# If an abelian group contains at most $n$ elements of order divisible by $n$, is it cyclic?

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 $$G \cong C_{p_1} \times \cdots \times C_{p_k} \cong C_{p_1 p_2 \cdots p_k} \cong C_{|G|}.$$

• Yes you are right
– Nope
Commented Feb 19, 2023 at 6:43
• @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 $x^2 \neq 1$. Commented Feb 19, 2023 at 8:27
• @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". Commented Feb 19, 2023 at 8:33
• Yes, I misread. Then, "Yes you are right". Commented Feb 19, 2023 at 8:56
• You can even prove that $k=1,$ i.e. $|G|$ is prime. Commented Feb 19, 2023 at 9:17