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Properties

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

Related objects

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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 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, 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