Exercise 2.3.20 (Order of $x$ if $x^{p^n} = 1$)

Question

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

Richard Ganaye · Accepted Answer

\begin{proof} 
If $x^{p^n} = 1$, then $|x|$ divides $p^n$ (Proposition 3). Since $p$ is prime, the divisors of $p^n$ are the integers $p^m$, where $m\leq n$. Therefore
$$|x| = p^m,\qquad 0 \leq m \leq n.$$

\end{proof}

Exercise 2.3.20 (Order of $x$ if $x^{p^n} = 1$)

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Exercise 2.3.20 (Order of $x$ if $x^{p^n} = 1$)

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