Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 9 down vote accepted

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 $i$, $0\leq i\leq n+1$. $\Box$

share|cite|improve this answer
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,

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.