**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.20**

Let $p$ be a prime and $n$ a positive integer. Show that if $x$ is an element of a group $G$ such that $x^{p^n} = 1$, then $|x| = p^m$ for some $1 \leq m \leq n$.

Solution: We prove a lemma (§2.3 Proposition3).

**Lemma**: Let $G$ be a group and $x \in G$ an element of finite order, say, $|x| = n$. If $x^m = 1$, then $n$ divides $m$.

Proof: Suppose to the contrary that $n$ does not divide $m$; then by the Division Algorithm there exist integers $q$ and $r$ such that $0 < r < |n|$ and $m = qn + r$. Then we have $$1 = x^m = x^{qn+r} = (x^n)^q + x^r = x^r.$$ But recall that by definition $n$ is the least positive integer with this property, so we have a contradiction. Thus $n$ divides $m$. $\square$

The main result then follows because every divisor of $p^n$ is of the form $p^m$.