Having difficulty understanding topological groups. Let $G$ be a group and $x,y\in G$. We say that a topology $\mathcal{T}$ is a group topology if the functions $$f: G\times G \rightarrow G,\quad (x,y)\mapsto xy$$ and $$g: G\to G,\quad x\mapsto x^{-1}$$ are continuous. We call the pair $(G,\mathcal{T})$ a topological group.
I am trying to understand the above definition by playing around with some basic examples. To create an example for myself, I tried to take the symmetric group $S_{3}$ and tried to turn it into as many topological groups as possible.
Right away, I've dismissed the discrete and trivial topology as intuitively obvious as topological groups and I feel like there are no other group topologies on $S_{3}$. Is this true?
I tried to constructively create a third group by setting $(123) \in S_{3}$ as an open set for my soon to be topology. Using the definition of a topology we know that
$$\{\emptyset, (123), S_{3}\}\subseteq \mathcal{T}.$$
Now, if I require $f$ to be continuous, I am already extremely lost to the extent that I don't know what to say. We have that $f$ is a function from $G\times G \rightarrow G$ as function on group elements; but the definition of continuity requires $f: X\rightarrow Y$ where $X$ and $Y$ are topological spaces. My inclination would be to ignore this technical detail by assuming that $G\times G$ and $G$ can act as topological spaces, which leads me back into a circle as to what the open sets of $G$ have to be. Do the open sets of $G$ as a topological space also have to be a topological group?  How would I begin to pick a topology on $G\times G$? The box topology is the most natural, but why not a more exotic one?
This seems to have opened up a much larger barrel of worms for me when all I did was conjecture to myself that the only two group topologies on $S_{3}$ are the discrete topology and the trivial one.
 A: One of the consequences of the definition of a topological group is that for any $g\in G$ fixed, the map
$$
h \mapsto gh
$$
is a homeomorphism. Thus, if any singleton set $\{h\}$ is open, then so is
$$
\{gh\} \quad\forall g\in G
$$
and so all singletons are open, ie. the topology is discrete.
IMHO, it is best to play around with examples such as $GL_n(\mathbb{R})$. For instance, one fact which I have always thought is very cool is that the connected component of the identity is a normal subgroup - determining this subgroup for linear groups is quite fun.
A: 
How would I begin to pick a topology on $G\times G$? The box topology is the most natural, but why not a more exotic one?

The product topology. In case of finitely many factors, that is the same as the box topology, but calling it the product topology has the advantage of indicating that that is the natural topology to endow a product with. And endowing a Cartesian product of topological spaces with the product topology makes it a product in the category of topological spaces (where the morphisms are continuous maps).
Whenever you encounter a product of topological spaces, unless it is explicitly mentioned which topology is used, it is tacitly understood that the product is endowed with the product topology.
Now, group topologies are somewhat special. In a topological group $G$, from the continuity of multiplication, we obtain that for every $g\in G$ the left translation $\gamma_g \colon x \mapsto gx$ and the right translation $\delta_g \colon x \mapsto xg$ are continuous. Since the translations are bijective and their inverses are also translations ($(\gamma_g)^{-1} = \gamma_{g^{-1}},\: (\delta_g)^{-1} = \delta_{g^{-1}}$), they are indeed homeomorphisms. Thus if you want to construct a group topology - on $S_3$ here, if you decide that a particular set shall be open, all of its (left and right) translates must be open too. In particular, if you have one open singleton, it follows that all singletons are open, and you have the discrete topology.
But, to construct topologies on finite groups, looking at closed sets rather than open sets has some advantages:
In a topological group $G$, if $H$ is a subgroup of $G$, then the continuity of multiplication and inversion shows that the closure $\overline{H}$ is again a subgroup. And if $H$ is a normal subgroup, then $\overline{H}$ is normal. Now, in any group, $\{e\}$ is a normal subgroup. And therefore, in every topological group, $\overline{\{e\}}$ is a normal subgroup.
Turning things around, in a group $G$, every normal subgroup $N$ is a candidate for $\overline{\{e\}}$ (and there always are group topologies on $G$ for which $N = \overline{\{e\}}$). In a finite group $G$, for every normal subgroup $N$, there is precisely one group topology on $G$ such that $N = \overline{\{e\}}$. And, in a finite group, every closed subgroup is also open. Since $S_3$ has three normal subgroups, there are precisely three group topologies on $S_3$.
