known presentation for the Alternating group A m +2 : i = 1 , ∀ i , ( a i a j ) 2 = 1 , i � = j � � a 1 , . . . , a m : a 3 (Think of the a i as being (1 , 2 , i + 2)). It was given by Carmichael in 1923. 2 / 21

Introduction Geometries and amalgams Sidki’s problem Some presentations The following is a well-known presentation for the Alternating group A m +2 : i = 1 , ∀ i , ( a i a j ) 2 = 1 , i � = j � � a 1 , . . . , a m : a 3 (Think of the a i as being (1 , 2 , i + 2)). It was given by Carmichael in 1923. In 1982, Sidki generalised this to: j ) 2 = 1 , i � = j , ∀ s � Y ( m , n ) := � a 1 , . . . , a m : a n i = 1 , ∀ i , ( a s i a s 2 / 21

Introduction Geometries and amalgams Sidki’s problem Some presentations The following is a well-known presentation for the Alternating group A m +2 : i = 1 , ∀ i , ( a i a j ) 2 = 1 , i � = j � � a 1 , . . . , a m : a 3 (Think of the a i as being (1 , 2 , i + 2)). It was given by Carmichael in 1923. In 1982, Sidki generalised this to: j ) 2 = 1 , i � = j , ∀ s � Y ( m , n ) := � a 1 , . . . , a m : a n i = 1 , ∀ i , ( a s i a s Conjecture: 2 / 21

Introduction Geometries and amalgams Sidki’s problem Some presentations The following is a well-known presentation for the Alternating group A m +2 : i = 1 , ∀ i , ( a i a j ) 2 = 1 , i � = j � � a 1 , . . . , a m : a 3 (Think of the a i as being (1 , 2 , i + 2)). It was given by Carmichael in 1923. In 1982, Sidki generalised this to: j ) 2 = 1 , i � = j , ∀ s � Y ( m , n ) := � a 1 , . . . , a m : a n i = 1 , ∀ i , ( a s i a s Conjecture: This presentation is finite. 2 / 21

Introduction Geometries and amalgams Sidki’s problem Known results There are some known results for small values of n and m . 3 / 21

Introduction Geometries and amalgams Sidki’s problem Known results There are some known results for small values of n and m . ◮ Y ( m , 2) is elementary abelian of order 2 m 3 / 21

Introduction Geometries and amalgams Sidki’s problem Known results There are some known results for small values of n and m . ◮ Y ( m , 2) is elementary abelian of order 2 m ◮ Y ( m , 3) is Carmichael’s presentation for the alternating group 3 / 21

Introduction Geometries and amalgams Sidki’s problem Known results There are some known results for small values of n and m . ◮ Y ( m , 2) is elementary abelian of order 2 m ◮ Y ( m , 3) is Carmichael’s presentation for the alternating group ◮ Y (1 , n ) is C n (!) 3 / 21

Introduction Geometries and amalgams Sidki’s problem Known results There are some known results for small values of n and m . ◮ Y ( m , 2) is elementary abelian of order 2 m ◮ Y ( m , 3) is Carmichael’s presentation for the alternating group ◮ Y (1 , n ) is C n (!) ◮ Y (2 , n ) is a metabelian group of order 2 n − 1 n by a result of Coxeter. It is equal to the augmentation ideal I 2 , n of F 2 C n extended by C n . 3 / 21

Introduction Geometries and amalgams Sidki’s problem Known results There are some known results for small values of n and m . ◮ Y ( m , 2) is elementary abelian of order 2 m ◮ Y ( m , 3) is Carmichael’s presentation for the alternating group ◮ Y (1 , n ) is C n (!) ◮ Y (2 , n ) is a metabelian group of order 2 n − 1 n by a result of Coxeter. It is equal to the augmentation ideal I 2 , n of F 2 C n extended by C n . ◮ If n | n ′ , then Y ( m , n ) is a quotient of Y ( m , n ′ ) 3 / 21

Introduction Geometries and amalgams Sidki’s problem A new presentation 4 / 21

Introduction Geometries and amalgams Sidki’s problem A new presentation Define: y ( m , n ) := � a , S m : a n = 1 , [(1 , 2) a s , (1 , 2)] = 1 ∀ s (1 , 2) 1+ a + ··· + a n − 1 = 1 , a ( i , i +1) = a − 1 for 2 ≤ i ≤ m − 1 � 4 / 21

Introduction Geometries and amalgams Sidki’s problem A new presentation Define: y ( m , n ) := � a , S m : a n = 1 , [(1 , 2) a s , (1 , 2)] = 1 ∀ s (1 , 2) 1+ a + ··· + a n − 1 = 1 , a ( i , i +1) = a − 1 for 2 ≤ i ≤ m − 1 � For n odd, Sidki showed that Y ( m , n ) ∼ = y ( m , n ). 4 / 21

