Properties

Label 36T7
Order \(36\)
n \(36\)
Cyclic No
Abelian No
Solvable Yes
Primitive No
$p$-group No
Group: $C_3^2:C_4$

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Group action invariants

Degree $n$ :  $36$
Transitive number $t$ :  $7$
Group :  $C_3^2:C_4$
Parity:  $-1$
Primitive:  No
Nilpotency class:  $-1$ (not nilpotent)
Generators:  (1,32,2,31)(3,30,4,29)(5,28,6,27)(7,26,8,25)(9,11,10,12)(13,17,14,18)(15,19,16,20)(21,35,22,36)(23,34,24,33), (1,3,2,4)(5,34,6,33)(7,35,8,36)(9,31,10,32)(11,30,12,29)(13,27,14,28)(15,26,16,25)(17,24,18,23)(19,21,20,22), (1,14,2,13)(3,15,4,16)(5,9,6,10)(7,12,8,11)(17,35,18,36)(19,34,20,33)(21,32,22,31)(23,29,24,30)(25,27,26,28)
$|\Aut(F/K)|$:  $36$

Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$
4:  $C_4$
6:  $S_3$ x 4
18:  $C_3^2:C_2$

Resolvents shown for degrees $\leq 10$

Subfields

Degree 2: $C_2$

Degree 3: $S_3$ x 4

Degree 4: $C_4$

Degree 6: $S_3$ x 4

Degree 9: $C_3^2:C_2$

Degree 12: $C_3 : C_4$ x 4

Degree 18: $C_3^2 : C_2$

Low degree siblings

There are no siblings with degree $\leq 10$
Data on whether or not a number field with this Galois group has arithmetically equivalent fields has not been computed.

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, 1, 1, 1, 1, 1, 1 $ $1$ $1$ $()$
$ 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2 $ $1$ $2$ $( 1, 2)( 3, 4)( 5, 6)( 7, 8)( 9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22) (23,24)(25,26)(27,28)(29,30)(31,32)(33,34)(35,36)$
$ 4, 4, 4, 4, 4, 4, 4, 4, 4 $ $9$ $4$ $( 1, 3, 2, 4)( 5,34, 6,33)( 7,35, 8,36)( 9,31,10,32)(11,30,12,29)(13,27,14,28) (15,26,16,25)(17,24,18,23)(19,21,20,22)$
$ 4, 4, 4, 4, 4, 4, 4, 4, 4 $ $9$ $4$ $( 1, 4, 2, 3)( 5,33, 6,34)( 7,36, 8,35)( 9,32,10,31)(11,29,12,30)(13,28,14,27) (15,25,16,26)(17,23,18,24)(19,22,20,21)$
$ 6, 6, 6, 6, 6, 6 $ $2$ $6$ $( 1, 7,33, 2, 8,34)( 3, 6,36, 4, 5,35)( 9,16,17,10,15,18)(11,13,19,12,14,20) (21,27,30,22,28,29)(23,26,32,24,25,31)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3 $ $2$ $3$ $( 1, 8,33)( 2, 7,34)( 3, 5,36)( 4, 6,35)( 9,15,17)(10,16,18)(11,14,19) (12,13,20)(21,28,30)(22,27,29)(23,25,32)(24,26,31)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3 $ $2$ $3$ $( 1, 9,30)( 2,10,29)( 3,12,31)( 4,11,32)( 5,13,24)( 6,14,23)( 7,16,22) ( 8,15,21)(17,28,33)(18,27,34)(19,25,35)(20,26,36)$
$ 6, 6, 6, 6, 6, 6 $ $2$ $6$ $( 1,10,30, 2, 9,29)( 3,11,31, 4,12,32)( 5,14,24, 6,13,23)( 7,15,22, 8,16,21) (17,27,33,18,28,34)(19,26,35,20,25,36)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3 $ $2$ $3$ $( 1,15,28)( 2,16,27)( 3,13,26)( 4,14,25)( 5,20,31)( 6,19,32)( 7,18,29) ( 8,17,30)( 9,21,33)(10,22,34)(11,23,35)(12,24,36)$
$ 6, 6, 6, 6, 6, 6 $ $2$ $6$ $( 1,16,28, 2,15,27)( 3,14,26, 4,13,25)( 5,19,31, 6,20,32)( 7,17,29, 8,18,30) ( 9,22,33,10,21,34)(11,24,35,12,23,36)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3 $ $2$ $3$ $( 1,17,21)( 2,18,22)( 3,20,24)( 4,19,23)( 5,12,26)( 6,11,25)( 7,10,27) ( 8, 9,28)(13,31,36)(14,32,35)(15,30,33)(16,29,34)$
$ 6, 6, 6, 6, 6, 6 $ $2$ $6$ $( 1,18,21, 2,17,22)( 3,19,24, 4,20,23)( 5,11,26, 6,12,25)( 7, 9,27, 8,10,28) (13,32,36,14,31,35)(15,29,33,16,30,34)$

Group invariants

Order:  $36=2^{2} \cdot 3^{2}$
Cyclic:  No
Abelian:  No
Solvable:  Yes
GAP id:  [36, 7]
Character table:   
      2  2  2  2  2  1  1  1  1  1  1  1  1
      3  2  2  .  .  2  2  2  2  2  2  2  2

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

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

A = -E(4)
  = -Sqrt(-1) = -i