5
$\begingroup$

Let $Q$ be a finite simple group which may be realized as the Galois group of some extension of $\mathbb{Q}$ (like for instance $PSL_2(\mathbb{F}_p)$ for $p\geq 5$, or the monster group) and let $G$ be an extension of $Q$ by a cyclic group. Then is it possible to see that $G$ is realizable as a Galois group over $\mathbb{Q}$?

$\endgroup$
  • $\begingroup$ You should clarify if you want just $G$ to be realizable as Galois group, or to be realizable with quotient a given realization. This second case is called the embedding problem en.m.wikipedia.org/wiki/Embedding_problem $\endgroup$ – Xarles Oct 29 '18 at 7:32
  • $\begingroup$ I'm not asking for a realization of the quotient map. $\endgroup$ – Anwesh Ray Oct 29 '18 at 7:42
1
$\begingroup$

It has occurred to me not too long after posting this question that the answer is obviously no, for instance the groups $\text{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ are known to be Galois over $\mathbb{Q}$ but the groups $\text{SL}_2(\mathbb{Z}/p\mathbb{Z})$ are not known to be Galois over $\mathbb{Q}$ for large $p$.

$\endgroup$
  • 1
    $\begingroup$ Have I got the wrong end of the stick? I thought that there weren't any groups that had been proved not to be the group of some Galois extension of the rationals. $\endgroup$ – Gerry Myerson Nov 16 '18 at 11:22
  • 1
    $\begingroup$ I meant that $\text{SL}_2(\mathbb{Z}/p\mathbb{Z})$ is not known to be the group of some Galois extension of the rationals not for lack of trying. $\endgroup$ – Anwesh Ray Nov 16 '18 at 16:36

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.