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Properties

Label 4T1
4T1 1 2 1->2 3 2->3 4 3->4 4->1
Degree $4$
Order $4$
Cyclic yes
Abelian yes
Solvable yes
Primitive no
$p$-group yes
Group: $C_4$

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Show commands: Magma

Copy content magma:G := TransitiveGroup(4, 1);
 

Group invariants

Abstract group:  $C_4$
Copy content magma:IdentifyGroup(G);
 
Order:  $4=2^{2}$
Copy content magma:Order(G);
 
Cyclic:  yes
Copy content magma:IsCyclic(G);
 
Abelian:  yes
Copy content magma:IsAbelian(G);
 
Solvable:  yes
Copy content magma:IsSolvable(G);
 
Nilpotency class:  $1$
Copy content magma:NilpotencyClass(G);
 

Group action invariants

Degree $n$:  $4$
Copy content magma:t, n := TransitiveGroupIdentification(G); n;
 
Transitive number $t$:  $1$
Copy content magma:t, n := TransitiveGroupIdentification(G); t;
 
CHM label:   $C(4) = 4$
Parity:  $-1$
Copy content magma:IsEven(G);
 
Primitive:  no
Copy content magma:IsPrimitive(G);
 
$\card{\Aut(F/K)}$:  $4$
Copy content magma:Order(Centralizer(SymmetricGroup(n), G));
 
Generators:  $(1,2,3,4)$
Copy content magma:Generators(G);
 

Low degree resolvents

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

Resolvents shown for degrees $\leq 47$

Subfields

Degree 2: $C_2$

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

LabelCycle TypeSizeOrderIndexRepresentative
1A $1^{4}$ $1$ $1$ $0$ $()$
2A $2^{2}$ $1$ $2$ $2$ $(1,3)(2,4)$
4A1 $4$ $1$ $4$ $3$ $(1,2,3,4)$
4A-1 $4$ $1$ $4$ $3$ $(1,4,3,2)$

Malle's constant $a(G)$:     $1/2$

Copy content magma:ConjugacyClasses(G);
 

Character table

1A 2A 4A1 4A-1
Size 1 1 1 1
2 P 1A 1A 2A 2A
Type
4.1.1a R 1 1 1 1
4.1.1b R 1 1 −1 −1
4.1.1c1 C 1 −1 −i i
4.1.1c2 C 1 −1 i −i

Copy content magma:CharacterTable(G);
 

Indecomposable integral representations

Complete list of indecomposable integral representations:

Name Dim $(1,2,3,4) \mapsto $
Triv $1$ $\left(\begin{array}{r}1\end{array}\right)$
$B$ $1$ $\left(\begin{array}{r}-1\end{array}\right)$
$C$ $2$ $\left(\begin{array}{rr}0 & -1\\1 & 0\end{array}\right)$
$E$ $2$ $\left(\begin{array}{rr}0 & 1\\1 & 0\end{array}\right)$
$(C,\text{Triv})$ $3$ $\left(\begin{array}{rrr}1 & 0 & 1\\0 & 0 & -1\\0 & 1 & 0\end{array}\right)$
$(C,B)$ $3$ $\left(\begin{array}{rrr}-1 & 0 & 1\\0 & 0 & -1\\0 & 1 & 0\end{array}\right)$
$(C,E)$ $4$ $\left(\begin{array}{rrrr}0 & 1 & 0 & 0\\1 & 0 & 0 & 1\\0 & 0 & 0 & -1\\0 & 0 & 1 & 0\end{array}\right)$
$(C,E)'$ $4$ $\left(\begin{array}{rrrr}0 & 1 & 0 & 1\\1 & 0 & 0 & -1\\0 & 0 & 0 & -1\\0 & 0 & 1 & 0\end{array}\right)$
$(C,\text{Triv}+B)$ $4$ $\left(\begin{array}{rrrr}1 & 0 & 0 & 1\\0 & -1 & 0 & 1\\0 & 0 & 0 & -1\\0 & 0 & 1 & 0\end{array}\right)$
The decomposition of an arbitrary integral representation as a direct sum of indecomposables is unique.

Regular extensions

$f_{ 1 } =$ $x^{4} −2 s \left(1+t^{2}\right)x^{2} +s^{2} t^{2} \left(1+t^{2}\right)$ Copy content Toggle raw display
The polynomial $f_{1}$ is generic for any base field $K$ of characteristic $\neq$ 2