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Can the Galois group of $x^5-5x^3+4x+1 \in \mathbb{Q}[x] $ be isomorphic to $S_5$?

I know that it has five real roots, then I think that it is impossible. I think that this Galois group cannot contain a 2-cycle. Any idea about how I can prove this?

pmtm
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    I know little about Galois Theory, but I understand you can have $S_5$ with one, three or five real roots. In fact given that $x^5-5x^3+\color{blue}{5}x+1=0$ is radical-soluble (actually, also reducible), I would not expect the given equation to be so also. – Oscar Lanzi Jul 23 '21 at 18:06
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    $$x^5-5x^3+4x+1=(x-2)(x-1)x(x+1)(x+2)+1$$ – razivo Jul 23 '21 at 18:08
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    Are you familiar Dedekind's theorem, locally described here. That will settle the matter quickly. I can disclose that after you factor this modulo the primes $2$, $7$ and $11$ you are able to deduce that the Galois group is, indeed, $S_5$. I have used the technique in a couple of answers already, so I will rather let someone new to the site do the honors. – Jyrki Lahtonen Jul 23 '21 at 18:22
  • Modulo $2$ factorization of this polynomial was done for example in this old answer of mine. Modulo the other two you can find zeros. Wait! There should be a prime that leads to a simpler argument (together with $p=2$). Give me a moment. – Jyrki Lahtonen Jul 23 '21 at 18:24
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    Hmm. It may not be so easy to check that modulo $p=79$ this polynomial has exactly three distinct zeros and hence an irreducible quadratic factor (and hence the Galois group indeed contains a 2-cycle). It is something you can test quickly with a spreadsheet though. Full factorization (modulo the other smaller primes I mentioned) needs a more powerful CAS. Have you access to one? – Jyrki Lahtonen Jul 23 '21 at 18:29
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    To spell out the plan of @JyrkiLahtonen a bit more, I note that $f$ is irreducible mod $2$, factors as a linear times quartic mod $7$, and as two linear time cubic mod $11$. (And three linear times quadratic mod $79$, which is very easy with CAS but annoying by hand). To use $2, 7, 11$, one would want to show that $S_5$ is generated by a $3$-cycle, a $4$-cycle, and a $5$-cycle. But this is true, as $5 \cdot 4 \cdot 3 = 60$ and the $4$ cycle is odd, and thus the subgroup generated by the cycles is at least half the size of the group and larger than $A_5$, and thus all of $S_5$. – davidlowryduda Jul 23 '21 at 20:19
  • @JyrkiLahtonen: since the polynomial is irreducible (as noted in one of the comments) the group is transitive. I also checked the discriminant (via pari/gp) which turns out to be a big prime. So the group is not contained in $A_5$. I wonder if we can figure out the Galois group using a listing of transitive subgroups of $S_5$. – Paramanand Singh Jul 27 '21 at 09:39
  • @ParamanandSingh $S_5$ has transitive subgroups of orders $10,20$ not contained in $A_5$. Namely the dihedral group $D_5$ and the holomorph $C_5\rtimes C_4$ often denoted $F_{20}$. This does mean that if you can show that $3\mid |G|$ then you are done. So finding a prime $p$ such that modulo $p$ the quintic factors either as $2\times3$ or $1\times1\times3$ means that you are done. Actually, cycle types $(5)$ and $(3,2)$ suffice with information about the discriminant because the latter is an odd permutation. – Jyrki Lahtonen Jul 29 '21 at 10:25
  • @JyrkiLahtonen : $p=11$ does it and the polynomial factors as $1\times 1\times 3$. – Paramanand Singh Jul 29 '21 at 11:47
  • Mod $11$ the polynomial factors as $(x-3)(x-6)(x^3-2x^2+3x-3)$. The identity can be checked by manually verifying it for $x=0,1,2,3,4,6$ (using expression given by @razivo). The third factor is irreducible mod $11$ as none of $0,1,2,\dots,10$ is a root. – Paramanand Singh Jul 29 '21 at 12:16
  • ^ ... cycle types $(5)$ and $(3,2)$ suffice without information about the discriminant... – Jyrki Lahtonen Jul 29 '21 at 13:57
  • @razivo can you explain your answer a little more? – Johnny Apple Dec 24 '23 at 09:33

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