Theorem (1): If $G$ is a finite cyclic group of order $n$ and $m \in \mathbb{N}$, then $G$ has a subgroup of order $m$ if and only if $m | n$. Moreover for each divisor $m$ of $n$, there is exactly one subgroup of order $m$ in $G$.

The above theorem states that for any finite cyclic group $G$ of order $n$, the subgroups of $G$ (and also their orders) are in a one-to-one correspondence with the set of divisors of $n$. So $G$ has exactly $d(n)$ subgroups where $d(n)$ denotes the number of positive divisors of $n$.

Now the following example was given in my class notes and I paraphrase them below.

Example: Consider $G = (\mathbb{Z}_{12}, +)$. We list all subgroups of $G$. Note that $G$ is cyclic because $G = \langle \bar{1} \rangle$ and $|G| = 12$. Note also that $12$ has $6$ positive divisors, those being $\{1, 2, 3, 4, 6, 12\}$, hence $G$ has exactly six subgroups those being $$\langle \bar{1} \rangle; \ \ \langle (\bar{1})^2 \rangle; \ \ \langle (\bar{1})^3 \rangle;\ \ \langle (\bar{1})^4 \rangle;\ \ \langle (\bar{1})^6 \rangle;\ \ \langle (\bar{1})^{12} \rangle\ \ $$

Now my question is that how did the authors of these class notes know that the subgroups would be of the form $\langle (\bar{1})^{d} \rangle$ where $d$ is a positive divisor of $12$? It leads me to make the following conjecture

Conjecture: Given a finite cyclic group $G = \langle x \rangle$ of order $n$, all subgroups of $G$ are of the form $\langle x^d \rangle$ where $d$ is a positive divisor of $n$.

  • $\begingroup$ Why the downvote? StackExchange allows you to answer your own questions while posting them Q&A style $\endgroup$ Sep 25, 2018 at 20:18
  • 1
    $\begingroup$ You are correct. StackExchange also allows you to downvote! (I did not downvote) $\endgroup$ Sep 25, 2018 at 20:19

1 Answer 1


The conjecture above is true. To prove it we need the following result:

Lemma: Let $G$ be a group and $x \in G$. If $o(x) = n$ and $\operatorname{gcd}(m, n) = d$, then $o(x^m) = \frac{n}{d}$

Here now is a proof of the conjecture.

Proof: Let $G = \langle x \rangle$ be a finite cyclic group of order $n$, then we have $o(x) = n$.

Choose a subgroup $H \leq G$, by theorem $(1)$ mentioned in the question above, $|H| = m$ where $m$ is some divisor of $n$. Since $m | n$ (and both $m$ and $n$ are positive integers), there exists a $d \in \mathbb{N}$ such that $md = n \iff \frac{n}{d}=m$. Note also that $d$ is a divisor of $n$.

By the above lemma and the fact that $\operatorname{gcd}(d, n) = d$ (since $d$ is a divisor of $n$) it follows that $o(x^d) = \frac{n}{d} = m$. Hence the subgroup $\langle x^d \rangle$ has order $m$. But since by theorem $(1)$ there is only one subgroup of order $m$ in $G$ we must have $H = \langle x^d \rangle$. Thus any subgroup of $G$ is of the form $\langle x^d \rangle$ where $d$ is a positive divisor of $n$. $\ \square$

The above conjecture and its subsequent proof allows us to find all the subgroups of a cyclic group once we know the generator of the cyclic group and the order of the cyclic group.

  • 1
    $\begingroup$ Alright, what is your question then? $\endgroup$
    – Mark
    Sep 25, 2018 at 20:15
  • $\begingroup$ @Mark While typing up this question I thought up a proof and I decided to post it $\endgroup$ Sep 25, 2018 at 20:17
  • $\begingroup$ For any $n\in \mathbb{N}$, there is, up to isomorphism, only one cyclic group of order $n$. So just use $\left(\mathbb{Z}_n,+\right)$ and the statement becomes obvious. Don't really need a formal proof imo $\endgroup$ Sep 25, 2018 at 20:20
  • $\begingroup$ Ok then. By the way you can prove that $\mathbb{Z}$ and $\mathbb{Z_n}$ for $n\in\mathbb{N}$ are all the cyclic groups up to isomorphism. Then it would be easier to find all the subgroups in my opinion. $\endgroup$
    – Mark
    Sep 25, 2018 at 20:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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