A proper closed subgroup of a torus has a nonzero annihilating character

Since $T$ is compact and connected, $Q=T\mathrm{/}H$ is compact and connected, and [F1] makes $\pi :T\to Q$ a smooth surjective submersion. As $T$ is abelian, $H$ is normal: the operations $(aH)(bH)=(ab)H$ and $(aH{)}^{-1}=a^{-1}H$ are independent of representatives and give an abelian group with identity $H$. By [F5], each point of $Q$ has an open neighborhood $U$ and a smooth section $s_U:U\to T$ with $\pi s_U={\mathrm{id}}_U$. On $U\times V$, multiplication is $(x,y)\mapsto \pi (s_U(x)s_V(y))$; on $U$, inversion is $x\mapsto \pi (s_U(x{)}^{-1})$. These formulas prove both operations smooth, so $Q$ is a Lie group. It is not a point because $H⊊T$; a zero-dimensional manifold is discrete, and connectedness would then make it a point. Hence $d=\dim Q\ge 1$.

Let $\mathfrak{q}=T_{e}Q$. The exponential map ${\exp }_Q:\mathfrak{q}\to Q$ has differential the identity at zero by [F2], so [F3] makes it a diffeomorphism from a neighborhood of zero onto a neighborhood of the identity. Since $Q$ is abelian, ${\exp }_Q(X+Y)={\exp }_Q(X){\exp }_Q(Y)$: the product of the one-parameter subgroups with velocities $X,Y$ is a one-parameter subgroup with velocity $X+Y$, and uniqueness of integral curves gives the identity. Thus ${\exp }_Q(\mathfrak{q})$ is a subgroup containing an open identity neighborhood, hence is open. Its other cosets are open, so its complement is open; connectedness of $Q$ gives ${\exp }_Q(\mathfrak{q})=Q$. The kernel $\mathrm{\Lambda }=\ker ({\exp }_Q)$ is discrete because the exponential is injective near zero. Translating its local inverse shows that ${\exp }_Q$ is open, so it induces a homeomorphism $\mathfrak{q}\mathrm{/}\mathrm{\Lambda }\cong Q$.

The linear span of $\mathrm{\Lambda }$ is all of $\mathfrak{q}$. Otherwise the quotient linear map $\mathfrak{q}\to \mathfrak{q}\mathrm{/}{\mathrm{span}}_{\mathbb{R}}\mathrm{\Lambda }$ factors through the compact space $Q=\mathfrak{q}\mathrm{/}\mathrm{\Lambda }$ and is onto a nonzero real vector space; that is impossible by [F4]. A discrete additive subgroup of a finite-dimensional real vector space that spans it is a lattice. To see this, a discrete subgroup has a neighborhood of zero containing no other subgroup point, so distinct subgroup points are uniformly separated and every bounded subset of it is finite. Choose a shortest nonzero vector ${\lambda }_1$; the intersection with $\mathbb{R}{\lambda }_1$ is $\mathbb{Z}{\lambda }_1$, since any smaller positive step would contradict shortestness. If projections of subgroup points to ${\lambda }_1^{\perp }$ approach zero, subtract nearest integer multiples of ${\lambda }_1$; the resulting points lie in a bounded set, hence in a finite set, so zero is isolated in the projected subgroup. The projection remains discrete and spans ${\lambda }_1^{\perp }$. Induction on $d$ gives a lattice basis for the projection; lifting that finite basis and adjoining ${\lambda }_1$ gives a $\mathbb{Z}$-basis ${\lambda }_1,\dots ,{\lambda }_d$ of $\mathrm{\Lambda }$ that is also a real basis of $\mathfrak{q}$.

Define the nonzero real linear functional $\mathrm{\ell }:\mathfrak{q}\to \mathbb{R}$ by $\mathrm{\ell }({\lambda }_1)=1$ and $\mathrm{\ell }({\lambda }_j)=0$ for $2\le j\le d$. Then $\mathrm{\ell }(\mathrm{\Lambda })\subseteq \mathbb{Z}$. Define ${\chi }_Q({\exp }_{Q}X):=\exp (2\pi i\mathrm{\ell }(X))$. This is well defined because two logarithms differ by an element of $\mathrm{\Lambda }$. It is a homomorphism by additivity of both the exponential map on the abelian Lie algebra and $\mathrm{\ell }$; it is smooth because the exponential map is a local diffeomorphism near zero and its translates are local diffeomorphisms everywhere. It is nontrivial: ${\exp }_Q({\lambda }_1\mathrm{/}2)\mathrm{\ne }e$, while ${\chi }_Q({\exp }_Q({\lambda }_1\mathrm{/}2))=-1$.

Let $\pi :T\to Q$ be the quotient homomorphism and set $\chi ={\chi }_Q\circ \pi$. Surjectivity of $\pi$ makes $\chi$ nontrivial, while $\pi (H)=e$ makes $\chi {\mathrm{\mid }}_H=1$. This proves the assertion.