Let $f(x)$ be irreducible in the rationals so that the order of $\operatorname{Gal}(f(x))$ is $255$. Show that $f(x)$ is solvable by radicals.

Question

I know that $f(x)$ is solvable by radicals $\iff \def\Gal{\text{Gal}}\Gal(f(x))$ is solvable, so I began by trying to find what $\Gal(f(x))$ is. Let $G = \Gal(f(x)).$

Since $|G|= 255 = (3)(5)(17)$, I tried to show that all of the Sylow p-subgroups are normal, so $G$ is isomorphic to $Z_{255}$ and is, therefore, solvable. However, not all of the subgroups are necessarily normal -- $H_{17}$ is and either $H_3$ or $H_5$ is, but not necessarily both. So $G$ is not necessarily isomorphic to $Z_{255}$.

Am I thinking about this correctly? I feel like I'm correct in thinking that I need to find $G$ to show that it's solvable, but I'm questioning whether Sylow p-subgroups are the way to go. Is there a simpler way to find what $G$ is?

Thank you in advance!

You can mod out the known normal subgroup $H_{17}$. The quotient group is of order fifteen, and therefore cyclic. — Jyrki Lahtonen, May 03 '18 at 17:06
I failed to find an example of a non-abelian group of order 255. Then I proved to myself that such things don't exist. Luckily I don't have to type it here because we've done it already. — Jyrki Lahtonen, May 03 '18 at 17:12

score 2 · Accepted Answer · answered May 03 '18 at 17:05

2

Groups of order $pqr$ for primes $p<q<r$ are solvable:

Group of order $pqr$, $p < q < r$ primes

Hence the Galois groups is solvable, so that $f$ is solvable by radicals.

answered May 03 '18 at 17:05

Dietrich Burde

130,978

Let $f(x)$ be irreducible in the rationals so that the order of $\operatorname{Gal}(f(x))$ is $255$. Show that $f(x)$ is solvable by radicals.

1 Answers1