Properties

Label 30T6
Order \(60\)
n \(30\)
Cyclic No
Abelian No
Solvable Yes
Primitive No
$p$-group No
Group: $C_3:F_5$

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Group action invariants

Degree $n$ :  $30$
Transitive number $t$ :  $6$
Group :  $C_3:F_5$
Parity:  $-1$
Primitive:  No
Nilpotency class:  $-1$ (not nilpotent)
Generators:  (1,28,23,19,15,11,7,3,29,25,21,17,13,9,5)(2,27,24,20,16,12,8,4,30,26,22,18,14,10,6), (1,4,7,16)(2,3,8,15)(5,12,23,18)(6,11,24,17)(9,20)(10,19)(13,27,25,22)(14,28,26,21)(29,30)
$|\Aut(F/K)|$:  $6$

Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$
4:  $C_4$
6:  $S_3$
12:  $C_3 : C_4$
20:  $F_5$

Resolvents shown for degrees $\leq 47$

Subfields

Degree 2: $C_2$

Degree 3: $S_3$

Degree 5: $F_5$

Degree 6: $S_3$

Degree 10: $F_5$

Degree 15: $C_{15} : C_4$

Low degree siblings

15T6

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

Group invariants

Order:  $60=2^{2} \cdot 3 \cdot 5$
Cyclic:  No
Abelian:  No
Solvable:  Yes
GAP id:  [60, 7]
Character table:   
     2  2  2  2  2   .  1  .  1   .
     3  1  1  .  .   1  1  1  1   1
     5  1  .  .  .   1  .  1  1   1

       1a 2a 4a 4b 15a 6a 5a 3a 15b
    2P 1a 1a 2a 2a 15a 3a 5a 3a 15b
    3P 1a 2a 4b 4a  5a 2a 5a 1a  5a
    5P 1a 2a 4a 4b  3a 6a 1a 3a  3a
    7P 1a 2a 4b 4a 15b 6a 5a 3a 15a
   11P 1a 2a 4b 4a 15b 6a 5a 3a 15a
   13P 1a 2a 4a 4b 15b 6a 5a 3a 15a

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

A = -E(4)
  = -Sqrt(-1) = -i
B = E(15)^7+E(15)^11+E(15)^13+E(15)^14
  = (1-Sqrt(-15))/2 = -b15