↑

Properties

Label 12T49
Order \(96\)
n \(12\)
Cyclic No
Abelian No
Solvable Yes
Primitive No
$p$-group No
Group: $\GL(2,Z/4)$

Related objects

Learn more about

Group action invariants

Degree $n$ :  $12$
Transitive number $t$ :  $49$
Group :  $\GL(2,Z/4)$
CHM label :  $[2]2S_{4}(6)_{2}$
Parity:  $-1$
Primitive:  No
Nilpotency class:  $-1$ (not nilpotent)
Generators:  (1,2,9,7)(3,11,5,12)(4,8,10,6), (1,9)(2,7)(3,5)(4,10)(6,8)(11,12), (1,2)(3,5)(4,6)(7,9)(8,10), (1,3,6,12)(2,4,7,10)(5,8,11,9)
$|\Aut(F/K)|$:  $2$

Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$ x 3
4:  $C_2^2$
6:  $S_3$
8:  $D_{4}$
12:  $D_{6}$
24:  $S_4$, $(C_6\times C_2):C_2$
48:  $S_4\times C_2$

Resolvents shown for degrees $\leq 47$

Subfields

Degree 2: None

Degree 3: $S_3$

Degree 4: None

Degree 6: $S_4\times C_2$

Low degree siblings

12T49, 12T50, 12T52, 16T186, 16T193, 24T153, 24T154, 24T155 x 2, 24T156, 24T157, 24T158, 24T159, 24T165, 24T166, 32T392

Siblings are shown with degree $\leq 47$

A number field with this Galois group has exactly one arithmetically equivalent field.

Conjugacy Classes

Cycle TypeSizeOrderRepresentative
$ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 $ $1$ $1$ $()$
$ 4, 4, 2, 1, 1 $ $12$ $4$ $( 2, 3,10,12)( 4,11, 7, 5)( 6, 8)$
$ 2, 2, 2, 2, 1, 1, 1, 1 $ $6$ $2$ $( 2, 4)( 3, 5)( 7,10)(11,12)$
$ 2, 2, 2, 2, 2, 1, 1 $ $12$ $2$ $( 2, 5)( 3, 7)( 4,12)( 6, 8)(10,11)$
$ 2, 2, 2, 2, 1, 1, 1, 1 $ $3$ $2$ $( 2,10)( 3,12)( 4, 7)( 5,11)$
$ 6, 6 $ $8$ $6$ $( 1, 2, 3, 9, 7, 5)( 4,12, 6,10,11, 8)$
$ 6, 6 $ $8$ $6$ $( 1, 2, 5, 6,10,12)( 3, 8, 4,11, 9, 7)$
$ 4, 4, 4 $ $12$ $4$ $( 1, 2, 8,10)( 3,12, 5,11)( 4, 9, 7, 6)$
$ 4, 4, 4 $ $12$ $4$ $( 1, 2, 9, 7)( 3,11, 5,12)( 4, 8,10, 6)$
$ 3, 3, 3, 3 $ $8$ $3$ $( 1, 2,11)( 3, 6,10)( 4, 5, 8)( 7,12, 9)$
$ 6, 6 $ $8$ $6$ $( 1, 2,12, 8, 4, 3)( 5, 9, 7,11, 6,10)$
$ 2, 2, 2, 2, 2, 2 $ $3$ $2$ $( 1, 6)( 2, 7)( 3,12)( 4,10)( 5,11)( 8, 9)$
$ 2, 2, 2, 2, 2, 2 $ $2$ $2$ $( 1, 6)( 2,10)( 3,11)( 4, 7)( 5,12)( 8, 9)$
$ 2, 2, 2, 2, 2, 2 $ $1$ $2$ $( 1, 9)( 2, 7)( 3, 5)( 4,10)( 6, 8)(11,12)$

Group invariants

Order:  $96=2^{5} \cdot 3$
Cyclic:  No
Abelian:  No
Solvable:  Yes
GAP id:  [96, 195]
Character table:    
      2  5  3  4  3  5  2  2  3  3  2  2  5  4  5
      3  1  .  .  .  .  1  1  .  .  1  1  .  1  1

        1a 4a 2a 2b 2c 6a 6b 4b 4c 3a 6c 2d 2e 2f
     2P 1a 2c 1a 1a 1a 3a 3a 2d 2f 3a 3a 1a 1a 1a
     3P 1a 4a 2a 2b 2c 2f 2e 4b 4c 1a 2e 2d 2e 2f
     5P 1a 4a 2a 2b 2c 6a 6c 4b 4c 3a 6b 2d 2e 2f

X.1      1  1  1  1  1  1  1  1  1  1  1  1  1  1
X.2      1 -1 -1  1  1  1 -1  1 -1  1 -1  1 -1  1
X.3      1 -1  1 -1  1  1  1 -1 -1  1  1  1  1  1
X.4      1  1 -1 -1  1  1 -1 -1  1  1 -1  1 -1  1
X.5      2  . -2  .  2 -1  1  .  . -1  1  2 -2  2
X.6      2  .  2  .  2 -1 -1  .  . -1 -1  2  2  2
X.7      2  .  .  . -2 -2  .  .  .  2  .  2  . -2
X.8      2  .  .  . -2  1  A  .  . -1 -A  2  . -2
X.9      2  .  .  . -2  1 -A  .  . -1  A  2  . -2
X.10     3 -1 -1  1 -1  .  . -1  1  .  . -1  3  3
X.11     3 -1  1 -1 -1  .  .  1  1  .  . -1 -3  3
X.12     3  1 -1 -1 -1  .  .  1 -1  .  . -1  3  3
X.13     3  1  1  1 -1  .  . -1 -1  .  . -1 -3  3
X.14     6  .  .  .  2  .  .  .  .  .  . -2  . -6

A = -E(3)+E(3)^2
  = -Sqrt(-3) = -i3