# Properties

 Label 25T4 Order $$50$$ n $$25$$ Cyclic No Abelian No Solvable Yes Primitive No $p$-group No Group: $D_{25}$

## Group action invariants

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

## Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$
10:  $D_{5}$

Resolvents shown for degrees $\leq 47$

## Subfields

Degree 5: $D_{5}$

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

## Group invariants

 Order: $50=2 \cdot 5^{2}$ Cyclic: No Abelian: No Solvable: Yes GAP id: [50, 1]
 Character table:  2 1 1 . . . . . . . . . . . . 5 2 . 2 2 2 2 2 2 2 2 2 2 2 2 1a 2a 5a 5b 25a 25b 25c 25d 25e 25f 25g 25h 25i 25j 2P 1a 1a 5b 5a 25f 25h 25j 25g 25i 25b 25e 25c 25a 25d 3P 1a 2a 5b 5a 25g 25i 25f 25h 25j 25e 25c 25a 25d 25b 5P 1a 2a 1a 1a 5a 5a 5a 5a 5a 5b 5b 5b 5b 5b 7P 1a 2a 5b 5a 25j 25g 25i 25f 25h 25d 25b 25e 25c 25a 11P 1a 2a 5a 5b 25d 25e 25a 25b 25c 25g 25h 25i 25j 25f 13P 1a 2a 5b 5a 25i 25f 25h 25j 25g 25a 25d 25b 25e 25c 17P 1a 2a 5b 5a 25h 25j 25g 25i 25f 25c 25a 25d 25b 25e 19P 1a 2a 5a 5b 25e 25a 25b 25c 25d 25i 25j 25f 25g 25h 23P 1a 2a 5b 5a 25f 25h 25j 25g 25i 25b 25e 25c 25a 25d X.1 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 1 X.3 2 . 2 2 A A A A A *A *A *A *A *A X.4 2 . 2 2 *A *A *A *A *A A A A A A X.5 2 . A *A B E F C D K G J I H X.6 2 . A *A C D B E F G J I H K X.7 2 . A *A D B E F C I H K G J X.8 2 . A *A E F C D B J I H K G X.9 2 . A *A F C D B E H K G J I X.10 2 . *A A G I K J H D F B C E X.11 2 . *A A H G I K J C E D F B X.12 2 . *A A I K J H G B C E D F X.13 2 . *A A J H G I K F B C E D X.14 2 . *A A K J H G I E D F B C A = E(5)^2+E(5)^3 = (-1-Sqrt(5))/2 = -1-b5 B = E(25)^3+E(25)^22 C = E(25)^8+E(25)^17 D = E(25)^7+E(25)^18 E = E(25)^12+E(25)^13 F = -E(25)^3-E(25)^7-E(25)^8-E(25)^12-E(25)^13-E(25)^17-E(25)^18-E(25)^22 G = E(25)^9+E(25)^16 H = E(25)^4+E(25)^21 I = E(25)^11+E(25)^14 J = -E(25)^4-E(25)^6-E(25)^9-E(25)^11-E(25)^14-E(25)^16-E(25)^19-E(25)^21 K = E(25)^6+E(25)^19