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Properties

Label 6T2
Order \(6\)
n \(6\)
Cyclic No
Abelian No
Solvable Yes
Primitive No
$p$-group No
Group: $S_3$

Related objects

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Group action invariants

Degree $n$ :  $6$
Transitive number $t$ :  $2$
Group :  $S_3$
CHM label :  $D_{6}(6) = [3]2$
Parity:  $-1$
Primitive:  No
Nilpotency class:  $-1$ (not nilpotent)
Generators:  (1,3,5)(2,4,6), (1,4)(2,3)(5,6)
$|\Aut(F/K)|$:  $6$

Low degree resolvents

|G/N|Galois groups for stem field(s)
2:  $C_2$

Resolvents shown for degrees $\leq 47$

Subfields

Degree 2: $C_2$

Degree 3: $S_3$

Low degree siblings

3T2

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 $ $1$ $1$ $()$
$ 2, 2, 2 $ $3$ $2$ $(1,2)(3,6)(4,5)$
$ 3, 3 $ $2$ $3$ $(1,3,5)(2,4,6)$

Group invariants

Order:  $6=2 \cdot 3$
Cyclic:  No
Abelian:  No
Solvable:  Yes
GAP id:  [6, 1]
Character table:    
     2  1  1  .
     3  1  .  1

       1a 2a 3a
    2P 1a 1a 3a
    3P 1a 2a 1a

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

Indecomposable integral representations

Complete list of indecomposable integral representations:

Name Dim $(1,3,5)(2,4,6) \mapsto $ $(1,2)(3,6)(4,5) \mapsto $
Triv $1$ $\left(\begin{array}{r}1\end{array}\right)$ $\left(\begin{array}{r}1\end{array}\right)$
Sign $1$ $\left(\begin{array}{r}1\end{array}\right)$ $\left(\begin{array}{r}-1\end{array}\right)$
$L$ $2$ $\left(\begin{array}{rr}1 & 0\\0 & 1\end{array}\right)$ $\left(\begin{array}{rr}0 & 1\\1 & 0\end{array}\right)$
$A$ $2$ $\left(\begin{array}{rr}0 & 1\\-1 & -1\end{array}\right)$ $\left(\begin{array}{rr}1 & 0\\-1 & -1\end{array}\right)$
$A'$ $2$ $\left(\begin{array}{rr}0 & 1\\-1 & -1\end{array}\right)$ $\left(\begin{array}{rr}-1 & 0\\1 & 1\end{array}\right)$
$(A,\textrm{Sign})$ $3$ $\left(\begin{array}{rrr}0 & 1 & 0\\-1 & -1 & 0\\1 & 0 & 1\end{array}\right)$ $\left(\begin{array}{rrr}1 & 0 & 0\\-1 & -1 & 0\\-1 & 0 & -1\end{array}\right)$
$(A',\textrm{Triv})$ $3$ $\left(\begin{array}{rrr}0 & 1 & 0\\-1 & -1 & 0\\1 & 0 & 1\end{array}\right)$ $\left(\begin{array}{rrr}-1 & 0 & 0\\1 & 1 & 0\\1 & 0 & 1\end{array}\right)$
$(A,L)$ $4$ $\left(\begin{array}{rrrr}0 & 1 & 0 & 0\\-1 & -1 & 0 & 0\\-1 & 0 & 1 & 0\\1 & 0 & 0 & 1\end{array}\right)$ $\left(\begin{array}{rrrr}1 & 0 & 0 & 0\\-1 & -1 & 0 & 0\\1 & 0 & 0 & 1\\-1 & 0 & 1 & 0\end{array}\right)$
$(A',L)$ $4$ $\left(\begin{array}{rrrr}0 & 1 & 0 & 0\\-1 & -1 & 0 & 0\\1 & 0 & 1 & 0\\1 & 0 & 0 & 1\end{array}\right)$ $\left(\begin{array}{rrrr}-1 & 0 & 0 & 0\\1 & 1 & 0 & 0\\1 & 0 & 0 & 1\\1 & 0 & 1 & 0\end{array}\right)$
$(A+A',L)$ $6$ $\left(\begin{array}{rrrrrr}0 & 0 & 1 & 0 & 0 & 0\\0 & 0 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 0 & 1 & 0\\0 & 0 & 0 & 0 & 0 & 1\\1 & 0 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & 0 & 0\end{array}\right)$ $\left(\begin{array}{rrrrrr}0 & 0 & 0 & 1 & 0 & 0\\0 & 0 & 1 & 0 & 0 & 0\\0 & 1 & 0 & 0 & 0 & 0\\1 & 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 1\\0 & 0 & 0 & 0 & 1 & 0\end{array}\right)$
The decomposition of an arbitrary integral representation as a direct sum of indecomposables is not unique, in general. It is unique up to the following isomorphisms:
Triv $\oplus$ $(A',L)$ $\cong$ $L$ $\oplus$ $(A',\textrm{Triv})$
Sign $\oplus$ $(A,L)$ $\cong$ $L$ $\oplus$ $(A,\textrm{Sign})$
Triv $\oplus$ $(A+A',L)$ $\cong$ $(A,L)$ $\oplus$ $(A',L)$