I more or less copied the following notes from my professor:
Everything about this proof makes perfect sense to me except the part in bold: Why is a subgroup (isomorphic) of ? It cannot be simply because acts on the four roots (because for instance any group of any order acts on any set trivially). All I can tell from that is that there is a homomorphism from into . So, how do I conclude that is a subgroup (isomorphic) of ?Find the Galois group of over .
Solution: has roots . Note that since acts on the four roots, is isomorphic to a subgroup of . The splitting field is , so . By Sylow, any two subgroups of of order must be isomorphic. So .
Also, I'm a bit confused as to how it can be that . For each is completely determined by and . Now, , which seems to imply . But then must be one of the four roots. So , a contradiction. Where did I go wrong, I wonder?
Any help would be much appreciated!