Reviewing a proof to show that a subgroup of a cyclic group is also cyclic

Question

I want to prove the following statement:

Let $H$ be a subgroup of a cyclic group $G$, here, finite. Then $H$ is also cyclic.

My question is about an attempt that I include below; in particular, about the final paragraph, specifically the highlighted sentence. I know there're other questions on this topic, but I've tried to prove it independently avoiding looking at them. I'm not interested in other proofs (unless they're directly relevant with the approach I've attempted).

---

Proof attempt:

$G$ is a cyclic group $\implies$ $G$ is made up of powers of an element $a$ (the generator):

$$G=\{a^0, a^1, a^2,...,a^{p-1}\}$$

$H$ will also only contain powers of $a$:

$$H=\{a^0, a^{i_1}, a^{i_2},...,a^{i_{m-1}}\}$$

noting that $a^0=e$ must be one of them.

Using Langrange's theorem, we know that the order of $H$ must divide the order of $G$. And then, the set:

$$\{a^{p/m}, \left(a^{p/m}\right)^2,\left(a^{p/m}\right)^3,...,\underbrace{\left(a^{p/m}\right)^m}_{=a^p=e}\}$$

is made of $m$ different elements, because $a^i\neq a^j \ \forall \ i,j \leq p \ \textrm{if} \ i\neq j$ (as the element $a$ generates $G$). That set, is also closed under multiplication, it's associative, has inverses, and an identity element. This is, that set is a group.

Such group will be a cyclic subgroup of $G$, and there'll be (at least) one such subgroup for every divisor of $p$. There won't be subgroups of any other order (Lagrange's theorem). But also, there can be only one group of each order (up to isomorphism). Therefore, these cyclic subgroups exhaust the subgroups of $G$, and this proves that all subgroups of $G$ are cyclic.

Answer 1

I am just a student so take my answer with a grain of salt, but below there is a different prove to your question. I hope this will help you or other student.

The group is $(G;*)$.

1- We writte $g'$ the element of $G$ s.t. $G = \langle g'\rangle $

2- $\forall h \in H \Rightarrow h \in G$ because $ H \subseteq G $. As $G$ is a cyclic group by assumption $\forall h \in H, \exists k \in \mathbb{Z} $ s.t $ g'^k=h $.
Now we choose $h'=g'^{k} \in H$ the element of $H$ that can be expressed as the smallest integer power of $g'$.

3- Any power $\mathbb{Z}$ of $h' = g'^{k}$ is in $H$ as by assumption $H$ is a group. Thus $\langle h'\rangle \subseteq H $.

4- It let us to prove that $H \subseteq \langle h'\rangle$.

• $\forall h \in H \Rightarrow h \in G \Rightarrow \exists l \in \mathbb{Z} $ s.t. $h=g^l$. And we can write $ l = kq + r$ and $ q,r \in \mathbb{Z}, 0 \leq r < k $ and $k$ as define in "3-".
• So $g^l = g^{kq}*g^r = (g^k)^q * g^r \Rightarrow (g^k)^{-q} * g^l = g^r $. Remember that $ g^k \in H \Rightarrow (g^k)^{-q} \in H $ too as $(g^k)^{-1} \in H $ by property of $H$ a sub group and because $(g^k)^{-q} = (g^k)^{-1} * ... * (g^k)^{-1} $ "q" times we can conclude $(g^k)^{-q}, g^l \in H$.
• But if $(g^k)^{-q}, g^l \in H \Rightarrow g^{-qk+l}=g^r \in H$ too.
• Beware that $0 \leq r < k$ and that here on one side $-qk+l \leq 0$ and on the other side $-qk+l = r \geq 0$. Hence the only solution is $r=0$ .
  In conclusion $g^l$ can be generated by $h'=g^k$. And this prove that $H \subseteq \langle h'\rangle$

Q.E.D.