Introduction Geometries and amalgams Sidki’s problem An embedding When m = 3, consider the following map: � α � � 1 � � 0 � 0 0 1 a �→ , (1 , 2) �→ , (2 , 3) �→ α − 1 0 1 1 1 0 where α is an element of order n in a field F of characteristic 2. 5 / 21

Introduction Geometries and amalgams Sidki’s problem An embedding When m = 3, consider the following map: � α � � 1 � � 0 � 0 0 1 a �→ , (1 , 2) �→ , (2 , 3) �→ α − 1 0 1 1 1 0 where α is an element of order n in a field F of characteristic 2. These matrices satisfy the desired relations, so this is a representation which embeds y (3 , n ) into SL 2 ( F ). 5 / 21

Introduction Geometries and amalgams Sidki’s problem An embedding When m = 3, consider the following map: � α � � 1 � � 0 � 0 0 1 a �→ , (1 , 2) �→ , (2 , 3) �→ α − 1 0 1 1 1 0 where α is an element of order n in a field F of characteristic 2. These matrices satisfy the desired relations, so this is a representation which embeds y (3 , n ) into SL 2 ( F ). Sidki goes on to show that Y (3 , n ) is SL 2 ( I 2 , n ) when n is odd (Notation: SL 2 ( I ) := ker( SL 2 ( R ) → SL 2 ( R / I ))). This is also finite. 5 / 21

Introduction Geometries and amalgams Sidki’s problem An embedding When m = 3, consider the following map: � α � � 1 � � 0 � 0 0 1 a �→ , (1 , 2) �→ , (2 , 3) �→ α − 1 0 1 1 1 0 where α is an element of order n in a field F of characteristic 2. These matrices satisfy the desired relations, so this is a representation which embeds y (3 , n ) into SL 2 ( F ). Sidki goes on to show that Y (3 , n ) is SL 2 ( I 2 , n ) when n is odd (Notation: SL 2 ( I ) := ker( SL 2 ( R ) → SL 2 ( R / I ))). This is also finite. The map above can be extended to give an embedding y ( m , n ) ֒ → SL 2 m − 2 ( F ). 5 / 21

Introduction Geometries and amalgams Sidki’s problem Some computational results For small values of m and n we can use a computer to construct and identify the group. These calculations are due to J. Neubuser and W. Felsch and the larger ones to O’Brien. 6 / 21

Introduction Geometries and amalgams Sidki’s problem Some computational results For small values of m and n we can use a computer to construct and identify the group. These calculations are due to J. Neubuser and W. Felsch and the larger ones to O’Brien. ◮ Y (3 , 5) ∼ = SL 2 (16) 6 / 21

Introduction Geometries and amalgams Sidki’s problem Some computational results For small values of m and n we can use a computer to construct and identify the group. These calculations are due to J. Neubuser and W. Felsch and the larger ones to O’Brien. ◮ Y (3 , 5) ∼ = SL 2 (16) ◮ Y (4 , 5) ∼ = Ω(5 , 4) 6 / 21

Introduction Geometries and amalgams Sidki’s problem Some computational results For small values of m and n we can use a computer to construct and identify the group. These calculations are due to J. Neubuser and W. Felsch and the larger ones to O’Brien. ◮ Y (3 , 5) ∼ = SL 2 (16) ◮ Y (4 , 5) ∼ = Ω(5 , 4) ◮ Y (5 , 5) ∼ = Ω − (6 , 4) 6 / 21

Introduction Geometries and amalgams Sidki’s problem Some computational results For small values of m and n we can use a computer to construct and identify the group. These calculations are due to J. Neubuser and W. Felsch and the larger ones to O’Brien. ◮ Y (3 , 5) ∼ = SL 2 (16) ∼ = Ω − (4 , 4) ◮ Y (4 , 5) ∼ = Ω(5 , 4) ◮ Y (5 , 5) ∼ = Ω − (6 , 4) 6 / 21

Introduction Geometries and amalgams Sidki’s problem Some computational results For small values of m and n we can use a computer to construct and identify the group. These calculations are due to J. Neubuser and W. Felsch and the larger ones to O’Brien. ◮ Y (3 , 5) ∼ = SL 2 (16) ∼ = Ω − (4 , 4) ◮ Y (4 , 5) ∼ = Ω(5 , 4) ◮ Y (5 , 5) ∼ = Ω − (6 , 4) = 4 6 : Ω − (6 , 4) ◮ Y (6 , 5) ∼ 6 / 21

Introduction Geometries and amalgams Sidki’s problem Some computational results For small values of m and n we can use a computer to construct and identify the group. These calculations are due to J. Neubuser and W. Felsch and the larger ones to O’Brien. ◮ Y (3 , 5) ∼ = SL 2 (16) ∼ = Ω − (4 , 4) ◮ Y (4 , 5) ∼ = Ω(5 , 4) ◮ Y (5 , 5) ∼ = Ω − (6 , 4) = 4 6 : Ω − (6 , 4) ◮ Y (6 , 5) ∼ ◮ Y (7 , 5) ∼ = Ω − (8 , 4) 6 / 21