This might be a bit silly question, but I'm having trouble understanding what the following theorem says. More precisely given a group and a subset of that group, I'm not sure how to use it. I suspect that I fully don't understand the notation.
Theorem: Let $\left(G,\cdot\right)$ be a group, $S\subseteq G$ a nonempty subset of $G$ such that $\forall a \in S: a^{-1} \in S$ then: $$\langle S \rangle = \{a_1\cdot a_2\cdot ...\cdot a_n;n\in\mathbb N, a_1,a_2,...a_n\in S\}.$$ Now let's say I'm working with $\left(\mathbb S_3,\circ \right), S=\{\left(1,2\right),\left(1,2,3\right),\left(1,3,2\right)\} \subseteq \mathbb S_3$ (I'm aware of the fact that it generates the whole group $\left(\mathbb S_3,\circ\right)$). How would I use the theorem? That is, what would $\langle S\rangle$ look like? I'm not sure what to compose with what and how many times. Any help would be appreciated.
The set $$ \{a_1a_2\dots a_n:n\in\mathbb{N},a_1,\dots,a_n\in S\} $$ surely contains $S$ and is a subset of every subgroup that contains $S$. So you just have to prove it is a subgroup. It is not empty and is closed under $xy^{-1}$, because $$ (a_1a_2\dots a_n)(b_1b_2\dots b_m)^{-1}= a_1a_2\dots a_nb_m^{-1}\dots b_2^{-1}b_1^{-1} $$ and all those terms belong to $S$. Therefore the set is indeed a subgroup.
Now define $S_0=S$, $S_1$ the set of products of two elements of $S_0$, and so on: $S_{k+1}$ is the set of products of two elements of $S_k$.
Since $1\in S_1$, we have that $S_k\subset S_{k+1}$, for $k\ge1$.
Then $\langle S\rangle$ is the union of all these sets. Note that, if the group $G$ is finite, at some step we will have $S_{k}=S_{k+1}$ and so $\langle S\rangle=S_k$.
What's $S_1$ in your case?