I usually see this question as "if $G$ has at most $n$ solutions to $x^n=1$, then $G$ is cyclic". But here, we have that if $G$ has at most $n$ elements of orders divisible by $n$, then $G$ is cyclic. I am wondering if the following proof works. 3 Let $G$ be such a group with order $|G| = p_1 ^{\alpha_1} \cdots p_k ^{\alpha_k}$ where each $p_i$ is a distinct prime. Every non-identity element of any $P_i \in \text{Syl}_{p_i}(G)$ has order divisible by $p_i$ by Lagrange's theorem, but by hypothesis, there are at most $p_i$ such elements, so $|P_i| \le p_i$. Since $P_i$ is not trivial, this means $P_i \cong C_{p_i}$. Furthermore, $P_i$ is normal because $G$ is abelian. Because this holds for all $i$, we see that $\alpha_i = 1$ and $$ G \cong C_{p_1} \times \cdots \times C_{p_k} \cong C_{p_1 p_2 \cdots p_k} \cong C_{|G|}. $$

  • $\begingroup$ Yes you are right $\endgroup$
    – Nope
    Commented Feb 19, 2023 at 6:43
  • $\begingroup$ @DerekHolt I don't think that is true. In $C_4 = \langle x \rangle$ the generator has order $4$ which is divisible by $2$ yet $x^2 \neq 1$. $\endgroup$
    – Isabella
    Commented Feb 19, 2023 at 8:27
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    $\begingroup$ @AnneBauval I mentioned something like that question in my first sentence. The lemma in the first answer says that $G$ will be cyclic if the number of elements with order dividing $d$ is less than $d$. My question concerns the same statement but switching "dividing" for "divisible by". $\endgroup$
    – Isabella
    Commented Feb 19, 2023 at 8:33
  • $\begingroup$ Yes, I misread. Then, "Yes you are right". $\endgroup$ Commented Feb 19, 2023 at 8:56
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    $\begingroup$ You can even prove that $k=1,$ i.e. $|G|$ is prime. $\endgroup$ Commented Feb 19, 2023 at 9:17


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