Properties

 Label 20T11 Order $$40$$ n $$20$$ Cyclic No Abelian No Solvable Yes Primitive No $p$-group No Group: $C_5:D_4$

Related objects

Group action invariants

 Degree $n$ : $20$ Transitive number $t$ : $11$ Group : $C_5:D_4$ Parity: $-1$ Primitive: No Nilpotency class: $-1$ (not nilpotent) Generators: (1,17,2,18)(3,16,4,15)(5,14,6,13)(7,12,8,11)(9,20,10,19), (1,4,6,7,10)(2,3,5,8,9)(11,13,16,17,20,12,14,15,18,19) $|\Aut(F/K)|$: $10$

Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$ x 3
4:  $C_2^2$
8:  $D_{4}$
10:  $D_{5}$
20:  $D_{10}$

Resolvents shown for degrees $\leq 47$

Subfields

Degree 2: $C_2$

Degree 4: $D_{4}$

Degree 5: $D_{5}$

Degree 10: $D_5$

Low degree siblings

20T7, 40T11

Siblings are shown with degree $\leq 47$

A number field with this Galois group has no arithmetically equivalent fields.

Conjugacy classes

 Cycle Type Size Order Representative $1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1$ $1$ $1$ $()$ $2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1$ $2$ $2$ $(11,12)(13,14)(15,16)(17,18)(19,20)$ $2, 2, 2, 2, 2, 2, 2, 2, 2, 2$ $1$ $2$ $( 1, 2)( 3, 4)( 5, 6)( 7, 8)( 9,10)(11,12)(13,14)(15,16)(17,18)(19,20)$ $10, 10$ $2$ $10$ $( 1, 3, 6, 8,10, 2, 4, 5, 7, 9)(11,13,16,17,20,12,14,15,18,19)$ $10, 5, 5$ $2$ $10$ $( 1, 3, 6, 8,10, 2, 4, 5, 7, 9)(11,14,16,18,20)(12,13,15,17,19)$ $10, 5, 5$ $2$ $10$ $( 1, 4, 6, 7,10)( 2, 3, 5, 8, 9)(11,13,16,17,20,12,14,15,18,19)$ $5, 5, 5, 5$ $2$ $5$ $( 1, 4, 6, 7,10)( 2, 3, 5, 8, 9)(11,14,16,18,20)(12,13,15,17,19)$ $10, 10$ $2$ $10$ $( 1, 5,10, 3, 7, 2, 6, 9, 4, 8)(11,15,20,13,18,12,16,19,14,17)$ $10, 5, 5$ $2$ $10$ $( 1, 5,10, 3, 7, 2, 6, 9, 4, 8)(11,16,20,14,18)(12,15,19,13,17)$ $10, 5, 5$ $2$ $10$ $( 1, 6,10, 4, 7)( 2, 5, 9, 3, 8)(11,15,20,13,18,12,16,19,14,17)$ $5, 5, 5, 5$ $2$ $5$ $( 1, 6,10, 4, 7)( 2, 5, 9, 3, 8)(11,16,20,14,18)(12,15,19,13,17)$ $2, 2, 2, 2, 2, 2, 2, 2, 2, 2$ $10$ $2$ $( 1,11)( 2,12)( 3,19)( 4,20)( 5,17)( 6,18)( 7,16)( 8,15)( 9,13)(10,14)$ $4, 4, 4, 4, 4$ $10$ $4$ $( 1,11, 2,12)( 3,19, 4,20)( 5,17, 6,18)( 7,16, 8,15)( 9,13,10,14)$

Group invariants

 Order: $40=2^{3} \cdot 5$ Cyclic: No Abelian: No Solvable: Yes GAP id: [40, 8]
 Character table:  2 3 2 3 2 2 2 2 2 2 2 2 2 2 5 1 1 1 1 1 1 1 1 1 1 1 . . 1a 2a 2b 10a 10b 10c 5a 10d 10e 10f 5b 2c 4a 2P 1a 1a 1a 5b 5b 5b 5b 5a 5a 5a 5a 1a 2b 3P 1a 2a 2b 10d 10f 10e 5b 10a 10b 10c 5a 2c 4a 5P 1a 2a 2b 2b 2a 2a 1a 2b 2a 2a 1a 2c 4a 7P 1a 2a 2b 10d 10e 10f 5b 10a 10c 10b 5a 2c 4a X.1 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 1 X.3 1 -1 1 1 -1 -1 1 1 -1 -1 1 1 -1 X.4 1 1 1 1 1 1 1 1 1 1 1 -1 -1 X.5 2 . -2 -2 . . 2 -2 . . 2 . . X.6 2 -2 2 A -A -A A *A -*A -*A *A . . X.7 2 -2 2 *A -*A -*A *A A -A -A A . . X.8 2 2 2 A A A A *A *A *A *A . . X.9 2 2 2 *A *A *A *A A A A A . . X.10 2 . -2 -*A B -B *A -A C -C A . . X.11 2 . -2 -*A -B B *A -A -C C A . . X.12 2 . -2 -A C -C A -*A -B B *A . . X.13 2 . -2 -A -C C A -*A B -B *A . . A = E(5)^2+E(5)^3 = (-1-Sqrt(5))/2 = -1-b5 B = -E(5)+E(5)^4 C = -E(5)^2+E(5)^3