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 $G/H$.


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 $G$.

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  • $\begingroup$ Does this theorem require choice? The proof I know uses Zorn’s lemma. $\endgroup$ – LSpice Dec 19 '19 at 15:03
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    $\begingroup$ @LSpice Yes, I think this fact uses the axiom of Choice. It is interesting if it is equivalent to (a weak form) of CH, maybe to the existence of a free ultrafilter? $\endgroup$ – Taras Banakh Dec 20 '19 at 8:17

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