Finding all the subgroups of a cyclic group

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 $$n$$.

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

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.

• Alright, what is your question then? – Mark Sep 25 '18 at 20:15
• @Mark While typing up this question I thought up a proof and I decided to post it – Perturbative Sep 25 '18 at 20:17
• 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 – Zubin Mukerjee Sep 25 '18 at 20:20
• 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. – Mark Sep 25 '18 at 20:22