↑

Properties

Label 16T44
Order \(32\)
n \(16\)
Cyclic No
Abelian No
Solvable Yes
Primitive No
$p$-group Yes
Group: $D_8:C_2$

Related objects

Learn more about

Group action invariants

Degree $n$ :  $16$
Transitive number $t$ :  $44$
Group :  $D_8:C_2$
Parity:  $1$
Primitive:  No
Nilpotency class:  $3$
Generators:  (1,7,16,9,2,8,15,10)(3,13,6,12,4,14,5,11), (1,8)(2,7)(3,14)(4,13)(5,11)(6,12)(9,16)(10,15), (1,14,2,13)(3,7,4,8)(5,9,6,10)(11,16,12,15)
$|\Aut(F/K)|$:  $4$

Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$ x 7
4:  $C_2^2$ x 7
8:  $D_{4}$ x 2, $C_2^3$
16:  $D_4\times C_2$

Resolvents shown for degrees $\leq 47$

Subfields

Degree 2: $C_2$ x 3

Degree 4: $C_2^2$, $D_{4}$ x 2

Degree 8: $D_4\times C_2$

Low degree siblings

16T44, 16T47, 32T26

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

Group invariants

Order:  $32=2^{5}$
Cyclic:  No
Abelian:  No
Solvable:  Yes
GAP id:  [32, 42]
Character table:    
      2  5  3  5  3  5  5  4  4  3  4  4  3  4  4

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

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

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