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1

$begingroup$

$newcommandQ{mathbb Q}$Suppose $G$ is a transitive finite permutation subgroup of $S_{2n}$ having fixpoint-free involution in and outside its center. Suppose furthermore that I know that there is a (totally) imaginary Galois number field $K$ whose Galois group is isomorphic to $G$. So I also want $[K:Q]=2n$. My question is the following:

If the complex conjugation is not in the center of $operatorname{Gal}(K/Q)$ then the maximal real subfield of $K$ is not Galois. But can I find another imaginary Galois number field $L$ with Galois group isomorphic to $G$ such that the complex conjugation is in the center of $G$? Vice versa, if the maximal real subfield of $operatorname{Gal}(K/Q)$ is Galois, can I find an imaginary number field $L$ with Galois group isomorphic to $G$ such that its maximal real subfield is not Galois?

field-theory galois-theory algebraic-number-theory

share|cite|improve this question

edited Mar 10 at 14:03

Bernard

123k741116

asked Mar 10 at 13:55

quantumquantum

538210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the smallest group matching your conditions?
  $endgroup$
  – franz lemmermeyer
  Mar 11 at 6:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Starting with degree 6 the smallest group with GAP description 6T11 matches this condition: galoisdb.math.upb.de/groups/view?deg=6&num=11.
  $endgroup$
  – quantum
  19 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To be precise: two examples would be the splitting fields of $x^6 + 3x^5 + 9x^4 + 13x^3 + 14x^2 + 8x + 2$ and $x^6 - 2x^5 + 2x^4 - 2x^3 + 2x^2 - x + 1$ respectively having Galois group 6T11 and one has maximal real subfield that is Galois while the other no.
  $endgroup$
  – quantum
  19 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

$newcommandQ{mathbb Q}$Suppose $G$ is a transitive finite permutation subgroup of $S_{2n}$ having fixpoint-free involution in and outside its center. Suppose furthermore that I know that there is a (totally) imaginary Galois number field $K$ whose Galois group is isomorphic to $G$. So I also want $[K:Q]=2n$. My question is the following:

If the complex conjugation is not in the center of $operatorname{Gal}(K/Q)$ then the maximal real subfield of $K$ is not Galois. But can I find another imaginary Galois number field $L$ with Galois group isomorphic to $G$ such that the complex conjugation is in the center of $G$? Vice versa, if the maximal real subfield of $operatorname{Gal}(K/Q)$ is Galois, can I find an imaginary number field $L$ with Galois group isomorphic to $G$ such that its maximal real subfield is not Galois?

field-theory galois-theory algebraic-number-theory

share|cite|improve this question

edited Mar 10 at 14:03

Bernard

123k741116

asked Mar 10 at 13:55

quantumquantum

538210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the smallest group matching your conditions?
  $endgroup$
  – franz lemmermeyer
  Mar 11 at 6:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Starting with degree 6 the smallest group with GAP description 6T11 matches this condition: galoisdb.math.upb.de/groups/view?deg=6&num=11.
  $endgroup$
  – quantum
  19 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To be precise: two examples would be the splitting fields of $x^6 + 3x^5 + 9x^4 + 13x^3 + 14x^2 + 8x + 2$ and $x^6 - 2x^5 + 2x^4 - 2x^3 + 2x^2 - x + 1$ respectively having Galois group 6T11 and one has maximal real subfield that is Galois while the other no.
  $endgroup$
  – quantum
  19 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

$newcommandQ{mathbb Q}$Suppose $G$ is a transitive finite permutation subgroup of $S_{2n}$ having fixpoint-free involution in and outside its center. Suppose furthermore that I know that there is a (totally) imaginary Galois number field $K$ whose Galois group is isomorphic to $G$. So I also want $[K:Q]=2n$. My question is the following:

If the complex conjugation is not in the center of $operatorname{Gal}(K/Q)$ then the maximal real subfield of $K$ is not Galois. But can I find another imaginary Galois number field $L$ with Galois group isomorphic to $G$ such that the complex conjugation is in the center of $G$? Vice versa, if the maximal real subfield of $operatorname{Gal}(K/Q)$ is Galois, can I find an imaginary number field $L$ with Galois group isomorphic to $G$ such that its maximal real subfield is not Galois?

field-theory galois-theory algebraic-number-theory

share|cite|improve this question

edited Mar 10 at 14:03

Bernard

123k741116

asked Mar 10 at 13:55

quantumquantum

538210

$endgroup$

share|cite|improve this question

edited Mar 10 at 14:03

Bernard

123k741116

asked Mar 10 at 13:55

quantumquantum

538210

share|cite|improve this question

edited Mar 10 at 14:03

Bernard

123k741116

asked Mar 10 at 13:55

quantumquantum

538210

edited Mar 10 at 14:03

Bernard

123k741116

edited Mar 10 at 14:03

Bernard

123k741116

asked Mar 10 at 13:55

quantumquantum

538210

asked Mar 10 at 13:55

quantumquantum

538210