# A Group with a Prime Power Order Elements Has Order a Power of the Prime.

## Problem 17

Let $p$ be a prime number. Suppose that the order of each element of a finite group $G$ is a power of $p$. Then prove that $G$ is a $p$-group. Namely, the order of $G$ is a power of $p$.

## Hint.

You may use Sylow’s theorem.
For a review of Sylow’s theorem, please check out the post Sylow’s Theorem (summary).

## Proof.

If $G$ is a trivial group, then the claim is trivial. So assume that $|G|>1$.

Seeking a contradiction, suppose that $|G|=p^nm$ for some $n,m \in \Z$ and $p$ and $m>1$ are relatively prime.
Let $l$ be a prime factor of $m$. Then by Sylow’s theorem, there exists a Sylow $l$-subgroup of $G$.
The order of a nontrivial element of this subgroup is divisible by the prime $l$ and this contradicts that each element has order power of $p$ since $l$ and $p$ are relatively prime.

## Comment.

If we assume Sylow’s theorem, then the proof of this problem is straightforward.
How about proving it more directly (without using Sylow’s theorem)?

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##### Any Subgroup of Index 2 in a Finite Group is Normal

Show that any subgroup of index $2$ in a group is a normal subgroup.

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