# A question about cyclic group and Euler’s totient function

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.
• 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?