Properties

Label 42T2
Order \(42\)
n \(42\)
Cyclic No
Abelian No
Solvable Yes
Primitive No
$p$-group No
Group: $C_2\times C_7:C_3$

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Group action invariants

Degree $n$ :  $42$
Transitive number $t$ :  $2$
Group :  $C_2\times C_7:C_3$
Parity:  $-1$
Primitive:  No
Nilpotency class:  $-1$ (not nilpotent)
Generators:  (1,18,24,2,17,23)(3,14,19,4,13,20)(5,16,21,6,15,22)(7,41,32,8,42,31)(9,37,33,10,38,34)(11,40,35,12,39,36)(25,30,27,26,29,28), (1,7,13,22,28,36,38,2,8,14,21,27,35,37)(3,10,15,23,30,32,39,4,9,16,24,29,31,40)(5,12,17,20,26,34,41,6,11,18,19,25,33,42)
$|\Aut(F/K)|$:  $42$

Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$
3:  $C_3$
6:  $C_6$
21:  $C_7:C_3$

Resolvents shown for degrees $\leq 10$

Subfields

Degree 2: $C_2$

Degree 3: $C_3$

Degree 6: $C_6$

Degree 7: $C_7:C_3$

Degree 14: $(C_7:C_3) \times C_2$

Degree 21: 21T2

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, 1, 1, 1, 1 $ $1$ $1$ $()$
$ 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2 $ $1$ $2$ $( 1, 2)( 3, 4)( 5, 6)( 7, 8)( 9,10)(11,12)(13,14)(15,16)(17,18)(19,20)(21,22) (23,24)(25,26)(27,28)(29,30)(31,32)(33,34)(35,36)(37,38)(39,40)(41,42)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3 $ $7$ $3$ $( 1, 3, 5)( 2, 4, 6)( 7,16,25)( 8,15,26)( 9,17,28)(10,18,27)(11,13,30) (12,14,29)(19,38,31)(20,37,32)(21,39,33)(22,40,34)(23,42,36)(24,41,35)$
$ 6, 6, 6, 6, 6, 6, 6 $ $7$ $6$ $( 1, 4, 5, 2, 3, 6)( 7,15,25, 8,16,26)( 9,18,28,10,17,27)(11,14,30,12,13,29) (19,37,31,20,38,32)(21,40,33,22,39,34)(23,41,36,24,42,35)$
$ 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3 $ $7$ $3$ $( 1, 5, 3)( 2, 6, 4)( 7,25,16)( 8,26,15)( 9,28,17)(10,27,18)(11,30,13) (12,29,14)(19,31,38)(20,32,37)(21,33,39)(22,34,40)(23,36,42)(24,35,41)$
$ 6, 6, 6, 6, 6, 6, 6 $ $7$ $6$ $( 1, 6, 3, 2, 5, 4)( 7,26,16, 8,25,15)( 9,27,17,10,28,18)(11,29,13,12,30,14) (19,32,38,20,31,37)(21,34,39,22,33,40)(23,35,42,24,36,41)$
$ 14, 14, 14 $ $3$ $14$ $( 1, 7,13,22,28,36,38, 2, 8,14,21,27,35,37)( 3,10,15,23,30,32,39, 4, 9,16,24, 29,31,40)( 5,12,17,20,26,34,41, 6,11,18,19,25,33,42)$
$ 7, 7, 7, 7, 7, 7 $ $3$ $7$ $( 1, 8,13,21,28,35,38)( 2, 7,14,22,27,36,37)( 3, 9,15,24,30,31,39) ( 4,10,16,23,29,32,40)( 5,11,17,19,26,33,41)( 6,12,18,20,25,34,42)$
$ 7, 7, 7, 7, 7, 7 $ $3$ $7$ $( 1,21,38,13,35, 8,28)( 2,22,37,14,36, 7,27)( 3,24,39,15,31, 9,30) ( 4,23,40,16,32,10,29)( 5,19,41,17,33,11,26)( 6,20,42,18,34,12,25)$
$ 14, 14, 14 $ $3$ $14$ $( 1,22,38,14,35, 7,28, 2,21,37,13,36, 8,27)( 3,23,39,16,31,10,30, 4,24,40,15, 32, 9,29)( 5,20,41,18,33,12,26, 6,19,42,17,34,11,25)$

Group invariants

Order:  $42=2 \cdot 3 \cdot 7$
Cyclic:  No
Abelian:  No
Solvable:  Yes
GAP id:  [42, 2]
Character table:   
      2  1  1  1   1  1   1   1   1   1   1
      3  1  1  1   1  1   1   .   .   .   .
      7  1  1  .   .  .   .   1   1   1   1

        1a 2a 3a  6a 3b  6b 14a  7a  7b 14b
     2P 1a 1a 3b  3b 3a  3a  7a  7a  7b  7b
     3P 1a 2a 1a  2a 1a  2a 14b  7b  7a 14a
     5P 1a 2a 3b  6b 3a  6a 14b  7b  7a 14a
     7P 1a 2a 3a  6a 3b  6b  2a  1a  1a  2a
    11P 1a 2a 3b  6b 3a  6a 14a  7a  7b 14b
    13P 1a 2a 3a  6a 3b  6b 14b  7b  7a 14a

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

A = E(3)^2
  = (-1-Sqrt(-3))/2 = -1-b3
B = -E(7)-E(7)^2-E(7)^4
  = (1-Sqrt(-7))/2 = -b7