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We say that some rational polynomial, [tex]f[/tex] has a Galois group isomorphic to the quaternions. We can then conclude that the polynomial has degree [tex]n \geq 8[/tex].

I have a few thoughts on this and I might be overlooking something simple...but letting [tex]K[/tex] be the splitting field, then [tex][K:\mathbb{Q}][/tex] divides [tex]n![/tex] so [tex]n \geq 4[/tex].

My other thought is that since the Galois group has finite number of subgroups then between K and the rational numbers are a finite number of intermediate fields, thus K is simple. With [tex]K \cong \mathbb{Q}(a)[/tex] for some root of a minimal polynomial whose degree must be 8 since [tex]\mathbb{Q}(a) \cong \mathbb{Q}[x]/\langle m_a \rangle [/tex].

Does [tex]m_a[/tex] divide [tex]f[/tex] or something?