I engaged in an exercise

Let $G$ be finite. Suppose that $\left\vert \{x\in G\mid x^n =1\}\right\vert \le n$ for all $n\in \mathbb{N}$. Then $G$ is cyclic.

And fortunately I got a perfect answer; but I am still stuck on one point which is mentioned in the forth paragraph of the answer saying that the hypothesis implies that $\psi(d)=\phi(d)$.

Why? Is that a property of Euler’s totient function? How to understand that? Any help will be sincerely appreciated!


No, it's nothing to do with Euler at this stage of the proof. It's a simple statement about inequalities. If you have some numbers $a \le A, b \le B , \ldots , z \le Z$ and you know $a + b + \cdots + z = A + B + \cdots + Z$ then all the inequalities must actually be equalities.

  • $\begingroup$ Thanks for your time. I’m sorry that I’m still confused. By the hypothesis, $\psi(d)\leq d$ but not $\psi(d)\leq\phi(d)$ and I think there is no $\sum_{d\mid m}\psi(d)=\sum_{d\mid m}\phi(d)$ until now. It may seem to me that your answer applies to the sixth paragraph perfectly but not the fourth one. Did I miss anything? $\endgroup$ – Benny Feb 25 '18 at 14:47
  • $\begingroup$ @Benny You are right that my argument concerns the sixth paragraph. I am as confused as you are about the problem in the fourth paragraph. The version of this argument I have seen is the second answer here: math.stackexchange.com/questions/837562/… - with different hypotheses than what puzzles you. I think you should try to get a better explanation on the answer you link to than at this new question of yours. $\endgroup$ – Ethan Bolker Feb 25 '18 at 15:18

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