A step in a proof about Extensions of characters (multiplicative group homomorphisms)

I do not understand this proof by Marcus:

$Lemma$: Let $G$ be a finite abelian group, $H$ a subgroup. Then every character of $H$ extends to |G/H| characters of $G$.

Just count characters. Every character of $H$ extends to at most $|G/H|$ characters of $G$. To see this, let $\chi_1,...,\chi_r$ be any $r$ such extensions; then the $\chi_1^{-1}\chi_i$ give $r$ distinct characters of $G/H$. On the other han, every one of the $|G|$ restricts to one of the $|H|$ characters of $H$. The result follows.

So the final part is just a brief combinatory problem and I think I get the point. The problem is with the other part. I don't understand what does "$\chi_1^{-1}\chi_i$ give $r$ distinct characters of $G/H$" mean, because I cannot see these characters defined on $G/H$ and the role of using $\chi_1^{-1}$. Any help?

Hint: when does a homomorphism $\chi: G \rightarrow F^{\ast}$ induce a homomorphism $\overline{\chi}:G/H \rightarrow F^{\ast}$ such that $\overline{\chi}(gH) = \chi(g)$ for all $g \in G$?
• But I can't see your point: Consider $\chi_1^{-1}\chi_i:G\to S^1$. I want that the kernel of this is $H$. But the kernel$=\{g\in G$ such that $\chi_1(g)=\chi_i(g)\}$ – Richard Jan 21 '17 at 14:30
• But this kernel contains $H$ which is normal in $G$ since $G$ is abelian and therefore we have an induced homomorphism of $G/H$. This works, doesn't it? Thank you ! – Richard Jan 21 '17 at 16:24