Properties

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

Related objects

Group action invariants

Degree $n$ :  $6$
Transitive number $t$ :  $2$
Group :  $S_3$
CHM label :  $D_{6}(6) = [3]2$
Parity:  $-1$
Primitive:  No
Generators:   (1,3,5)(2,4,6), (1,4)(2,3)(5,6)
$|\Aut(F/K)|$:  $6$
Low degree resolvents:
 2: 2T1

Subfields

Degree 2: $C_2$

Degree 3: $S_3$

Low degree siblings

3T2
A number field with this Galois group has no arithmetically equivalent fields.

Conjugacy Classes

 Cycle Type Size Order Representative $1, 1, 1, 1, 1, 1$ $1$ $1$ $()$ $2, 2, 2$ $3$ $2$ $(1,2)(3,6)(4,5)$ $3, 3$ $2$ $3$ $(1,3,5)(2,4,6)$

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 $\oplus$ $(A+A',L)$ $\cong$ $(A,L)$ $\oplus$ $(A',L)$