A finite group $G$ of order $n$ is said to be realizable (over $\mathbb{Q}$) if there exists a Galois extension $L/\mathbb{Q}$ such that $\mathrm{Gal}(L/\mathbb{Q})=G$. I'm curious what classes of groups are known to be realizable?

Shafarevich proved that solvable groups are realizable. Thus the following results "nontrivially" imply realizability:

If $n$ is odd, $G$ is solvable.

If $n$ has prime factorization $n=\prod p_i$, then $G$ is solvable.

If $n$ is of the form $p^\alpha q^\beta$ where $p,q$ are prime and $\alpha,\beta \geq 0$, $G$ is solvable.

It is worthwhile to point out that:

If $G_1,\ldots,G_m$ are solvable then so is $G_1 \times \cdots \times G_m$.

What other classes are known to be realizable? I've googled on it but can't find a big list or whatever.

(The smallest group I've found through Groupprops not to be immediately identifiable as realizable is $\mathrm{SL}(2,\mathbb{F}_5)$.)

Edit: If no one objects I'll continuously update this question whenever I find something interesting.

  • $\begingroup$ This is a problem of inverse Galois theory. I know nothing about inverse problems, but I'm fairly sure that it is an inappropriate tag. Taking the liberty of retagging a fine question. I am half-expecting somebody to give a link to a MO-question soon :-) $\endgroup$ – Jyrki Lahtonen Jul 27 '13 at 20:28
  • $\begingroup$ It's inverse GT, yes, hence an inverse problem. I don't get it :) $\endgroup$ – Erik Vesterlund Jul 27 '13 at 20:29
  • 2
    $\begingroup$ Have you (or I?) read the tag wiki of inverse problems. I think that inverse-problems form a class of analytical problems. That's kinda restrictive, but if analysts want to keep that game (or name) to themselves, I'm not gonna argue. $\endgroup$ – Jyrki Lahtonen Jul 27 '13 at 20:30
  • 2
    $\begingroup$ Dear Erik, Are you sure that $SL(2,\mathbb F_p)$ is known to be realizable as a Galois group over $\mathbb Q$ for large values of $p$? This contradicts the statement here, for example. Regards, $\endgroup$ – Matt E Jul 28 '13 at 4:52
  • 1
    $\begingroup$ @MattE I double-checked my reference and noticed my error: The precise statement is that for the case $SL(m,F_q)$ there exist extensions $\mathbb{Q} \subset K \subset L$ s.t $K/\mathbb{Q}$ is abelian and $L/K$ is Galois with $SL(m,F_q)$ as its group. Thanks for noticing my mistake. $\endgroup$ – Erik Vesterlund Jul 28 '13 at 5:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.