Proving there aren't any more subgroups for $S_3$ I'm currently doing an exercise to find all the subgroups of $S_3$, with a hint given that there are exactly $6$ and then to prove that no more subgroups exist. Take $$() \equiv e, (12) \equiv x, (13) \equiv y, (23) \equiv z, (123) \equiv \alpha, (132) \equiv \beta$$ Then for the subgroups we have:
As with any group we have the group itself $S_3$ and the trivial subgroup $\{e\}$. Now by inspection of the multiplication table we have $\{e, \alpha, \beta\}, \{e, x\}, \{e, y\}, \{e, z\}$ These are all subgroups because they fulfill the requirements for a subgroup. Namely that they are closed under products and inverses.

Now, I want to show that no more subgroups can exist. I was going to go about showing this by showing that each possible permutation (a set with any elements except the identity, a set with only transpositions but more than $1$ transposition, etc...) doesn't fulfill the requirements above. This is obviously quite tedious and I'm wondering if there's a more systematic way to show this.
For Context: The only facts we have been presented with thus far about the existence of subgroups have been for $\mathbb{Z}$ and $\mathbb{Z_n}$. We have not encountered the theorem that states that for a finite group, the order of any subgroup must divide the order of the group. Which we can see here since $\lvert S_3 \rvert = 6$ and the only positive integers which divide $6$ are $1$, $2$, $3$ and $6$, which are the orders of the subgroups listed. So we aren't "allowed" to use that reasoning.
 A: Notice that both $3$-cycles together with the identity give you the subgroup of order $3$. And each of the transpositions together with the identity produce their own subgroup of order $2$. In each of these cases, you can show that adding any other element generates the entire group $G = S_3$:

*

*If $H < G$ contains at least one of $\{\alpha, \beta\}$ and at least one of $\{x, y, z\}$, then $H = G$.

*If $H < G$ contains at least two of $\{x, y, z\}$, then $H = G$.


The key concept in all of this is to consider what a set of elements generates, meaning all elements that are guaranteed to be in the subgroup as a consequence of closure and inverses. For instance, if your subgroup contains the $3$-cycle $\alpha$, then it has to contain $\beta$ since $\beta = \alpha^2$ (also $\beta = \alpha^{-1}$). In symbols, the subgroup generated by an element, say $\alpha$, is
$$
\langle \alpha \rangle = \{ \alpha^n \mid n \in \mathbb{Z} \}, 
$$
and in this group, it happens that $\alpha^3 = e$, so $\alpha^m = \alpha^n$ if and only if $m \equiv n \pmod{3}$. Hence,
$$
\langle \alpha \rangle = \{ e, \alpha, \alpha^2 \}.
$$
The bulleted claims above can be written more succinctly in the language of generators, e.g. $\langle \alpha, x \rangle = G$, etc.
