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.