# Galois Group of Polynomial

• Jun 7th 2008, 07:08 PM
ThePerfectHacker
Galois Group of Polynomial
I had this homework problem, I think I got it correct, just want a check from somebody.

Find the Galois group of $x^5 - 11$ over $\mathbb{Q}$.
I get that the splitting field is $K = \mathbb{Q}(\sqrt[5]{11},\zeta)$ where $\zeta = e^{2\pi i/5}$.

The next step is to calculate the dimension of this field. Here I used: let $f(x)$ be irreducible over $F$ with degree $n$ and $[K:F]=m$, then $f(x)$ is irreducible over $K$ if $\gcd(n,m)=1$. Since $x^4+x^3+x^2+x+1$ is minimal polynomial for $\zeta$ it follows from above that this polynomial is irreducible over $\mathbb{Q}(\sqrt[5]{11})$. Thus, $[K:\mathbb{Q}(\sqrt[5]{11})] = 4$. And thus, $[K:\mathbb{Q}] = 5\cdot 4 = 20$.

Now let $\tau: \sqrt[5]{11}\mapsto \zeta \sqrt[5]{11} \mbox{ and }\zeta \mapsto \zeta$, $\sigma: \sqrt[5]{11}\mapsto \sqrt[5]{11} \mbox{ and }\zeta \mapsto \zeta^2$.
Thus, $\tau^5 = 1$ and $\sigma^4 = 1$, also $\sigma \tau = \tau^2 \sigma$.
Finally, $\{ \tau^{i}\sigma^j | 0\leq i \leq 4 \mbox{ and }0\leq j \leq 3\}$ is a set of distinct $\mathbb{Q}$-automorphism.
There are $20$ all together. Thus we have found all elements of $\mbox{Gal}(K/\mathbb{Q})$.

My remaining question is what type of group is this?
I think this is the Frobenius group.
And what is the subgroup diamgram of this group?
(Wikipedia did not have it).
• Jun 7th 2008, 09:26 PM
NonCommAlg
Quote:

Originally Posted by ThePerfectHacker
I had this homework problem, I think I got it correct, just want a check from somebody.

Find the Galois group of $x^5 - 11$ over $\mathbb{Q}$.
I get that the splitting field is $K = \mathbb{Q}(\sqrt[5]{11},\zeta)$ where $\zeta = e^{2\pi i/5}$.

The next step is to calculate the dimension of this field. Here I used: let $f(x)$ be irreducible over $F$ with degree $n$ and $[K:F]=m$, then $f(x)$ is irreducible over $K$ if $\gcd(n,m)=1$. Since $x^4+x^3+x^2+x+1$ is minimal polynomial for $\zeta$ it follows from above that this polynomial is irreducible over $\mathbb{Q}(\sqrt[5]{11})$. Thus, $[K:\mathbb{Q}(\sqrt[5]{11})] = 4$. And thus, $[K:\mathbb{Q}] = 5\cdot 4 = 20$.

or even easier: if $\gcd([F(a):F],[F(b):F])=1,$ then $[F(a,b):F]=[F(a):F][F(b):F].$

Quote:

My remaining question is what type of group is this?
I think this is the Frobenius group.
that's right.

Quote:

And what is the subgroup diagram of this group?
(Wikipedia did not have it).
answering this question will take some time and effort! you may find the answer (at least partially) in a decent group theory book.
• Jun 7th 2008, 10:07 PM
ThePerfectHacker
Quote:

Originally Posted by NonCommAlg
answering this question will take some time and effort! you may find the answer (at least partially) in a decent group theory book.

I guess it seems I have to the hard work. The subgroups must have orders: 1,2,4,5,10,20. The ones with 1,2,5 are easy to find because they are cyclic. The one with 4 is a little harder but still easy, either cyclic, or generated by two order two elements. That I can do. The one that is harder is 10. The cyclic one simply does not exist. This means if it has one then it must have D_5 subgroup. Or possibly a few of these. Can you provide me with the # of D_5 subgroups.

This is Mine 99:):)th Post!!!