Question 2.5.12 on page 23 of Ayman Badawi.
Probably a typo and they want $2\cdot \frac{m+1}2$ there.
A perhaps simpler argument: The existence part gives us a map $f\colon G\to G$ such that $f(a)^2=a$ for all $a$. Then $f$ is clearly injective, hence bijective, hence surjective. So for any $x$ with $x^2=a$ we have $x=f(y)$ for some $y$ and have $y=f(y)^2=x^2=a$ and hence $x=f(a)$.
Probably that's just a mistake. I think what the autor wanted to write is: $$b^{1+Ord(a)}=(b^2)^{(Ord(a)+1)/2}.$$