Properties

Label 28T12
Order \(84\)
n \(28\)
Cyclic No
Abelian No
Solvable Yes
Primitive No
$p$-group No
Group: $C_7:C_{12}$

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Group action invariants

Degree $n$ :  $28$
Transitive number $t$ :  $12$
Group :  $C_7:C_{12}$
Parity:  $-1$
Primitive:  No
Nilpotency class:  $-1$ (not nilpotent)
Generators:  (1,8,25,24,18,28,2,7,26,23,17,27)(3,13,15,22,11,9,4,14,16,21,12,10)(5,20,6,19), (1,9,13)(2,10,14)(5,26,21)(6,25,22)(7,19,11)(8,20,12)(15,23,28)(16,24,27)
$|\Aut(F/K)|$:  $4$

Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$
3:  $C_3$
4:  $C_4$
6:  $C_6$
12:  $C_{12}$
42:  $F_7$

Resolvents shown for degrees $\leq 47$

Subfields

Degree 2: $C_2$

Degree 4: $C_4$

Degree 7: $F_7$

Degree 14: $F_7$

Low degree siblings

There are no siblings 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, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 $ $1$ $1$ $()$
$ 3, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1 $ $7$ $3$ $( 3,19,24)( 4,20,23)( 5, 9,18)( 6,10,17)( 7,27,11)( 8,28,12)(13,26,21) (14,25,22)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1 $ $7$ $3$ $( 3,24,19)( 4,23,20)( 5,18, 9)( 6,17,10)( 7,11,27)( 8,12,28)(13,21,26) (14,22,25)$
$ 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2 $ $1$ $2$ $( 1, 2)( 3, 4)( 5, 6)( 7, 8)( 9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22) (23,24)(25,26)(27,28)$
$ 6, 6, 6, 6, 2, 2 $ $7$ $6$ $( 1, 2)( 3,20,24, 4,19,23)( 5,10,18, 6, 9,17)( 7,28,11, 8,27,12) (13,25,21,14,26,22)(15,16)$
$ 6, 6, 6, 6, 2, 2 $ $7$ $6$ $( 1, 2)( 3,23,19, 4,24,20)( 5,17, 9, 6,18,10)( 7,12,27, 8,11,28) (13,22,26,14,21,25)(15,16)$
$ 4, 4, 4, 4, 4, 4, 4 $ $7$ $4$ $( 1, 3, 2, 4)( 5,27, 6,28)( 7,25, 8,26)( 9,24,10,23)(11,22,12,21)(13,19,14,20) (15,18,16,17)$
$ 12, 12, 4 $ $7$ $12$ $( 1, 3,10,28,26,19, 2, 4, 9,27,25,20)( 5,16,17,23,13,11, 6,15,18,24,14,12) ( 7,22, 8,21)$
$ 12, 12, 4 $ $7$ $12$ $( 1, 3,14, 8, 5,24, 2, 4,13, 7, 6,23)( 9,16,17,28,21,19,10,15,18,27,22,20) (11,25,12,26)$
$ 4, 4, 4, 4, 4, 4, 4 $ $7$ $4$ $( 1, 4, 2, 3)( 5,28, 6,27)( 7,26, 8,25)( 9,23,10,24)(11,21,12,22)(13,20,14,19) (15,17,16,18)$
$ 12, 12, 4 $ $7$ $12$ $( 1, 4,10,27,26,20, 2, 3, 9,28,25,19)( 5,15,17,24,13,12, 6,16,18,23,14,11) ( 7,21, 8,22)$
$ 12, 12, 4 $ $7$ $12$ $( 1, 4,14, 7, 5,23, 2, 3,13, 8, 6,24)( 9,15,17,27,21,20,10,16,18,28,22,19) (11,26,12,25)$
$ 7, 7, 7, 7 $ $6$ $7$ $( 1, 5, 9,13,18,21,26)( 2, 6,10,14,17,22,25)( 3, 7,11,16,19,24,27) ( 4, 8,12,15,20,23,28)$
$ 14, 14 $ $6$ $14$ $( 1, 6, 9,14,18,22,26, 2, 5,10,13,17,21,25)( 3, 8,11,15,19,23,27, 4, 7,12,16, 20,24,28)$

Group invariants

Order:  $84=2^{2} \cdot 3 \cdot 7$
Cyclic:  No
Abelian:  No
Solvable:  Yes
GAP id:  [84, 1]
Character table:   
      2  2  2  2  2   2   2  2   2   2  2   2   2  1   1
      3  1  1  1  1   1   1  1   1   1  1   1   1  .   .
      7  1  .  .  1   .   .  .   .   .  .   .   .  1   1

        1a 3a 3b 2a  6a  6b 4a 12a 12b 4b 12c 12d 7a 14a
     2P 1a 3b 3a 1a  3b  3a 2a  6a  6b 2a  6a  6b 7a  7a
     3P 1a 1a 1a 2a  2a  2a 4b  4b  4b 4a  4a  4a 7a 14a
     5P 1a 3b 3a 2a  6b  6a 4a 12b 12a 4b 12d 12c 7a 14a
     7P 1a 3a 3b 2a  6a  6b 4b 12c 12d 4a 12a 12b 1a  2a
    11P 1a 3b 3a 2a  6b  6a 4b 12d 12c 4a 12b 12a 7a 14a
    13P 1a 3a 3b 2a  6a  6b 4a 12a 12b 4b 12c 12d 7a 14a

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

A = E(3)^2
  = (-1-Sqrt(-3))/2 = -1-b3
B = -E(4)
  = -Sqrt(-1) = -i
C = -E(12)^7