Suppose that for any prime $p$ such that $p$ divides $|G|$, $G$ a finite abelian group, there are strictly fewer than $p$ elements of $G$ with order $p$.

Can we use this fact to show that $G$ is cyclic? I'm actually trying to show that this condition is equivalent to cyclic-ness, but I proved the converse indirectly. I suppose the first step would be to write $|G| = {p_1}^{e_1}{p_2}^{e_2}\cdots{p_r}^{e_r}$ for distinct primes $p_i$ and try using the Fundamental Theorem, but I got nowhere doing that.

Any help would be appreciated.


Let $|G| = {p_1}^{e_1}{p_2}^{e_2}\cdots{p_r}^{e_r}$ with $p_i \ne p_j$ for $i \ne j$.

For $p_i$, from the Fundamental Theorem, the component of $G$ consisting of $p_i$ is the product of groups with orders multiplying to $p_i^{e_i}$.

Each group in the product has $p-1$ elements of order $p_i$ (why?). If the product consists of more than one groups, then there would be at least $2p-2$ elements of order $p_i$ (why?), contradicting the assumption.

Therefore, we know that $G = \Bbb Z_{p_1^{e_1}} \times \Bbb Z_{p_2^{e_2}} \times \cdots \times \Bbb Z_{p_n^{e_n}}$, so it is cyclic.

  • $\begingroup$ What is 'the component of $G$ consisting of $p_i$'? Is this the subgroup of $G$ of order ${p_i}^{e_i}$? I don't really see why any cyclic group in the product you mention needs to have order exactly $p_i$ for any $i$. Or don't we require that? $\endgroup$ – SPS Sep 9 '17 at 15:13
  • $\begingroup$ @SPS For example, an abelian group of order $60$ could be isomorphic to $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_3 \times \Bbb Z_5$ or $\Bbb Z_4 \times \Bbb Z_3 \times \Bbb Z_5$. The part without $\Bbb Z_3$ and $\Bbb Z_5$ is what I'm referring to "the component of $G$ consisting of $2$. $\endgroup$ – Kenny Lau Sep 9 '17 at 15:15
  • $\begingroup$ OK, but can I ask how we know that in such a component, which has order ${p_i}^{e_i}$, there is a factor with order $p_i$? Because otherwise that factor needn't have any elements of order $p_i$. I understand that a group of order $p_i$ must be cyclic and therefore must have ${p_i}-1$ elements of order $p_i$, but we need such a group first, don't we? $\endgroup$ – SPS Sep 9 '17 at 15:27
  • $\begingroup$ I mean, whether $K=\Bbb Z_2\times\Bbb Z_2$ or $K=\Bbb Z_4$, there is $2-1$ element in every factor of $K$ which has order $2$. $\endgroup$ – Kenny Lau Sep 9 '17 at 15:43

From fundamental theorem of abelian groups any finite abelian group $G\cong \Bbb Z_{{p_1}^{q_1}}× \Bbb Z_{{p_2}^{q_2}}×.....× \Bbb Z_{{p_t}^{q_t}}$; using standard notation Now the given condition $\Rightarrow $ we can't allow repitation of primes in $p_1,p_2,...,p_t$ .So $ \Bbb Z_{{p_1}^{q_1}{p_2}^{q_2}....{p_t}^{q_t}}$


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