# Do we know that 'most' finite groups are Galois groups of number fields?

The inverse Galois problem is a classical problem in mathematics and asks whether every finite group can be realized as the Galois group of a finite field extension of the rational numbers. The analogous question has been answered for function fields of a single variable over an algebraically closed field of characteristic zero. In the number field case, Shafarevich showed that any finite solvable group can be realized as the Galois group of a field extension over $\mathbb{Q}$.

In particular, we note that every $p$-group is solvable. More specifically, any group whose order is $2^k$ for some $k \in \mathbb{N}$ is solvable. Thus, any group of order $2^k$ is the Galois group of some number field over $\mathbb{Q}$. In a conversation with Bjorn Poonen in July, he mentioned that if there was a 'finite group' race, where he would list all the groups (up to isomorphism) of order $2^k \leq x$ and another person listed every other group whose order is at most $x$, then he would surely list more groups; and in fact, he said that his list will have a proportion approaching 100% as $x \rightarrow \infty$.

Applying Poonen's heuristic to the inverse Galois problem, then order groups by their order, a proportion approaching 100% of all finite groups are Galois groups of number fields over the rationals.

Has this been proved? Here, I am satisfied with a much weaker notion of 'most' which is the statement that a proportion exceeding $1/2$ of all groups of order up to $x$ are Galois groups as $x \rightarrow \infty$.

Edit: To clarify, I am asking whether statement 'a proportion exceeding $1/2$ of all finite groups are Galois groups over $\mathbb{Q}$' has been proved or not. This statement is implied by the statement `most finite groups have order $2^k$ for some $k$' by Shafarevich's theorem.

• Are you asking whether the fact that most groups are $2$-groups has been proved? Yes, it has. – Felipe Voloch Nov 27 '14 at 19:44
• Since one can take a direct product of groups of order $2^k$ with say $C_3$, it doesn't seem correct to say that most groups are $2$-groups. However, I would guess that most groups are solvable: the orders of nonsolvable groups seems to be understood, see oeis.org/A056866 . – Lucia Nov 27 '14 at 19:55
• @FelipeVoloch I don't believe that has been proved.It has been around as a conjecture for a long time. There is a stronger conjecture that most groups are nilpotent $2$-groups of class $2$. But the methods available to estimate the numbers are not precise enough to prove the result. – Derek Holt Nov 27 '14 at 20:38
• @DerekHolt: I see my error. It was that $2^{k+1} < 3\times 2^k$ and the groups of order $2^{k+1}$ swamp the groups of order $3\times 2^k$. – Lucia Nov 27 '14 at 20:50
• For more information, see the answer by Nick Gill to mathoverflow.net/questions/151491 – Derek Holt Nov 27 '14 at 21:18