1. ## 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

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

2. Originally Posted by TheEmptySet
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 $F$.

3. Originally Posted by ThePerfectHacker
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

TheEmptySet

4. Originally Posted by ThePerfectHacker
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 $$ is a cyclic group with $=P$ and $ \le Z(G)$ so it is in the center and $ \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.

5. Originally Posted by TheEmptySet
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 $$ is a cyclic group with $=P$ and $ \le Z(G)$ so it is in the center and $ \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 $p^{n-1}\cdot p = p^n$.

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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).