Properties

Label 4T2
Order \(4\)
n \(4\)
Cyclic No
Abelian Yes
Solvable Yes
Primitive No
$p$-group Yes
Group: $V_4$

Related objects

Group action invariants

Degree $n$ :  $4$
Transitive number $t$ :  $2$
Group :  $V_4$
CHM label :  $E(4) = 2[x]2$
Parity:  $1$
Primitive:  No
Generators:   (1,2)(3,4), (1,4)(2,3)
$|\Aut(F/K)|$:  $4$
Low degree resolvents:  
2: 2T1, 2T1, 2T1

Subfields

Degree 2: $C_2$, $C_2$, $C_2$

Low degree siblings

There is no other low degree representation.
A number field with this Galois group has no arithmetically equivalent fields.

Conjugacy Classes

Cycle TypeSizeOrderRepresentative
$ 1, 1, 1, 1 $ $1$ $1$ $()$
$ 2, 2 $ $1$ $2$ $(1,2)(3,4)$
$ 2, 2 $ $1$ $2$ $(1,3)(2,4)$
$ 2, 2 $ $1$ $2$ $(1,4)(2,3)$

Group invariants

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

       1a 2a 2b 2c
    2P 1a 1a 1a 1a

X.1     1  1  1  1
X.2     1 -1 -1  1
X.3     1 -1  1 -1
X.4     1  1 -1 -1

Indecomposable integral representations

Partial list of indecomposable integral representations:

Name Dim $(1,2)(3,4) \mapsto $ $(1,4)(2,3) \mapsto $
Triv $1$ $\left(\begin{array}{*{1}{r}}1\end{array}\right)$ $\left(\begin{array}{*{1}{r}}1\end{array}\right)$
$A_1$ $1$ $\left(\begin{array}{*{1}{r}}1\end{array}\right)$ $\left(\begin{array}{*{1}{r}}-1\end{array}\right)$
$A_2$ $1$ $\left(\begin{array}{*{1}{r}}-1\end{array}\right)$ $\left(\begin{array}{*{1}{r}}1\end{array}\right)$
$A_3$ $1$ $\left(\begin{array}{*{1}{r}}-1\end{array}\right)$ $\left(\begin{array}{*{1}{r}}-1\end{array}\right)$
$B_1$ $2$ $\left(\begin{array}{*{2}{r}}-1 & 0\\0 & -1\end{array}\right)$ $\left(\begin{array}{*{2}{r}}0 & 1\\1 & 0\end{array}\right)$
$B_2$ $2$ $\left(\begin{array}{*{2}{r}}0 & 1\\1 & 0\end{array}\right)$ $\left(\begin{array}{*{2}{r}}-1 & 0\\0 & -1\end{array}\right)$
$B_3$ $2$ $\left(\begin{array}{*{2}{r}}0 & 1\\1 & 0\end{array}\right)$ $\left(\begin{array}{*{2}{r}}0 & -1\\-1 & 0\end{array}\right)$
$A_2B_1$ $2$ $\left(\begin{array}{*{2}{r}}1 & 0\\0 & 1\end{array}\right)$ $\left(\begin{array}{*{2}{r}}0 & 1\\1 & 0\end{array}\right)$
$A_3B_1$ $2$ $\left(\begin{array}{*{2}{r}}1 & 0\\0 & 1\end{array}\right)$ $\left(\begin{array}{*{2}{r}}0 & -1\\-1 & 0\end{array}\right)$
$A_1B_2$ $2$ $\left(\begin{array}{*{2}{r}}0 & 1\\1 & 0\end{array}\right)$ $\left(\begin{array}{*{2}{r}}1 & 0\\0 & 1\end{array}\right)$
$A_3B_2$ $2$ $\left(\begin{array}{*{2}{r}}0 & -1\\-1 & 0\end{array}\right)$ $\left(\begin{array}{*{2}{r}}1 & 0\\0 & 1\end{array}\right)$
$A_1B_3$ $2$ $\left(\begin{array}{*{2}{r}}0 & 1\\1 & 0\end{array}\right)$ $\left(\begin{array}{*{2}{r}}0 & 1\\1 & 0\end{array}\right)$
$A_2B_3$ $2$ $\left(\begin{array}{*{2}{r}}0 & -1\\-1 & 0\end{array}\right)$ $\left(\begin{array}{*{2}{r}}0 & -1\\-1 & 0\end{array}\right)$
$J$ $3$ $\left(\begin{array}{*{3}{r}}0 & -1 & 1\\0 & -1 & 0\\1 & -1 & 0\end{array}\right)$ $\left(\begin{array}{*{3}{r}}0 & 1 & -1\\1 & 0 & -1\\0 & 0 & -1\end{array}\right)$
$J'$ $3$ $\left(\begin{array}{*{3}{r}}0 & 0 & 1\\-1 & -1 & -1\\1 & 0 & 0\end{array}\right)$ $\left(\begin{array}{*{3}{r}}0 & 1 & 0\\1 & 0 & 0\\-1 & -1 & -1\end{array}\right)$
The decomposition of an arbitrary integral representation as a direct sum of indecomposables is unique.