Group action invariants
| Degree $n$ : | $11$ | |
| Transitive number $t$ : | $3$ | |
| Group : | $C_{11}:C_5$ | |
| CHM label : | $F_{55}(11)=11:5$ | |
| Parity: | $1$ | |
| Primitive: | Yes | |
| Nilpotency class: | $-1$ (not nilpotent) | |
| Generators: | (1,3,9,5,4)(2,6,7,10,8), (1,2,3,4,5,6,7,8,9,10,11) | |
| $|\Aut(F/K)|$: | $1$ |
Low degree resolvents
|G/N| Galois groups for stem field(s) 5: $C_5$ Resolvents shown for degrees $\leq 47$
Subfields
Prime degree - none
Low degree siblings
There are no siblings 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 $ | $1$ | $1$ | $()$ |
| $ 5, 5, 1 $ | $11$ | $5$ | $( 2, 4,10, 6, 5)( 3, 7, 8,11, 9)$ |
| $ 5, 5, 1 $ | $11$ | $5$ | $( 2, 5, 6,10, 4)( 3, 9,11, 8, 7)$ |
| $ 5, 5, 1 $ | $11$ | $5$ | $( 2, 6, 4, 5,10)( 3,11, 7, 9, 8)$ |
| $ 5, 5, 1 $ | $11$ | $5$ | $( 2,10, 5, 4, 6)( 3, 8, 9, 7,11)$ |
| $ 11 $ | $5$ | $11$ | $( 1, 2, 3, 4, 5, 6, 7, 8, 9,10,11)$ |
| $ 11 $ | $5$ | $11$ | $( 1, 3, 5, 7, 9,11, 2, 4, 6, 8,10)$ |
Group invariants
| Order: | $55=5 \cdot 11$ | |
| Cyclic: | No | |
| Abelian: | No | |
| Solvable: | Yes | |
| GAP id: | [55, 1] |
| Character table: |
5 1 1 1 1 1 . .
11 1 . . . . 1 1
1a 5a 5b 5c 5d 11a 11b
2P 1a 5d 5c 5a 5b 11b 11a
3P 1a 5c 5d 5b 5a 11a 11b
5P 1a 1a 1a 1a 1a 11a 11b
7P 1a 5d 5c 5a 5b 11b 11a
11P 1a 5a 5b 5c 5d 1a 1a
X.1 1 1 1 1 1 1 1
X.2 1 A /A /B B 1 1
X.3 1 B /B A /A 1 1
X.4 1 /B B /A A 1 1
X.5 1 /A A B /B 1 1
X.6 5 . . . . C /C
X.7 5 . . . . /C C
A = E(5)^4
B = E(5)^3
C = E(11)^2+E(11)^6+E(11)^7+E(11)^8+E(11)^10
= (-1-Sqrt(-11))/2 = -1-b11
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