I'm trying to do my homework (the overall problem is about $A$ a finite abelian group and showing $A \cong A^\vee \cong A^{\vee\vee}$, but you can ignore that), and I think I have a fundamental lack of understanding of how to extend a character $A \to \mathbb{C}^\times$.
For example, more precisely, let's say $A$ and $B$ are finite abelian groups, and $A \hookrightarrow B$ (or even $A \leq B$, whatever is clearer). If we have a character $\chi_A: A \to \mathbb{C}^\times$, how do we find a character $\chi_B: B \to \mathbb{C}^\times$ such that $\chi_B |_A = \chi_A$?
Any examples or recommendations for resources I can look up will be appreciated!
Here are my thoughts so far: Let's say you have $\{1,\sigma\} = \mathbb{Z}/ 2 \mathbb{Z} \leq V_4$, the Klein 4-group, and I have a character $\sigma \mapsto -1$; any extension will just take the generator of the other copy of $\mathbb{Z}/2 \mathbb{Z}$ in $V_4$ and map it to either -1 or 1. But that only works if $A$ is the product of a subset of the direct summands of $B$ (which are all cyclic). If $A = \mathbb{Z} / 2 \mathbb{Z}$ and $B = \mathbb{Z} / 4 \mathbb{Z}$, we wouldn't be able to do this; we would need to cleverly pick out the image of a generator of order 4 such that the image of the element of order 2 matches up with $\chi_A$. When $A$ and $B$ have more summands, would I be able to still use this process, for each cyclic piece of $A$ that is a subgroup of a cyclic piece of $B$?