# Properties

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

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

## 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