# Properties

 Label 16T4 Order $$16$$ n $$16$$ Cyclic No Abelian Yes Solvable Yes Primitive No $p$-group Yes Group: $C_4^2$

# Related objects

## Group action invariants

 Degree $n$ : $16$ Transitive number $t$ : $4$ Group : $C_4^2$ Parity: $1$ Primitive: No Nilpotency class: $1$ Generators: (1,7,5,11)(2,8,6,12)(3,9,15,13)(4,10,16,14), (1,13,2,14)(3,8,4,7)(5,9,6,10)(11,15,12,16) $|\Aut(F/K)|$: $16$

## Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$ x 3
4:  $C_4$ x 6, $C_2^2$
8:  $C_4\times C_2$ x 3

Resolvents shown for degrees $\leq 47$

## Subfields

Degree 2: $C_2$ x 3

Degree 4: $C_4$ x 6, $C_2^2$

Degree 8: $C_4\times C_2$ x 3

## 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 $1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1$ $1$ $1$ $()$ $2, 2, 2, 2, 2, 2, 2, 2$ $1$ $2$ $( 1, 2)( 3, 4)( 5, 6)( 7, 8)( 9,10)(11,12)(13,14)(15,16)$ $4, 4, 4, 4$ $1$ $4$ $( 1, 3, 6,16)( 2, 4, 5,15)( 7, 9,12,14)( 8,10,11,13)$ $4, 4, 4, 4$ $1$ $4$ $( 1, 4, 6,15)( 2, 3, 5,16)( 7,10,12,13)( 8, 9,11,14)$ $2, 2, 2, 2, 2, 2, 2, 2$ $1$ $2$ $( 1, 5)( 2, 6)( 3,15)( 4,16)( 7,11)( 8,12)( 9,13)(10,14)$ $2, 2, 2, 2, 2, 2, 2, 2$ $1$ $2$ $( 1, 6)( 2, 5)( 3,16)( 4,15)( 7,12)( 8,11)( 9,14)(10,13)$ $4, 4, 4, 4$ $1$ $4$ $( 1, 7, 5,11)( 2, 8, 6,12)( 3, 9,15,13)( 4,10,16,14)$ $4, 4, 4, 4$ $1$ $4$ $( 1, 8, 5,12)( 2, 7, 6,11)( 3,10,15,14)( 4, 9,16,13)$ $4, 4, 4, 4$ $1$ $4$ $( 1, 9, 2,10)( 3,12, 4,11)( 5,13, 6,14)( 7,15, 8,16)$ $4, 4, 4, 4$ $1$ $4$ $( 1,10, 2, 9)( 3,11, 4,12)( 5,14, 6,13)( 7,16, 8,15)$ $4, 4, 4, 4$ $1$ $4$ $( 1,11, 5, 7)( 2,12, 6, 8)( 3,13,15, 9)( 4,14,16,10)$ $4, 4, 4, 4$ $1$ $4$ $( 1,12, 5, 8)( 2,11, 6, 7)( 3,14,15,10)( 4,13,16, 9)$ $4, 4, 4, 4$ $1$ $4$ $( 1,13, 2,14)( 3, 8, 4, 7)( 5, 9, 6,10)(11,15,12,16)$ $4, 4, 4, 4$ $1$ $4$ $( 1,14, 2,13)( 3, 7, 4, 8)( 5,10, 6, 9)(11,16,12,15)$ $4, 4, 4, 4$ $1$ $4$ $( 1,15, 6, 4)( 2,16, 5, 3)( 7,13,12,10)( 8,14,11, 9)$ $4, 4, 4, 4$ $1$ $4$ $( 1,16, 6, 3)( 2,15, 5, 4)( 7,14,12, 9)( 8,13,11,10)$

## Group invariants

 Order: $16=2^{4}$ Cyclic: No Abelian: Yes Solvable: Yes GAP id: [16, 2]
 Character table:  2 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 1a 2a 4a 4b 2b 2c 4c 4d 4e 4f 4g 4h 4i 4j 4k 4l 2P 1a 1a 2c 2c 1a 1a 2b 2b 2a 2a 2b 2b 2a 2a 2c 2c 3P 1a 2a 4l 4k 2b 2c 4g 4h 4f 4e 4c 4d 4j 4i 4b 4a X.1 1 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 -1 -1 X.3 1 1 -1 -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 -1 1 1 X.5 1 -1 -1 1 -1 1 A -A -A A -A A A -A 1 -1 X.6 1 -1 -1 1 -1 1 -A A A -A A -A -A A 1 -1 X.7 1 -1 1 -1 -1 1 A -A A -A -A A -A A -1 1 X.8 1 -1 1 -1 -1 1 -A A -A A A -A A -A -1 1 X.9 1 -1 A -A 1 -1 -1 1 -A A -1 1 -A A A -A X.10 1 -1 -A A 1 -1 -1 1 A -A -1 1 A -A -A A X.11 1 -1 A -A 1 -1 1 -1 A -A 1 -1 A -A A -A X.12 1 -1 -A A 1 -1 1 -1 -A A 1 -1 -A A -A A X.13 1 1 A A -1 -1 A A -1 -1 -A -A 1 1 -A -A X.14 1 1 -A -A -1 -1 -A -A -1 -1 A A 1 1 A A X.15 1 1 A A -1 -1 -A -A 1 1 A A -1 -1 -A -A X.16 1 1 -A -A -1 -1 A A 1 1 -A -A -1 -1 A A A = -E(4) = -Sqrt(-1) = -i