0
$\begingroup$

The fundamental theorem of cyclic groups says that a for a cyclic group $G$ where $|G|=n$, for every divisor $d$ of $n$, there is a unique subgroup of $G$ with order $d$. My question is that whether these are the only subgroups of any cyclic groups? In other words, is it true that $G$ contains exactly $m$ subgroups where $m$ is the number of divisors of $|G|$?

$\endgroup$

marked as duplicate by Dietrich Burde abstract-algebra Oct 22 '18 at 20:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2
$\begingroup$

Yes. You know that for each divisor $d$ of $n$ there is a unique subgroup of order $d$. And Lagrangre's theorem tells us that if $k$ is not a divisor of $n$ then there can't be a subgroup of order $k$. Hence there can't be subgroups of other orders.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.