I am trying to digest the concept of a topological group. So far it makes sense to me that some structures (where the topology is already given), may have a continuous operation that also satisfies the axioms of a group.
This led me to wonder if the converse is true. Suppose we have a group, can we define a topology such that the group operation is also continuous? My intuition says such a construction (if possible) would be similar to the weak topology on a family of maps.
Yes, in a trivial way: you can always take the discrete topology.
More generally, let $G$ be any topological group, not necessarily Hausdorff, and let $G_e = \overline{ \{ e \} }$ denote the closure of the identity. Then you can show that $G_e$ is a closed normal subgroup, and the quotient $G/G_e$ is Hausdorff. We get a short exact sequence
$$1 \to G_e \to G \to G/G_e \to 1$$
where $G/G_e$ is the universal Hausdorff topological group to which $G$ maps. If $G$ is furthermore finite, then any Hausdorff topology on a finite set is discrete, so $G/G_e$ has the discrete topology, and since $G_e$ by definition has the property that $e$ is dense it's not hard to show that $G_e$ has the indiscrete topology. This means the topology on $G$ is generated by the cosets of $G_e$. So topologies on a finite group correspond precisely to normal subgroups $G_e$, and of these only the discrete topology (corresponding to $G_e = \{ e \}$) is Hausdorff.
In the infinite case these topologies associated to normal subgroups continue to exist but also various more interesting things can happen. Probably the two most common types of topological groups are Lie groups and profinite groups, both of which are either finite or uncountable. Nontrivial topologies on countable groups seem uncommon, to me at least.