Subgroups of $\mathbb F_{p^n}$

Question

Is it possible to give a discription of the possible subgroups (with respect to $+$) of the finite field $\mathbb F_{p^n}$ (obviously, $p$ is a prime number). Of course, if $n = 1$, $(\mathbb F_p,+)$ is simple, so we may as well assume $n > 1$. The overrall question I'm mostly interested in is: For any $k$ with $1 < k < n$, does there exist a subgroup of order $p^k$?

Answer 1

If I'm not mistaken in understanding of your question. Let's $G$ be an abelian group of order $p^n$. Does a subgroup of order $p^k$ exist? Yes, it exists:

1. If $G$ is cyclic then subgroup of order $p^k$ is generated by element $g = p^{n-k}$. It is an implication of the fact that $g$ has order $p^k$.

2. Every abelian group of order $p^n$ can be represented as a sum of $m$ cyclic groups. In each of them you can take subgroup of order $s_i, i \in \{1, \dots, m\}$ such that $s_1 + \dots + s_m = k$. Then you have to take their direct sum and get a subgroup of a desired order.

Answer 2

To generalize Jihad's answer:

Let $G$ be any group of order $p^n$. Then $G$ has a subgroup of order $p^k$ for every $k=0,\dots,n$. Proof by induction:

Case 1: $G$ is abelian. This is Jihad's case.

Case 2: $G$ is not abelian. Nonetheless (by a standard argument based on the class equation) it has nontrivial center. Say $|Z(G)|=p^m$ with $0<m<n$. Then $G$ has subgroups of order $p^k$ for $0\leq k\leq m$ by case 1, and $G/Z(G)$ has subgroups of order $p^k$ for $0\leq k \leq n-m$ (where $p^{n-m}$ is the order of $G/Z(G)$) by induction; and the corresponding subgroups in $G$ are of orders $p^k$ for $m\leq k \leq n$.