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I'm tackling the following question:

Does there exist an irreducible polynomial $g(x) \in Q[x]$ of degree 5 with Galois group over Q isomorphic to $S_3×S_3$?

I think the answer is true. I figured out that $S_3×S_3$ is generated by the cycles $(1,2,3) , (1,2), (4,5,6), (4,5)$ but I cant find a polynomial, whose Galois group contains them.

Any help is appreciated!

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    $\begingroup$ If $g$ is irreducible of degree $d$, then $d$ divides the order of the Galois group. So the answer is... $\endgroup$ – Andreas Caranti Aug 29 '19 at 9:37
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Hint: We know that the size of the Galois group of a polynomial of degree $n$ divides $n!$.

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  • $\begingroup$ $S3×S3$ has 36 elements. 36 does not divide $120 = 5!$ and thus there can't exist such a Galois group, right? $\endgroup$ – Godsbane Aug 29 '19 at 9:47
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There are multiple ways to infer that $S_3×S_3$ cannot be isomorphous to the Galois group of an irreducible quintic polynomial.

First as noted elsewhere the order of this group ($36$) does not divide the order of $S_5 (120)$. We might also say that $S_3×S_3$ has two threefold symmetry elements independent of other symmetries whereas $S_5$ has only one, therefore by contradiction $S_3×S_3\not\in S_5$. These two interpretations are of course r wr lated; the second threefold element of $S_3×S_3$ which $S_5$ doesn't have causes the invisibility requirement to fail. If we had given up the second threefold element, specifying $S_3×Z_2$ instead of $S_3×S_3$, we would have met both the divisibility and containment requirements; $S_3×Z_2$ is a possible isomorphism for quintic polynomials.

Note that the last sentence does not contain the word "ireeducible". In an irreducible polynomial of prime degree $p$, the Galois group must contain a $p$-fold symmetry element, and $S_3×S_3$ fails this requirement for $p=5$. In fact even $S_3×Z_2$ fails also. In the latter case, this does not stop $S_3×Z_2$ from being isomorphous to the Galois group of quintic polynomials, but it does mean that such quintic polynomials must be reducible.

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