Group of order 10 has an element of order 5, without using Cauchy's or Sylow's theorems This is almost a duplicate of the following questions (but, read further):
Group of order $63$ has an element of order $3$, without using Cauchy's or Sylow's theorems
Show any group of order $275$ has an element of order $5$.
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!

Since $|G|=10$, therefore for $g \in G$ the possible orders are $1,2,5,10$. Suppose 


*

*There is an element $a \in G$ such that $|a|=10$, then $|a^2|=5$, thus we have an element of order $5$.

*Suppose there is an element of order $5$, then we have nothing to prove.

*So what if all the elements have order $2$? Then you can show that the group is abelian (small exercise). Now consider distinct elements $a,b \in G$ such that $|a|=|b|=2$, then $|ab|=2$, hence $H=\{e,a,b,ab\}$ will form a  subgroup of $G$. But $|H|=4$ and $4 \not| 10$. It violates Lagrange's theorem. So such a $H$ cannot exist. That means not all the elements can have order $2$.


I hope this resolves it within the scope of things you know at this stage.
