# Question 2.5.12 on page 23 of Ayman Badawi.

What I don't understand in this solution is the last line, where he writes the following: $$b=bb^{Ord(a)}=b^{1+Ord(a)}=(b^2)^{Ord(a)+1}$$ I don't understand this last identity, why is $$b^2=b$$?

From what does this follow?

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: $$b^{1+Ord(a)}=(b^2)^{(Ord(a)+1)/2}.$$