1. ## Galois extension

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

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

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

according to another part of the fundemental theorem I get

$\displaystyle [K:E_1]=|H_1|$ and $\displaystyle [E_1:F]=|G:H_1|$

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

$\displaystyle [K:F]=|G|=p^{n}$ 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

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

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

TheEmptySet

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

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

So G has a normal subgroup of order p.

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

Base case n=1 $\displaystyle |G|=p$

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

Asume true for n

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

Show n+1

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

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

$\displaystyle H/N \underline\vartriangleleft G/N$ Now by the 4th isomorphism theorem(lattice) $\displaystyle H \underline\vartriangleleft G$ and $\displaystyle |H|=p^n$ and this completes the induction.

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

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

Base case n=1 $\displaystyle |G|=p$

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

Asume true for n

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

Show n+1

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

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

$\displaystyle H/N \underline\vartriangleleft G/N$ Now by the 4th isomorphism theorem(lattice) $\displaystyle H \underline\vartriangleleft G$ and $\displaystyle |H|=p^n$ 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 $\displaystyle \pi : G\to G/N$ to be the natural projection $\displaystyle \pi (x) = xN$. Now we know there is $\displaystyle H\triangleleft G/N$ with order $\displaystyle p^{n-1}$ so $\displaystyle \pi^{-1}(H)$ is a normal subgroup of $\displaystyle G$ with order $\displaystyle p^{n-1}\cdot p = p^n$.

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