G/H need not be a quotient Lie group

Every finite discrete group is a zero-dimensional Lie group: singleton charts take values in ${\mathbb{R}}^0$, and every map between discrete manifolds is smooth. Thus $G$ is a Lie group and its subgroup $H$ is closed. By [A1], the three-element left-coset space $G\mathrm{/}H$ has its quotient smooth-manifold structure.

The subgroup $H$ is not normal. Indeed, conjugating its nonidentity element by the $3$-cycle gives $(123)(12)(123{)}^{-1}=(23)\mathrm{\notin }H$.

Suppose a group law on this set $G\mathrm{/}H$ made the usual coset map $q(g)=gH$ a group homomorphism. Its kernel would be exactly $H$, because $q(g)=H$ if and only if $gH=H$, equivalently $g\in H$. Every homomorphism kernel is normal: if $h$ is in the kernel, then $q(gh{g}^{-1})=q(g)eq(g{)}^{-1}=e$. Hence $H$ would be normal, contradicting step 1.2.

Therefore the smooth homogeneous space $S_3\mathrm{/}H$ admits no group structure for which the coset map is a homomorphism. The quotient theorem [F1] is sharp: closedness supplies the manifold, whereas normality is necessary for the quotient group law. This finite witness has neither endpoint nor positive-dimensional issue, and its algebraic obstruction is choice-free; ${\mathrm{A}\mathrm{C}}_{\omega }$ is used only to invoke the library's general homogeneous-space supplier [A1].