# Irreducible Polynomial of degree $~5~$ isomorphic to the direct Product of $S_3$

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!

• If $g$ is irreducible of degree $d$, then $d$ divides the order of the Galois group. So the answer is... – Andreas Caranti Aug 29 '19 at 9:37

Hint: We know that the size of the Galois group of a polynomial of degree $$n$$ divides $$n!$$.
• $S3×S3$ has 36 elements. 36 does not divide $120 = 5!$ and thus there can't exist such a Galois group, right? – Godsbane Aug 29 '19 at 9:47
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.