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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?

Thanks in advance.

  • 1
    $\begingroup$ Please don't rely on pictures of text $\endgroup$
    – Shaun
    Dec 28, 2021 at 14:08
  • $\begingroup$ Which Badawi book are you referring to? $\endgroup$
    – Shaun
    Dec 28, 2021 at 14:15

2 Answers 2


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


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