Why isn't there interest in nontrivial, nondiscrete topologies on finite groups?

Recall that a topology is completely determined by the closure relation. Let $G$ be a finite group with some topology and let $G_0$ be the closure of the identity. Then $G_0$ is a normal subgroup. Moreover, the closure of any finite subset of $G$ is the union of the corresponding translates of $G_0$ (since this is a closed subset contained in the closure). Consequently, the topology on $G$ is completely determined by $G_0$. In other words, studying nontrivial topologies on a finite group is equivalent to studying normal subgroups.

In general, if $G$ is a topological group and $G_0$ the closure of the identity, then $G_0$ is a normal subgroup and $G/G_0$ is the universal Hausdorff topological group into which $G$ maps. (Recall that as Hagen mentions in the comments, for finite spaces Hausdorff is equivalent to discrete.) So the study of topological groups reduces almost immediately to the study of Hausdorff topological groups.

Evidently I was missing some things. All is well now, though.

In response to Qiaochu's answer I wanted to flesh out the details. Let $(G,\tau_G)$ be a finite topological group, $1:=\{1_G\}$ the trivial subgroup, $\mathrm{cl}(\cdot)$ the topological closure operator, and $\mathrm{ncr}(\cdot)$ the group-theoretic normal core operator. So $\mathrm{cl}(A)=\bigcap \{F~\textrm{closed}:A\subseteq F\}$ and $\mathrm{ncr}(A)=\bigcap_{g\in G}g^{-1}Ag$.

Lemma 1. Let $(X,\tau_X)$ be a topological space. The pairs $A,\,\mathrm{cl}_X(A)$ determine $\tau_X$ and vice-versa.

Proof. The vice-versa direction is clear by $\mathrm{cl}$'s intersection formula. Conversely, a subset $A\subseteq X$ is closed if and only if $\mathrm{cl}(A)=A$ (if $A$ is closed then $A\subseteq \bigcap F\subseteq A\implies \mathrm{cl}(A)=A$, and conversely if equality holds then $A$ is an intersection of closed sets hence is closed). Hence we can determine the open sets of $X$ precisely in terms of closure: $A\subseteq X$ is open iff $X\setminus A=\mathrm{cl}(X\setminus A)$.

Lemma 2. The topology $\tau_G$ on $G$ is determined by $S:=\mathrm{cl}(1)$.

Proof. Let $X\subseteq G$ be a subset. Any closed set $F$ containing $X$ contains any singleton subset of $X$, so also contains the closure $\mathrm{cl}(x)$ for all $x\in X$, hence will contain the finite union $Y:=\cup_{x\in X}\mathrm{cl}(x)$; since the union is finite, $Y$ is also closed. Note $X\subseteq Y$ since $x\in\mathrm{cl}(x)$ always. So we conclude that $\mathrm{cl}(X)=Y$ because $Y$ exhibits the universal property of the closure of $X$. Moreover,

$$\begin{array}{cl} \mathrm{cl}(x) & = \bigcap \{F~\textrm{closed}:x\in F\} \\ & = \bigcap x\{x^{-1}F~\textrm{closed}:1_G\in x^{-1}F\} \\ & =x\bigcap\{E~\textrm{closed}:1_G\in E\} \\ & = x\,\mathrm{cl}(1)=xS \end{array}$$ $$\therefore~~ \mathrm{cl}(X)=\bigcup_{x\in X}\mathrm{cl}(x)=\bigcup_{x\in X}xS=XS.$$

All closures are therefore uniquely determined by $S$, and hence so with $\tau_G$ by Lemma 1.

Lemma 3. $S=\mathrm{cl}(1)\trianglelefteq G$ is a normal subgroup.

Proof of normality. Since left and right translation are continuous and $S$ is closed containing $1_G$, any conjugate $g^{-1}Sg$ must be closed and contain $1_G$ hence $S\subseteq g^{-1}Sg=S^g$ for all $g$. Therefore $$S\subseteq \bigcap_{g\in G}S^g=\mathrm{ncr}(S) \subseteq S\implies S=\mathrm{ncr}(S).$$ Since the group-theoretic normal core is normal, $S$ is a normal subset of $G$.

Proof of subgroup. Note that $S$ is a closed set containing $x$ for any $x\in S$, hence $xS\subseteq\mathrm{cl}(x)\subseteq S$, and since $x^{-1}S\subseteq S\implies S\subseteq xS$ by left-multiplication, $xS=S$ for all $x\in S$, which gives closure under multiplication as well as inverses (since $1_G\in S=xS\implies 1_G=xy$ for some $y\in S$).

Theorem. The only nontrivial topologies on a finite group are lifted from discrete topologies on factor groups. That is, a topology on $G$ must have as base the coset space $G/N$ for a $N\trianglelefteq G$.

Proof. Let $\tau_G$ be a topology on $G$ and let $N=S=\mathrm{cl}_G(1)$. A subset $X\subseteq G$ is open iff $G\setminus X$ is closed iff $G\setminus X=\mathrm{cl}(G\setminus X)=(G\setminus X)S=\cup_{y\in G\setminus X}yN$ is a union of left cosets of $N$ (and indeed since $N$ is closed and translation is continuous, any union of left cosets of $N$ is closed). But since cosets partition $G$, if $G\setminus X$ is a union of cosets, so is $X$. Hence $G/N$ is a base for $\tau_G$.

Remark. Suppose $(G,\tau_G)$ is Hausdorff. Then for each nonidentity element $g\in G$, there are disjoint open sets $1_G\in U_g$ and $g\in V_g$. Then $1_G\in\cap_{g\ne 1_G}U_g$ is open and cannot contain a non-identity element hence $\{1_G\}$ is open, and subsequently by continuity any singleton and by union any subset is open, so $\tau_G=\mathcal{P}(G)$ is in fact the discrete topology.