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I'm trying to prove that every group which has order equal to a power of a prime has an element of order $p$.

The first doubt I have is that if $|G| = p^0$ for some prime $p$, then $G$ is just the trivial group consisting only of the identity element, which doesn't have prime order. So I believe what this theorem is really saying is a "positive" power of a prime. Is that a correct interpretation? I'm going to proceed in my attempt under the assumption that $|G| = p^n$ for some $n > 0$.

EDIT: I made a mistake in my original argument and wrongly asserted that $G$ was cyclic. Here is my updated attempt based on some helpful feedback.

Let $G$ be a group of order $p^n$ for some prime $p$ and $n > 0$. Then $|G| > 1$, so $G$ contains a non-trivial element $g$. Because $g$ generates a cyclic subgroup of order $|\langle g \rangle| = |g|$, the order of $g$ must divide $p^n$ by Lagrange's theorem. So $|g| = p^k$ for some $1 \leq k \leq n$. Then the cyclic subgroup generated by $g$ is $$ \langle g \rangle = \left\{e, g, g^2, \ldots, g^{p^{k-1}} \right\}. $$ Consider $g^{p^{k-1}}$. Then $$ \left(g^{p^{k-1}}\right)^p = g^{p^k} = e, $$ so $g^{p^{k-1}}$ has order $p$.

How does this look?

Brad G.
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  • That argument almost works. The first part, showing the order of an element divides the order of the group, also confirms your clarification of the assertion that the statement should be "every group whose order is a positive power of p contains an element or order p". You claim G has to be cyclic, which isn't true, but why do you need that? – krm2233 Jul 10 '23 at 02:47
  • Why do you conclude that $G$ is cyclic of order $p^n$? This is false. The Klein 4-group is a group of order $2^2$ but it is not cyclic. – Dave S. Jul 10 '23 at 02:47

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Yes, you must assume that the group is non-trivial to get an element of prime order. You seem to have gotten started quite nicely, but you cannot conclude that your entire group $G$ is cyclic; it is certainly not true that every group of prime power order is cyclic! Your element $g$ need not have order equal to $p^n$ (the order of $G$), but may have order $p^k$ for some non-zero $k < n$. However, you can apply the remainder of your argument to the cyclic subgroup $\langle g\rangle$ of $G$ and get a suitable power of $g$ whose order is $p$.

James
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