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Let $Q_8$ be the 8-element quaternion group. What is the minimum degree of an irreducible polynomial over a field which has a Galois group isomorphic to the $ Q_8 $ group?

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Let $f$ be an irreducible polynomial of degree $d$. Then its Galois group is a (transitive) subgroup of $S_d$, the symmetric group on $d$ letters. At this site, there is a proof that the quaternion group is not a subgroup of $S_d$ for $d<8$, so the answer to the question is, at least 8. But examples with degree 8 are given in the comments, so the minimal degree is 8.

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Assume a monic polynomial $f(x) \in \mathbb Q[x]$ has Galois group $Q_8$, and its splitting field is L. Choose a root $\alpha$ of $f(x)$ and consider $\mathbb Q(\alpha)$, which must be Galois over $\mathbb Q$, as all subgroups of $Q_8$ are normal. Hence it contains every conjugation of $\alpha$ i.e the roots of $f$. Therefore it is just the splitting field L, so we have $8=[L:\mathbb Q]=[\mathbb Q(\alpha):\mathbb Q]=\operatorname{deg} f$.

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