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$G$ is a group of order $p^k$ where $p$ is prime and $k$ is a positive integer. I want to show that every element of $G$ must be of order $p^i$, for some $i$ with $0\le i\le k$.

I have some ideas but no concrete proof yet.

  • Is this like saying that for a group with only prime elements, it will be cyclic? Since any $a\in G$ is $a=p^i$, some $i$. So I need to show $G$ is cyclic?

  • Maybe Lagrange's theorem will be useful here?

  • There must be an element in $G$ that has order $p$ right? Not sure how to show this though.

  • $|G|=p^k$, for $p$ prime. Need to show that any $a\in G$ is $a=p^i$, for $0\le i\le k$. Since every element of $G$ is a power of a prime, it must generate a subgroup for $G$?

I know there are some similar posts on here, but I'm not completely sure if they are actually related and if I'm thinking correctly so far. I found this: Show that every group of prime order is cyclic , but I'm not sure if it's the same idea and if I need to do something more specific to ensure I'm answering my question fully.

Dietrich Burde
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eddie
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    Yes, Lagrange is enough. Let $x\in G$ and $H=\langle x\rangle$. Then $|H|$ divides $|G|=p^k$, so that $|H|=p^n$ for some $n$. Here $H$ is a cyclic group. So $x$ has order $p^n$. You should write $p$ instead of $g$ for a prime number. – Dietrich Burde Mar 21 '22 at 16:16
  • @DietrichBurde I'm not completely sure how to actually use the Lagrange theorem. This is my first time doing a question with it and I'm not completely sure how to put it all together properly. I have some ideas but I'm having a hard time putting it together. – eddie Mar 21 '22 at 16:19
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    @DietrichBurde Thanks, I figured it out!! :) – eddie Mar 21 '22 at 16:47
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    Careful, $a\in G$, but $a=p^i$ doesn't make sense for a prime number $p$. The elements of $G$ need not be numbers. The numbers $p^i$ denote the order of elements. – Dietrich Burde Mar 21 '22 at 17:34

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