This is almost a duplicate of the following questions (but, read further):
But, I wasn't able to apply the proof to my case of a group of order 10 and an element of order 5.
In my case, if I try a proof by contradiction, I'm left with the fact that every element in $G$ is of order 2, but if I divide them into subgroups, every subgroup is of order 2, and has one unique element.
I also tried other methods to prove the theorem, but couldn't come up with anything meaningful.
I'm a beginner in the field, and am not familiar with theorems such as Cauchy's or Sylow's theorem (I'm aware of Lagrange's theorem). Therefore, please try to provide a simple answer, thanks!