Let $H$ be an open subgroup of a locally compact Hausdorff abelian group $G$. Assume that $G/H$ is a finitely generated abelian group. Let $\chi: H \rightarrow \mathbb{C}^{\ast}$ be a continuous homomorphism. Does $\chi$ extend to a continuous homomorphism into $\mathbb{C}^{\ast}$ defined on all of $G$?


If $\chi$ maps $H$ into the circle $S^1$, then $\chi$ does extend to a continuous homomorphism on all of $G$, also mapping into $S^1$. This follows from Pontryagin duality, and in fact this is true when $H$ is a closed, not necessarily open subgroup, and with no assumption about $G/H$.



Every quasicharacter of an open subgroup extends to a quasicharacter on the whole group
The answer is affirmative and follows from the classical 


**Theorem (Baer).** Any homomorphism $h:H\to Y$ from a subgroup $H$ of an Abelian group $G$ to a divisible Abelian group $Y$ extends to a homomorphism $\bar h:G\to Y$.


Observe that the multiplicative group $\mathbb C^\*$ is Abelian and divisible. So, by Baer's Theorem, each continuous homomorphism $h:H\to \mathbb C^\*$ defined on an open subgroup $H$ of an Abelian topological group $G$ extends to a homomorphism $\bar h:G\to\mathbb C^\*$. The continuity of $\bar h$ follows from the continuity of its restrition $\bar h|H=h$ to the open subgroup $H$ on $G$. 


