It is well known that there exists some subfield $K$ of a cyclotomic field with $G\cong\mathbf{Gal}(K/\mathbb{Q})$ for any finite abelian group $G$. So if there exists a finite group $G$ which can not appear as a Galois group over $\mathbb{Q}$, this group must be non-abelian. But I can not find a explicit one.
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13This is known as the Inverse Galois Problem and it remains unsolved – lulu Nov 05 '24 at 06:34
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See also this MO-post for the smallest finite group, where this is not known. Also, this post. – Dietrich Burde Nov 05 '24 at 09:28
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A nice survey on the inverse Galois problem is given also at wikipedia. In general, the conjecture is still open. The results are as follows so far.
1.) All finite exceptional simple groups arise as a Galois group over $\Bbb Q$, except possibly the Mathieu group $M_{23}$ of order $3^3\cdot 2^5\cdot 4\cdot 7\cdot 11 \cdot 23= 10200960$.
2.) All $13$ non-abelian simple groups smaller than $PSL(2,25)$ of order $7800$ are known to be realizable over $\mathbb {Q}$. However, already $PSL(2,27)$ is open.
3.) All finite solvable groups can arise (Shafarevich).
4.) All groups $S_n$ and $A_n$ can arise (Hilbert).
Dietrich Burde
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2Any individual $PSL_2(F)$, where $F$ is a finite field in which $-1$ is a square, shouldn’t be too difficult to realize by searching for a suitable modular form. But the remaining cases such as $PSL(2,27)$ look a lot harder, because their lifts to $GL_2$ have to be even and therefore cannot come as easily from geometry. – Aphelli Nov 05 '24 at 12:54
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This is an open problem, "the inverse Galois problem", as mentioned.
There's various partial results, such as yes (they are realizable) for permutation groups of degree up to $5.$
Apparently it dates back to the $19$th century.
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1If I recall correctly, every $S_n$, every $GL_2(\mathbb{F}_p)$ and every solvable group are also known to be realizable. – Aphelli Nov 05 '24 at 07:51
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