Consider a group $G$ of order $10$.

Then $G$ can be abelian as well non-abelian.

What is the number of non-trivial elements of $G$ of order $2$?

**Answer:**

If $G$ is abelian, $G$ can be cyclic as well non-cyclic.

If $G$ is cyclic ,then it is isomorphic to $\mathbb{Z}\_{10}$,which has only $1$ element of order $2$ i.e., $\bar 5$ .

If $G$ is non-cyclic,then also it has $1$ element of order $2$.

But if $G$ be non-abelian,then the situation becomes different.

By Sylow principle, $G$ has $5$ Sylow $2$-subgroups and hence there are $5$ elements of order $2$.

My question is-

**Edit:** How to turn the non-abelian case when the group will abelian using Sylow theorem?? 

In non-abelian case we get $5$ sylow $2$-subgroup. 
How this turn into $1$ sylow $2$-subgroup is $G$ becomes abelian ??