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I need to know if every group whose order is a power of a prime $p$ contains an element of order $p$? Should I proceed by picking an element $g$ of the group and proving that there is an element in $\langle g \rangle$ that has order $p$?

- The order of a conjugacy class is bounded by the index of the center
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- Is $SL(2, 3) $ a subgroup of $SL(2, p)$ for $ p>3$?
- Prove that there is no element of order $8$ in $SL(2,3)$
- Primitive Permutation Group with Subdegree 3

There are some results that are much stronger than that. Cauchy’s theorem states that every finite group whose order is divided by some prime $p$ has a subgroup of order $p$. And from Sylow’s theorem it can be deduced (although not immediately) that if the order of the group is $p^n$ then there is one subgroup of order $p^k$ for every $k=0,1,..,n$.

One more thing, a subgroup of order $p$ must be cyclic, that is, there has to be an element of order $p$ in it. That is because by Langrange’s theorem the order of every element must divide the $p$ and since it is prime then the order must be $1$ or $p$. Any element different from the identity will do the trick.

*Note*: Lagrange’s theorem alone is not enough to prove this since it only states that if the order of the group is $p^n$ then every subgroup is of the form $p^k$. That is because what Lagrange’s theorem says is that the order of every subgroup must divide the order of the group. So you actually need a little bit more I think.

*Late note due to nice comments*: While the result doesn’t follow directly from Lagrange’s theorem statement. It *can be derived* from it as it is nicely shown to you in other answers. So you actually can avoid appealing to a stronger result such as Cauchy’s theorem since you are in a finite $p$-group (what I mean by saying that it doesn’t follow directly is that Lagrange’s theorem makes no reference to $p$-groups, so there is math involved).

This follows immediately from Lagrange theorem, you don’t need any stronger result.

If the order of the group is $p^k$ with $k \neq 0$, then by Lagrange Theorem, the order of any element divides $p^k$.

Pick some $x \in G, x \neq e$. Then the order of $x$ is $p^m$ with $1 \leq m \leq k$. Let

$$y:=x^{p^{m-1}} \,.$$

Prove that the order of $y$ is $p$.

Not the group of order $p^0=1$!

Other than that, first prove that if the order of a group element $x$ is $mn$, then the order of $x^m$ is $n$. Then you can either show directly that if $x\in G$ and $|G|$ is finite, $x^{|G|}$ is the identity, or apply Lagrange’s theorem to $\langle x \rangle$.

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