↑

Properties

Label 16T26
Order \(32\)
n \(16\)
Cyclic No
Abelian No
Solvable Yes
Primitive No
$p$-group Yes
Group: $D_4:C_4$

Related objects

Learn more about

Group action invariants

Degree $n$ :  $16$
Transitive number $t$ :  $26$
Group :  $D_4:C_4$
Parity:  $1$
Primitive:  No
Nilpotency class:  $3$
Generators:  (1,3,2,4)(5,16,6,15)(7,13,8,14)(9,11,10,12), (1,5)(2,6)(3,4)(7,15)(8,16)(9,14)(10,13)(11,12)
$|\Aut(F/K)|$:  $4$

Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$ x 3
4:  $C_4$ x 2, $C_2^2$
8:  $D_{4}$ x 2, $C_4\times C_2$
16:  $D_{8}$, $QD_{16}$, $C_2^2:C_4$

Resolvents shown for degrees $\leq 47$

Subfields

Degree 2: $C_2$

Degree 4: $C_4$, $D_{4}$ x 2

Degree 8: $D_{8}$, $QD_{16}$, $C_2^2:C_4$

Low degree siblings

16T26, 32T12

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

Group invariants

Order:  $32=2^{5}$
Cyclic:  No
Abelian:  No
Solvable:  Yes
GAP id:  [32, 9]
Character table:    
      2  5  3  5  3  3  4  3  4  4  4  4  4  5  5

        1a 2a 2b 2c 4a 8a 4b 8b 4c 4d 8c 8d 2d 2e
     2P 1a 1a 1a 1a 2b 4c 2b 4c 2d 2d 4c 4c 1a 1a
     3P 1a 2a 2b 2c 4b 8c 4a 8d 4c 4d 8a 8b 2d 2e
     5P 1a 2a 2b 2c 4a 8d 4b 8c 4c 4d 8b 8a 2d 2e
     7P 1a 2a 2b 2c 4b 8b 4a 8a 4c 4d 8d 8c 2d 2e

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      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
X.5      1 -1 -1  1  A -A -A  A -1  1  A -A  1 -1
X.6      1 -1 -1  1 -A  A  A -A -1  1 -A  A  1 -1
X.7      1  1 -1 -1  A  A -A -A -1  1 -A  A  1 -1
X.8      1  1 -1 -1 -A -A  A  A -1  1  A -A  1 -1
X.9      2  . -2  .  .  .  .  .  2 -2  .  .  2 -2
X.10     2  .  2  .  .  .  .  . -2 -2  .  .  2  2
X.11     2  . -2  .  .  B  . -B  .  .  B -B -2  2
X.12     2  . -2  .  . -B  .  B  .  . -B  B -2  2
X.13     2  .  2  .  .  C  .  C  .  . -C -C -2 -2
X.14     2  .  2  .  . -C  . -C  .  .  C  C -2 -2

A = -E(4)
  = -Sqrt(-1) = -i
B = -E(8)-E(8)^3
  = -Sqrt(-2) = -i2
C = -E(8)+E(8)^3
  = -Sqrt(2) = -r2