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 $\displaystyle x^5 - 11$ over $\displaystyle \mathbb{Q}$.

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

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

Now let $\displaystyle \tau: \sqrt[5]{11}\mapsto \zeta \sqrt[5]{11} \mbox{ and }\zeta \mapsto \zeta$, $\displaystyle \sigma: \sqrt[5]{11}\mapsto \sqrt[5]{11} \mbox{ and }\zeta \mapsto \zeta^2$.

Thus, $\displaystyle \tau^5 = 1$ and $\displaystyle \sigma^4 = 1$, also $\displaystyle \sigma \tau = \tau^2 \sigma$.

Finally, $\displaystyle \{ \tau^{i}\sigma^j | 0\leq i \leq 4 \mbox{ and }0\leq j \leq 3\}$ is a set of distinct $\displaystyle \mathbb{Q}$-automorphism.

There are $\displaystyle 20$ all together. Thus we have found all elements of $\displaystyle \mbox{Gal}(K/\mathbb{Q})$.

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).