# Quintic polynomial over rationals , with Galois group of order more than $24$ , is not solvable by radicals?

Let $f(x) \in \mathbb Q[x]$ be a quintic polynomial, $E$ be the splitting field of $f(x)$ over $\mathbb Q$; if $|\text{Gal}(E/\mathbb Q)|>24$ then is it true that $f(x)$ is not solvable by radicals ?

I can see that considering $G:=\text{Gal}(E/\mathbb Q)|$ as a subgroup of $S_p$, the condition on $G$ implies $|G|=30,40,60,120$; if $|G|=60$ or $120$ then they correspond to the non-solvable groups $S_5$ or $A_5$, so $f(x)$ is not solvable then. So only remains $|G|=30, 40$, but since any group of order $30, 40$ is solvable, to prove the claim in affirmative we must show that Galois group of any quintic polynomial cannot be $30$ or $40$. I don't know how to proceed from here. Or is there any other approach?

Please help. Thanks in advance.

• If the result is true, you have to study the structure of these groups of order 30 and 40. The situation reminds me of the fact that the quaternion group of order 8 cannot be the Galois group of the splitting field of a quartic polynomial. – franz lemmermeyer Apr 25 '17 at 19:27

## 2 Answers

Your condition $[E:\mathbb Q] > 24$ clearly implies that the quintic is irreducible.

By this result (due to Evariste Galois himself) an irreducible polynomial of prime degree is solvable if and only if its splitting field is obtained by adjoining two roots. If you adjoin two roots of a quintic, you get a field extension of degree $\leq 20$. Hence the condition $[E:\mathbb Q] > 24$ yields that the polynomial is not solvable.

It clearly suffices to show that $S_5$ does not have a subgroup of order $30$ or $40$.

To see that $S_5$ has no subgroup of order $30$, we can proceed as follows: Easy combinatorics show that $S_5$ contains $20$ 3-cycles, hence $10$ 3-Sylow subgroups, thus the normalizer of a 3-Sylow subgroup has index $10$, hence $12$ elements. Now let $G$ be a group of 30 elements. The only possibilities for the number 3-Sylow subgroups are $1$ and $10$. If there is one 3-Sylowsubgroup, the normalizer of it must be the whole of $G$. Now if $G$ embedds into $S_5$, then the normalizer of 3-Sylow subgroup in $G$ is a subgroup of the normalizer of the same 3-Sylow subgroup in $S_5$, but this implies that $30 | 12$ by Lagrange's theorem which is absurd. Suppose there are $10$ 3-Sylow subgroups, then there are $20$ elements of order $3$ in $G$. This forces the number of $5$-Sylow subgroups to be $1$, for else $G$ would have too many elements. Now if there is only one 5-Sylow subgroup, then the normalizer of this 5-Sylow subgroup is $G$, but the normalizer of a 5-Sylow subgroup in $S_5$ has $20$ elements, so it is impossible that $G$ embedds into $S_5$.

To see that a group $G$ of order $40$ does not embedd into $S_5$ is even easier, because Sylow's theorems force the number of 5-Sylow subgroups to be $1$, so the normalizer of the 5-Sylow subgroup is $G$, but the normalizer of a 5-Sylow subgroup in $S_5$ has 20 elements, which would imply $40 | 20$ by Lagrange's theorem.