Galois group of $x^7-2$

Question

Determine the fixed field of a non-trivial element of least order and the element of highest order and its action on the roots of $x^7-2$ in the galois group.

With a little bit of research I was able to find out that $\text{Gal}(x^7-2)\cong\mathbb{Z}_6\ltimes\mathbb{Z}_7$. Unfortunately I'm not really familiar with the semi direct product so I fail to classify that group any further (if that's possible at all). Is it of order $42$? And if so, how do I find a fixed field of degree $21$ over $\mathbb Q$ (which corresponds to an element of order 2)? Also I don't know how to find out which element is of highest order?

Edit: I would guess that I can take $\mathbb{Q}\left(\cos(\frac{2\pi}{7}),\sqrt[7]{2}\right)$ as the fixed field of the complex conjugation?

Possible duplicate of Galois Group of $X^p - 2$, p-an odd prime — lulu, Jul 23 '17 at 10:04
Yes, this thread is what I meant by research. As far as I can see it, my questions are not being discussed there. — user424862, Jul 23 '17 at 10:11
One element of order two in the Galois group is complex conjugation; what are the real elements of your splitting field? — Angina Seng, Jul 23 '17 at 10:15
Correct. The fixed field of complex conjugation is what you described, and also of degree $21$. For a description of the elements Tom Oldfield's answer may be more useful than mine. Alternatively, knowing that $3$ is a primitive root modulo $7$ you can show that as permutations of the set ${0,1,2,3,4,5,6}$ the Galois group is generated by $(0123456)$ and $(132645)$ corresponding to $x\mapsto x+1$ and $x\mapsto 3x$. — Jyrki Lahtonen, Jul 24 '17 at 05:38
(cont'd) The cyclic group of order six has seven conjugates in the full group, and they intersect trivially, so between them they cover the 35 elements that don't belong to the normal subgroup of order 7. — Jyrki Lahtonen, Jul 24 '17 at 05:41

Galois group of $x^7-2$

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