how can you make bx = xb without commutative law?
(you make apply b^(-1) since b = b^(-1)
This is a WAY different question than what you originally asked. It's asking you to prove that if b is the only element of order 2, then it commutes with every element of G.
Let $\displaystyle x \in G$ and consider $\displaystyle xbx^{-1}$. What is the order of this element? What does that tell you about its relation to the element b?