Properties

Label 16T42
Order \(32\)
n \(16\)
Cyclic No
Abelian No
Solvable Yes
Primitive No
$p$-group Yes
Group: $C_4\wr C_2$

Related objects

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Group action invariants

Degree $n$ :  $16$
Transitive number $t$ :  $42$
Group :  $C_4\wr C_2$
Parity:  $1$
Primitive:  No
Nilpotency class:  $3$
Generators:  (1,3,5,15)(2,4,6,16)(7,8)(9,10)(11,12)(13,14), (1,10)(2,9)(3,12)(4,11)(5,14)(6,13)(7,16)(8,15)
$|\Aut(F/K)|$:  $8$

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:  $C_2^2:C_4$

Resolvents shown for degrees $\leq 47$

Subfields

Degree 2: $C_2$ x 3

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

Degree 8: $D_4$, $C_4\wr C_2$ x 2

Low degree siblings

8T17 x 2, 16T28, 32T14

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

Group invariants

Order:  $32=2^{5}$
Cyclic:  No
Abelian:  No
Solvable:  Yes
GAP id:  [32, 11]
Character table:   
      2  5  4   4   4   4  5  4  5   4  3  3  3  3  5

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

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

A = -E(4)
  = -Sqrt(-1) = -i
B = -1-E(4)
  = -1-Sqrt(-1) = -1-i
C = 2*E(4)
  = 2*Sqrt(-1) = 2i