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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1$\begingroup$ Note that if $Q_8$ does not act transitively on $4$ elements, because $\mathbf S_4$ has no subgroup isomorphic to it. Also relevant is this example of degree 8. $\endgroup$– MyselfDec 30, 2014 at 0:56
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1$\begingroup$ Richard Dean showed $x^8 - 72 x^6 + 180 x^4 - 144 x^2 + 36$ has $Q_8$ as its Galois group. More information at en.wikipedia.org/wiki/Quaternion_group#Galois_group also starting page 67 of alg-geo.epfl.ch/travdipl/GaloisInverse.pdf $\endgroup$– Gerry MyersonDec 30, 2014 at 1:13
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$\begingroup$ Another construction at projecteuclid.org/download/pdf_1/euclid.pja/1195512561 and another at feryll.blogspot.com/2013/10/quaternion-galois-group-14227.html $\endgroup$– Gerry MyersonDec 30, 2014 at 1:19
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2$\begingroup$ I don't understand why this question has so many votees to close, honestly. Someone should gather Gerry's comments and write an answer. $\endgroup$– Mariano Suárez-ÁlvarezDec 30, 2014 at 7:37
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$\begingroup$ @Mariano, I don't know whether any of the sites I linked to have a proof that degree 8 is best possible, so I'm not sure that even taken together they constitute an answer. $\endgroup$– Gerry MyersonDec 31, 2014 at 23:30
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2 Answers
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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.
answered Jan 2, 2015 at 16:08
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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 $8=[L:\mathbb Q]=[\mathbb Q(\alpha):\mathbb Q]=\operatorname{deg} f$.