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Math Help - Galois extension

  1. #1
    Behold, the power of SARDINES!
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    Galois extension

    Suppose K is a Galois extension of F of degree p^n for some prime p and some n \ge 1. Show there are Galois extensions of F contained in K of degrees p and p^{n-1}.

    So what I want is Fields F \subset E_1 \subset K or F \subset E_2 \subset K

    From the fundemental theorem of Galois I get the two diagrams relating the groups to the fields

    Galois extension-capture.jpg

    according to another part of the fundemental theorem I get

    [K:E_1]=|H_1| and [E_1:F]=|G:H_1|

    I get that E_1 is Galios over F iff H_1 is a normal subgroup of of G.

    [K:F]=|G|=p^{n} For some reason I think that G must be cyclic but I can't prove why. If that Is the case I am done.

    Thanks
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    Quote Originally Posted by TheEmptySet View Post
    Suppose K is a Galois extension of F of degree p^n for some prime p and some n \ge 1. Show there are Galois extensions of F contained in K of degrees p and p^{n-1}.
    You need to know some group theory. If G is a finite p-group, so |G| = p^n then there exist subgroup H,K so that |H|=p,|K|=p^{n-1}. This follows from the Sylow theorems. Now it can also be shown that K must automatically be normal (another group theory result). Since K is a normal subgroup it follows that its fixed field, which is over degree p, must be a normal (and therefore Galois) extension over F.
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    Quote Originally Posted by ThePerfectHacker View Post
    You need to know some group theory. If G is a finite p-group, so |G| = p^n then there exist subgroup H,K so that |H|=p,|K|=p^{n-1}. This follows from the Sylow theorems. Now it can also be shown that K must automatically be normal (another group theory result). Since K is a normal subgroup it follows that its fixed field, which is over degree p, must be a normal (and therefore Galois) extension over F.
    I did know some last term . Thanks I need to review the sylow theorems anyway. Thanks again

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    Quote Originally Posted by ThePerfectHacker View Post
    You need to know some group theory. If G is a finite p-group, so |G| = p^n then there exist subgroup H,K so that |H|=p,|K|=p^{n-1}. This follows from the Sylow theorems. Now it can also be shown that K must automatically be normal (another group theory result). Since K is a normal subgroup it follows that its fixed field, which is over degree p, must be a normal (and therefore Galois) extension over F.

    Okay I think I have a proof. Please tell me if it looks okay. I didn't see how I could use the sylow theorems. If there is a way I would like to know.

    Since |G|=P^n via the class equation this group must have a non trivial center Z(G) \ne 1 and Z(G) is ablelian and since it is a subgroup its order must divide P^n so |Z(G)|=p^m So this group must have an element of order P call it x. Then <x> is a cyclic group with <x>=P and <x> \le Z(G) so it is in the center and <x> \underline\vartriangleleft G.

    So G has a normal subgroup of order p.

    to get the group of order p^{n-1} I will use induction.

    Base case n=1 |G|=p

    case n=2 |G|=p^2 would be covered by the above argument

    Asume true for n

    |G|=p^n has a normal subgroup H of order |H|=p^{n-1}

    Show n+1

    |G|=p^{n+1} so by the first argument this has a normal subgroup of order P call it N.

    So then |G/N|=p^n by the induction hypothsis this has a normal subgroup H of order |H|=p^{n-1} so

    H/N \underline\vartriangleleft G/N Now by the 4th isomorphism theorem(lattice) H \underline\vartriangleleft G and |H|=p^n and this completes the induction.
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    Quote Originally Posted by TheEmptySet View Post
    Since |G|=P^n via the class equation this group must have a non trivial center Z(G) \ne 1 and Z(G) is ablelian and since it is a subgroup its order must divide P^n so |Z(G)|=p^m So this group must have an element of order P call it x. Then <x> is a cyclic group with <x>=P and <x> \le Z(G) so it is in the center and <x> \underline\vartriangleleft G.



    to get the group of order p^{n-1} I will use induction.

    Base case n=1 |G|=p

    case n=2 |G|=p^2 would be covered by the above argument

    Asume true for n

    |G|=p^n has a normal subgroup H of order |H|=p^{n-1}

    Show n+1

    |G|=p^{n+1} so by the first argument this has a normal subgroup of order P call it N.

    So then |G/N|=p^n by the induction hypothsis this has a normal subgroup H of order |H|=p^{n-1} so

    H/N \underline\vartriangleleft G/N Now by the 4th isomorphism theorem(lattice) H \underline\vartriangleleft G and |H|=p^n and this completes the induction.
    This is what I would do (since I am not sure what you did in the end, but I like your approach). Define \pi : G\to G/N to be the natural projection \pi (x) = xN. Now we know there is H\triangleleft G/N with order p^{n-1} so \pi^{-1}(H) is a normal subgroup of G with order p^{n-1}\cdot p = p^n.

    ---

    There is another way. If G is finite group and H a p-group then (G:H) \equiv (N(H):H)(\bmod p).
    That immediately shows that not only does there exist a normal subgroup of order p^{n-1} but shows any normal subgroup of that order must be normal!
    (Since |N(H)| = |G| in this case).
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