As background, I am trying to do exercise 3.10 in Deitmar's "Principles of Harmonic Analysis." I can do most of the problem but I'm stuck on the third part proving surjectivity.
Given a locally compact abelian group $G,$ a closed subgroup $H,$ and a character $\chi: H \rightarrow S^1,$ I need to construct an extension of $\chi$ to all of $G.$ Any extension will do, but it's not clear to me that one should be able to construct an extension. For example, we can't just define an extension to be 1 outside of $H,$ since elements outside of $H$ can multiply to something within $H.$
Can anyone help me out? Thanks!
We can use the following results, from Rudin's book Fourier Analysis on Groups.
Denote $$A(H):=\{\gamma\in\widehat G,\forall h\in H,\gamma(h)=0\}$$ the annihilator of $H$. It's a closed subgroup of $\widehat G$, which ensures us that $\widehat G/A(H)$ is Hausdorff and locally compact.
Sketch of proof: Let $\pi$ the projection $\pi\colon G\to G/H$. We define an one-to-one correspondence between $A(H)$ and $\widehat{\left(\frac GH\right)}$ by $$\gamma(x):=\phi(\pi(x)).$$ Furthermore, it's compatible with the group structure.
By Pontryagin duality theorem, and using the fact that for $x_0\notin H$ we can find $\gamma\in \widehat G$ such that $\gamma(x_0)\neq 1$, we deduce the second assertion. To show it's an homeomorphism, we can use the fact that projection is continuous and open.