# How to get a polynomial corresponding to a solvable Galois group of order 60?

Of course, we know the $$A_5$$ of $$60$$ order is an unsolvable group. But as the wiki here, there are also $$12$$ solvable groups in the same $$60$$ order still:

Then I have generated many many irreducible polynomials whose Galois group is $$60$$ order with maple program to check, but none of them is solvable. What's wrong?

Is there a solvable polynomial whose Galois group is $$60$$ order? If it exists, can you please give any example?

• It doesn't work with degree $5$, because $S_5$ has no transitive subgroup of order $60$ except for $A_5$. But it works for degree $15$, see the table here. For example the solvable group $C_3⋊F_5$, where $F_5$ is the Frobenius group of order $20$. Commented May 20, 2022 at 12:39

The irreducible polynomial $$x^{15}-30x^{10}-3708x^5-2$$ has Galois group $$C_3\rtimes F_5,$$ which is solvable of order $$60$$. Here $$F_5$$ is the Frobenius group of order $$20$$.
• Yes, the transitive groups of order $60$ are just known. In this case, irreducible polynomials are also known, see the status of the Inverse Galois Problem, what is known and what is unknown. Commented May 21, 2022 at 8:30