Galois group: 3T2

• Label: 3T2
• Degree: $3$
• Order: $6$
• Cyclic: no
• Abelian: no
• Solvable: yes
• Primitive: yes
• $p$-group: no
• Group: $S_3$

Group invariants

Abstract group:		$S_3$
Order:		$6=2 \cdot 3$
Cyclic:		no
Abelian:		no
Solvable:		yes
Nilpotency class:		not nilpotent

Group action invariants

Degree $n$:		$3$
Transitive number $t$:		$2$
CHM label:		$S3$
Parity:		$-1$
Primitive:		yes
$\card{\Aut(F/K)}$:		$1$
Generators:		$(1,3)$, $(1,2)$

Low degree resolvents

$\card{(G/N)}$	Galois groups for stem field(s)
$2$:	$C_2$

Resolvents shown for degrees $\leq 47$

Subfields

Prime degree - none

Low degree siblings

6T2

Siblings are shown with degree $\leq 47$

A number field with this Galois group has no arithmetically equivalent fields.

Conjugacy classes

Label	Cycle Type	Size	Order	Index	Representative
1A	$1^{3}$	$1$	$1$	$0$	$()$
2A	$2,1$	$3$	$2$	$1$	$(1,3)$
3A	$3$	$2$	$3$	$2$	$(1,2,3)$

Malle's constant $a(G)$:     $1$

Character table

		1A	2A	3A
	Size	1	3	2
	2 P	1A	1A	3A
	3 P	1A	2A	1A
	Type
6.1.1a	R	$1$	$1$	$1$
6.1.1b	R	$1$	$-1$	$1$
6.1.2a	R	$2$	$0$	$-1$

Indecomposable integral representations

Complete list of indecomposable integral representations:

Name	Dim	$(1,2,3) \mapsto $	$(1,2) \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 $\oplus$ $(A+A',L)$	$\cong$	$(A,L)$ $\oplus$ $(A',L)$

Regular extensions

$f_{ 1 } =$	$x^{3}+t x+t$
	The polynomial $f_{1}$ is generic for any base field $K$