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The following question was in one of my assignment:
Question 1: $G$ a finite group that for each number $d$ the number of elements that $x^d=1_G$ is at most $d$. Prove that $G$ cyclic.

I solve this question by getting its higher element order. I was thinking on this question with the a little narrowing its assumption:
Question 2: $G$ a finite group that for each number $d$ the number of elements with order $d$ is at most $d$. Prove that $G$ cyclic.

I am not sure even it is correct for all $|G|=n$. I have tried lots of inequalities but I could not reach any thing. In the way that we use Sylows Theorem and the $|G|=n$, odd number, we can see $G$ is a cyclic group as following:
Assume $|G|=n=p_1^{\beta_1}p_2^{\beta_2}\dots p_k^{\beta_k}$ then we have subgroups with the $p_i^{\beta_i}$ order.
Lemma 1: A group with $p^{\beta}$ order with the assumption that for each $p^{\alpha}$ that $\alpha \le \beta$. the number of elements with this order is less that $p^{\alpha}$, then the group is cyclic.
Proof: Assume that there is not any element with the $p^{\beta}$ order. Thus each element has order $p^{\alpha}$ that $\alpha \lt \beta$, and we have at most $p^{\alpha}$ this elements. $$p^{\beta} \le \sum_{i=0}^{\beta-1}p^{i}=\frac{p^{\beta}-1}{p-1}<p^{\beta}$$ As a result, we should have at least one element with the $p^{\beta}$ order.$\blacksquare$

Now we should show that each of this sylow subgroups are normal, and with the use of above lemma we will get that the result. For being normal proof see that each sylow subgroup is a cyclic and there are $p^{\beta -1}(p-1)$ members that have $p^{\beta}$ order. If we have another sylow subgroup then we have at least $2.p^{\beta -1}(p-1)$ elements of $p^{\beta}$. With the assumption that all $|G|$ is odd, and the number of elemnts of order $p^{\beta}$ is less than $p^{\beta}$, we can conclude that it could not happen. As a result, there are only one p-sylow subgroup. and this means that each of them are Normal. With the use of Lemma 1, each of them are cyclic too. So, $G$ is the product of cyclic sylow subgroups, then it is cyclic too.

My Question:

  1. Is my proof Okay, or I am missing something.
  2. Is there any basic solution. Like using inequalities or counting numbers?
  3. Any other proof for first or the second question will be great.

Thanks.

Janbazif
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    How are you accounting for the possibility that $p=2$? I guess if that's the case and the $2$-Sylow subgroup is not normal, then $n_2(G) \geq 3$ and that gets you where you need to be. Other than that gap, the proof seems correct, though my personal preference would be to fill in some gaps in the exposition -- it took me some time to see why your inequalities were true. – Robert Shore Aug 17 '21 at 21:17
  • Yes thanks this will fill the gap. I was more searching for more basic solution like counting . Did you find one? – Janbazif Aug 17 '21 at 22:16
  • I'm pretty sure there's a counting proof. This is how you prove that the multiplicative group of a finite field is cyclic (because $x^n-1$ never has more than $n$ roots in a field). – Robert Shore Aug 17 '21 at 23:13
  • Could you give some hints how should I count for the prove of the question? – Janbazif Aug 17 '21 at 23:38
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    Look at the third comment in the accepted answer to this question: https://math.stackexchange.com/questions/837562/why-is-the-multiplicative-group-of-a-finite-field-cyclic – Robert Shore Aug 18 '21 at 08:44

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