I have a question... if for every finite simple group, we can construct a Galois extension over $\mathbb{Q}$ with that Galois group, does it follow that we can construct Galois extensions over $\mathbb{Q}$ with any finite group as Galois group?

Is the above known to be true or false? I was just using naive reasoning thinking finite simple groups are the "building blocks" of all finite groups (Jordan-Hölder theorem), so I'm just guessing that the inverse Galois problem reduces to the problem for finite simple groups.

Is there a proper investigation of the above?


1 Answer 1


The inverse Galois problem lies much deeper than the problem of realizing finite simple groups. Let me try to give two reasons:

  1. It is very easy to realize a cyclic group of order $p$ as a Galois group over $\mathbb{Q}$. But in order to realize any $p$-group much more is needed. One needs to study embedding problems: Given an exact sequence $$ 1\to \mathbb{Z}/p\mathbb{Z}\to \tilde{G}\to G \to 1 $$ of $p$-groups and a realization $G\cong Gal(L/\mathbb{Q})$, when $L$ can be embedded into a field $F$, such that $Gal(F/\mathbb{Q})\cong \tilde{G}$ and the restriction map $Gal(F/\mathbb{Q})\to Gal(L/\mathbb{Q})$ coincides with the map of the exact sequence.

    This was done by Richardt and Scholtz for odd $p$, and it is much more difficult than realizing cyclic groups. A good reference is Serre's book "Topics in Galois theory".

    This approach was later extended by Shafarevich to realize all solvable groups. Again solving embedding problems is very difficult, while realizing the simple groups here (which are cyclic) is very easy.

  2. Here is a sketch of how to find a field $K$ with the property that every simple group can be realized as a Galois group although not every finite group:

    Let $\Gamma=\prod_{S} S$ (where the product runs over all finite simple groups) and $\Lambda$ the universal Frattini cover of $\Gamma$. Then $\Lambda$ is a projective profinite group, hence it is realizable as the absolute Galois group of a field $K$ (by a Theorem of Lubotsky and van-der Dries).

    The set of finite Galois groups over $K$ coincides with the set of (continuous) quotients of $\Lambda$. By the theory of universal Frattini covers, the latter coincides with the set of finite quotients of $\Gamma$, which are products of simple groups. (In particular $Sym(5)$, say, is not a Galois group over $K$.)

    EDIT: As Jack Smith noted, $\Lambda$ has more finite quotients, namely all the finite Frattini covers of products of finite simple groups. But one can readily show that these are not all finite groups (since for example the kernel of a Frattini cover is nilpotent, but I skip details since this doesn't directly relate to the question...)

  • 2
    $\begingroup$ In 2, why aren't finite quotients of $\Lambda$ allowed as continuous quotients? For instance, why not the perfect group of order 120, $\operatorname{SL}(2,5)$? $\endgroup$ Jul 3, 2014 at 16:17
  • $\begingroup$ @JackSchmidt. Sure, you are right. I was referring to Lemma 22.6.3 in Fried-Jarden "Field Arithmetic", 3rd edition, that determines the continuous quotients of the universal Frattini cover of a profinite group, and I over looked what the lemma says. I added an edit in the answer. Thanks for pointing this out. $\endgroup$
    – Lior B-S
    Jul 3, 2014 at 21:59

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