Properties

Label 12T128
Order \(288\)
n \(12\)
Cyclic No
Abelian No
Solvable Yes
Primitive No
$p$-group No
Group: $A_4\wr C_2$

Related objects

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Group action invariants

Degree $n$ :  $12$
Transitive number $t$ :  $128$
Group :  $A_4\wr C_2$
CHM label :  $[1/4E(4)^{3}:3]S(3)$
Parity:  $1$
Primitive:  No
Nilpotency class:  $-1$ (not nilpotent)
Generators:  (1,10)(3,12)(4,7)(6,9), (1,7,10)(2,5,11)(3,6,9), (1,5)(2,10)(4,8)(7,11), (1,5,9)(2,6,10)(3,7,11)(4,8,12)
$|\Aut(F/K)|$:  $1$

Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$
3:  $C_3$
6:  $S_3$, $C_6$
18:  $S_3\times C_3$

Resolvents shown for degrees $\leq 47$

Subfields

Degree 2: None

Degree 3: $S_3$

Degree 4: None

Degree 6: None

Low degree siblings

8T42, 12T126, 12T129, 16T708, 18T112, 18T113, 24T692, 24T694, 24T695, 24T702, 24T703, 24T704, 32T9306, 36T316, 36T318, 36T456, 36T457, 36T458, 36T459

Siblings are shown with degree $\leq 47$

A number field with this Galois group has no arithmetically equivalent fields.

Conjugacy Classes

Cycle TypeSizeOrderRepresentative
$ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 $ $1$ $1$ $()$
$ 3, 3, 3, 1, 1, 1 $ $16$ $3$ $( 4, 7,10)( 5,11, 8)( 6, 9,12)$
$ 3, 3, 3, 1, 1, 1 $ $16$ $3$ $( 4,10, 7)( 5, 8,11)( 6,12, 9)$
$ 2, 2, 2, 2, 1, 1, 1, 1 $ $12$ $2$ $( 2, 3)( 5,12)( 6,11)( 8, 9)$
$ 6, 3, 2, 1 $ $48$ $6$ $( 2, 3)( 4, 7,10)( 5, 6, 8,12,11, 9)$
$ 6, 3, 2, 1 $ $48$ $6$ $( 2, 3)( 4,10, 7)( 5, 9,11,12, 8, 6)$
$ 2, 2, 2, 2, 1, 1, 1, 1 $ $9$ $2$ $( 2, 5)( 3,12)( 6, 9)( 8,11)$
$ 3, 3, 3, 3 $ $8$ $3$ $( 1, 2, 3)( 4, 5, 9)( 6,10, 8)( 7,11,12)$
$ 3, 3, 3, 3 $ $8$ $3$ $( 1, 2, 3)( 4, 8,12)( 5, 6, 7)( 9,10,11)$
$ 3, 3, 3, 3 $ $32$ $3$ $( 1, 2, 3)( 4,11, 6)( 5,12,10)( 7, 8, 9)$
$ 4, 4, 2, 2 $ $36$ $4$ $( 1, 2, 4,11)( 3, 6)( 5, 7, 8,10)( 9,12)$
$ 6, 6 $ $24$ $6$ $( 1, 2, 6, 7,11, 9)( 3, 4, 5,12,10, 8)$
$ 6, 6 $ $24$ $6$ $( 1, 2, 6,10,11,12)( 3, 4, 8, 9, 7, 5)$
$ 2, 2, 2, 2, 2, 2 $ $6$ $2$ $( 1, 4)( 2, 5)( 3, 9)( 6,12)( 7,10)( 8,11)$

Group invariants

Order:  $288=2^{5} \cdot 3^{2}$
Cyclic:  No
Abelian:  No
Solvable:  Yes
GAP id:  [288, 1025]
Character table:   
      2  5  1  1  3   1   1  5   2   2  .  3   2   2  4
      3  2  2  2  1   1   1  .   2   2  2  .   1   1  1

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

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

A = E(3)^2
  = (-1-Sqrt(-3))/2 = -1-b3
B = 2*E(3)
  = -1+Sqrt(-3) = 2b3
C = 3*E(3)^2
  = (-3-3*Sqrt(-3))/2 = -3-3b3