# Galois extensions of degrees $p$ and $p^{n-1}$ given a Galois extension of $p^n$

Suppose $K$ is a Galois extension of a field $F$ of degree $p^n$ for a $p$ a prime.

I want to see if there are Galois extensions of degrees $p$ and $p^{n-1}$ over $F$.

If $G=\text{Gal}(K/F)$, then $|G|=p^n$. If $G$ is abelian, I know there are subgroups of order $p^i$ for $0\leq i\leq n$, so there are subgroups of orders $p^{n-1}$ and $p$, and their corresponding fixed fields are of degrees $p$ and $p^{n-1}$ over $F$, and are Galois extensions since the subgroups are obviously normal in $G$.

But if $G$ is not abelian, is this still true?

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## 2 Answers

A group of order $p^n$ always has subgroups of order $p^{n-1}$ (in fact, all maximal subgroups are of order $p^{n-1}$), and they are always normal; and it always has subgroups of order $p$ that are normal (in fact central).

To see this, we use the class equation. Recall that if $G$ is a finite group and $Z(G)$ is the center of the group (the set of all elements $g\in G$ such that $gx=xg$ for all $x\in G$), then $$|G| = |Z(G)| + \sum [G:C(x_i)],$$ where $x_1,\ldots,x_n$ are representatives of the conjugacy classes of $G$ with more than one element. If $|G|=p^n$, then $[G:C(x_i)]$ is a multiple of $p$ for every $i$, so considering the equation modulo $p$ we conclude that $|Z(G)|\equiv 0 \pmod{p}$. Since $Z(G)$ has at least one element (the identity), it must be nontrivial.

Since $Z(G)$ is nontrivial, and is abelian, it has a subgroup of order $p$. This subgroup is normal in $G$. This gives you a normal subgroup of order $p$, and hence a field of degree $p^{n-1}$ over $F$ that is Galois.

To show that all maximal subgroups are of order $p^{n-1}$ and that they are all normal, we proceed by induction on $n$. If $n=1$ or $n=2$, then $G$ is abelian and we know the result is true. Assume the result is true for groups of order $p^{n-1}$, and let $G$ be of order $p^n$.

Let $N$ be a subgroup of $Z(G)$ of order $p$, and let $H$ be a maximal subgroup of $G$. If $N\subseteq H$, then consider $G/N$. This has order $p^{n-1}$, and $H/N$ is maximal (by the Lattice Isomorphism Theorem); hence $H/N$ is of order $p^{n-2}$ and normal in $G/N$, so $|H|=|N|\times|H/N| = p^{n-1}$, and is normal in $G$.

If $N$ is not contained in $H$, then $HN$ is a subgroup of $G$ that contains $H$ (since $N$ is central, it is normal, so $HN$ is a subgroup), hence $HN=G$. Since $|G|=p^n = |HN| = |H|\,|N|/|H\cap N|$, and $|H\cap N| = 1$ ($N$ is cyclic of order $p$ and not contained in $H$), then $|H|=p^{n-1}$. Moreover, given any $g\in G$, we can write $g=hn$ with $h\in H$ and $n\in N$. Then $gHg^{-1} = hnHn^{-1}h^{-1} = hHnn^{-1}h^{-1} = hHh^{-1} =H$ (with the second equality because $n$ is central). Thus, $H$ is normal in $G$.

Hence, every maximal subgroup of $G$ has order $p^{n-1}$ and is normal.

In conclusion, you can always find normal subgroups of $G$ of order $p$ and of order $p^{n-1}$, when $|G|=p^n$.

Added. You can now use this to show that $G$ always has subgroups of order $p^i$ that are normal for every $i$, $0\leq i\leq n$.

Proceed by induction on $n$. The result holds for groups of order $p$ and $p^2$. Assume the result holds for any group of order $p^n$, and let $G$ have order $p^{n+1}$. Let $N$ be a normal subgroup of order $p$, and consider $G/N$. Then $G/N$ has subgroups $\overline{H_i}$ of order $p^i$ that are normal in $G/N$, $i=0,\ldots,n$. These correspond to subgroups $H_i$ of $G$ that contain $N$, of order $p^{i+1}$, and that are normal in $G$. So $G$ has subgroups of order $p,\ldots,p^{n+1}$ that are normal; together with the trivial group of order $p^0$, this gives you subgroups of order $p^i$ that are normal for every $i$, $0\leq i\leq n+1$. $\Box$

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Thanks, this was easy to follow. – Turk Nov 24 '11 at 22:40

I recall to have seen this question in the Dummit & Foote's Abstract Algebra, and if you go to page 188 of this book you will see that Theorem 1 (3) gives you that $\mathrm{Gal}(K/F)$ has a normal subgroup of order $p^k$ for $0 \le k \le n$, thus giving you the Galois extensions you were looking for. The proofs are in the book.

Hope that helps,

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Thanks mate, I'll try to find this. – Turk Nov 24 '11 at 22:40