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Properties

Label 27T11
Order \(54\)
n \(27\)
Cyclic No
Abelian No
Solvable Yes
Primitive No
$p$-group No
Group: $He_3:C_2$

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Group action invariants

Degree $n$ :  $27$
Transitive number $t$ :  $11$
Group :  $He_3:C_2$
Parity:  $1$
Primitive:  No
Nilpotency class:  $-1$ (not nilpotent)
Generators:  (1,18,4,22,25,20)(2,16,5,23,26,21)(3,17,6,24,27,19)(7,8,9)(10,15,12,14,11,13), (1,27)(2,25)(3,26)(7,19)(8,20)(9,21)(10,17)(11,18)(12,16)(13,24)(14,22)(15,23)
$|\Aut(F/K)|$:  $3$

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 3: $C_3$, $S_3$

Degree 9: $S_3\times C_3$, $C_3^2 : C_6$, $C_3^2 : S_3 $

Low degree siblings

9T11, 9T13, 18T20, 18T21, 18T22

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, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 $ $1$ $1$ $()$
$ 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1 $ $9$ $2$ $( 4,26)( 5,27)( 6,25)( 7,23)( 8,24)( 9,22)(10,21)(11,19)(12,20)(13,16)(14,17) (15,18)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 3 $ $3$ $3$ $( 1, 2, 3)( 4, 5, 6)( 7,12,14)( 8,10,15)( 9,11,13)(16,22,19)(17,23,20) (18,24,21)(25,26,27)$
$ 6, 6, 6, 6, 3 $ $9$ $6$ $( 1, 2, 3)( 4,27, 6,26, 5,25)( 7,20,14,23,12,17)( 8,21,15,24,10,18) ( 9,19,13,22,11,16)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 3 $ $3$ $3$ $( 1, 3, 2)( 4, 6, 5)( 7,14,12)( 8,15,10)( 9,13,11)(16,19,22)(17,20,23) (18,21,24)(25,27,26)$
$ 6, 6, 6, 6, 3 $ $9$ $6$ $( 1, 3, 2)( 4,25, 5,26, 6,27)( 7,17,12,23,14,20)( 8,18,10,24,15,21) ( 9,16,11,22,13,19)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 3 $ $2$ $3$ $( 1, 5,27)( 2, 6,25)( 3, 4,26)( 7,11,15)( 8,12,13)( 9,10,14)(16,20,24) (17,21,22)(18,19,23)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 3 $ $6$ $3$ $( 1, 7,16)( 2, 8,17)( 3, 9,18)( 4,10,19)( 5,11,20)( 6,12,21)(13,22,25) (14,23,26)(15,24,27)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 3 $ $6$ $3$ $( 1, 8,19)( 2, 9,20)( 3, 7,21)( 4,11,22)( 5,12,23)( 6,10,24)(13,18,27) (14,16,25)(15,17,26)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 3 $ $6$ $3$ $( 1, 9,22)( 2, 7,23)( 3, 8,24)( 4,12,16)( 5,10,17)( 6,11,18)(13,20,26) (14,21,27)(15,19,25)$

Group invariants

Order:  $54=2 \cdot 3^{3}$
Cyclic:  No
Abelian:  No
Solvable:  Yes
GAP id:  [54, 5]
Character table:    
      2  1  1  1   1  1   1  .   .   .  .
      3  3  1  2   1  2   1  3   2   2  2

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

X.1      1  1  1   1  1   1  1   1   1  1
X.2      1 -1  1  -1  1  -1  1   1   1  1
X.3      1 -1  A  -A /A -/A  1   A  /A  1
X.4      1 -1 /A -/A  A  -A  1  /A   A  1
X.5      1  1  A   A /A  /A  1   A  /A  1
X.6      1  1 /A  /A  A   A  1  /A   A  1
X.7      2  .  2   .  2   .  2  -1  -1 -1
X.8      2  .  B   . /B   .  2 -/A  -A -1
X.9      2  . /B   .  B   .  2  -A -/A -1
X.10     6  .  .   .  .   . -3   .   .  .

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