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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 $n$.

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

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    $\begingroup$ Alright, what is your question then? $\endgroup$ – Mark Sep 25 '18 at 20:15
  • $\begingroup$ @Mark While typing up this question I thought up a proof and I decided to post it $\endgroup$ – Perturbative Sep 25 '18 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 $\endgroup$ – Zubin Mukerjee Sep 25 '18 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 '18 at 20:22

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