Existence of elements of order $p^k$ in a finite group

Question

By Cauchy's theorem, if a prime number $p$ divides the order of a group $G$, then there is an element $g$ in $G$ whose order is $p$. In addition, if $|G|$ admits a prime decomposition with a factor $p^k$, then Sylow's theorem tells us that a subgroup of order $p^k$ must exist.

My question is the following. If $|G|$ admits a prime decomposition with a factor $p^k$, does this imply that an element of order $p^k$ must exist in $G$?

Update: As explained in Mindlack's answer (among others), the implication does not hold and simple counterexamples exist. In the comments, Mindlack discusses special cases where such an element exists.

Answer 1

No, because that would imply that $G$ has a cyclic Sylow subgroup, which is not always true. A simple example: $G=\mathbb{F}_p^2$ for addition.

Answer 2

No. $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ has order $2^2$ but no elements of order $2^2$.

Answer 3

No, $G= C_p \times \cdots \times C_p$ ($k$ factors) has order $p^k$ but all nontrivial elements have order $p$.