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Math Help - Please help clarify this Galois proof

  1. #1
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    Please help clarify this Galois proof

    I more or less copied the following notes from my professor:

    Find the Galois group of x^4-2 over \mathbb{Q}.

    Solution: x^4-2 has roots \pm\sqrt[4]{2},\pm i\sqrt[4]{2}. Note that since G acts on the four roots, G is isomorphic to a subgroup of S_4. The splitting field is \mathbb{Q}(\sqrt[4]{2},i), so |G|=8. By Sylow, any two subgroups of S_4 of order 8 must be isomorphic. So G\cong D_8.
    Everything about this proof makes perfect sense to me except the part in bold: Why is G a subgroup (isomorphic) of S_4? It cannot be simply because G 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 G into S_4. So, how do I conclude that G is a subgroup (isomorphic) of S_4?

    Also, I'm a bit confused as to how it can be that |G|=8. For each \alpha\in G is completely determined by \alpha(i) and \alpha(\sqrt[4]{2}). Now, 1=\alpha(1)=\alpha(-i^2)=-\alpha(i)^2, which seems to imply \alpha(i)=i. But then \alpha(\sqrt[4]{2}) must be one of the four roots. So |G|\leq 4, a contradiction. Where did I go wrong, I wonder?

    Any help would be much appreciated!
    Last edited by hatsoff; December 5th 2010 at 07:16 AM.
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  2. #2
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    Quote Originally Posted by hatsoff View Post
    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 G a subgroup (isomorphic) of S_4? It cannot be simply because G 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 G into S_4. So, how do I conclude that G is a subgroup (isomorphic) of S_4?

    Any help would be much appreciated!

    As you remark the part in bold is not completely correct. For it to be it should be that the Galois group acts transitively on

    the four roots, and then it'd follow that the group is isomorphic to a subgroup of S_4.

    There's another way to show that G\cong D_8: first, it's easy to see that G cannot be abelian and that

    its order is divisible by four and divides 8, so it is either D_8\,\,or\,\,Q_8. Nevertheless, this extension has a non-normal

    subextension, \mathbb{Q}(\sqrt[4]{2})/\mathbb{Q} , from where it follows that there is a non-normal subgroup of G

    of index 4, and since all the subgroups Q_8 are normal then we're done.

    Tonio
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  3. #3
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    Quote Originally Posted by hatsoff View Post
    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 G a subgroup (isomorphic) of S_4? It cannot be simply because G 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 G into S_4. So, how do I conclude that G is a subgroup (isomorphic) of S_4?

    Also, I'm a bit confused as to how it can be that |G|=8. For each \alpha\in G is completely determined by \alpha(i) and \alpha(\sqrt[4]{2}). Now, 1=\alpha(1)=\alpha(-i^2)=-\alpha(i)^2, which seems to imply \alpha(i)=i. But then \alpha(\sqrt[4]{2}) must be one of the four roots. So |G|\leq 4, a contradiction. Where did I go wrong, I wonder?

    Any help would be much appreciated!

    \alpha(i)^2=-1\Longrightarrow \alpha(i)=\pm i , since i^2=(-i)^2=-1

    Tonio
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  4. #4
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    Okay, I see now what is going on. Thanks a bunch for the help!
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