Q:

Number of homomorphisms/characters

Proposition. Let $G$ be a finite abelian group of order $n,$ $H$ its subgroup and $F$ a field such that polynomial $x^n-1$ has exactly $n$ roots in $F.$ Then for every homomorphism $\chi:H\to F^\times$ there are $[G:H]$ homorphisms $\chi':G\to F^\times$ such that $\left.\chi'\right|_H=\chi.$
Proposition. Let $G$ be an abelian group, $H$ its subgroup of finite index and $F$ algebraically closed field. Then for every homomorphism $\chi:H\to F^\times$ there are $[G:H]$ homorphisms $\chi':G\to F^\times$ such that $\left.\chi'\right|_H=\chi.$

Both propositions hold and can be prooved basically the same way. My question is, can I combine them into one more general, where

I don't need to assume finiteness of $G,$ only finiteness of the index $[G:H]$,
and I don't need to assume $F$ to be algebraically closed, but something weaker, similar to the first proposition?

Edit. Ok, I'm trying to prove the following.

Let $G$ an abelian group, $H$ its subgroup of finite index $k=[G:H]$ and $n$ positive integer such that $k\mid n.$ Moreover, let $F$ be a field such that polynomial $x^n-1$ has exactly $n$ roots in $F$. Then for every homomorphism $\chi:H\to F^\times$ there are $[G:H]$ homorphisms $\chi':G\to F^\times$ such that $\left.\chi'\right|_H=\chi.$

Let $1<k\le n,a\in G\setminus H$ and $m=\operatorname{ord}_{G/H}(aH)$, then $m\mid k\mid n.$ Apparently, $m=[\langle H,a\rangle:H]$ and $a^m\in H.$ Also, let $\chi:H\to F^\times,\tilde\chi:\langle H,a\rangle\to F^\times$ be homomorphisms, $\left.\tilde\chi\right|_H=\chi.$ Then $\bigl(\tilde\chi(a)\bigr)^m=\tilde\chi\left(a^m\right)=\chi\left(a^m\right).$ So, I would like to set $\tilde\chi(a)$ as a $m$-th root of $\chi\left(a^m\right).$ But do I know that these roots exists in $F$? I guess it would be helpful if $\operatorname{cod}(\chi)$ is a (sub)set of roots of unity, but I failed to show that. I tried to factorise $\chi'=\dot\chi\circ q,$ where $q:G\to G/H$ is the quotient map and $\dot\chi:G/H\to F^\times$ is the induced map and I would be done since $G/H$ is finite. But in this case I need to assume $H\subseteq\ker\chi'.$ So, what can I do now?
Edit2. The cancelled text is not true, consider $H=\mathbb Z\to\mathbb C^\times,\ell\mapsto\operatorname{e}^\ell.$

A:

The last statement is wrong.
Let $p>2$ be a prime, then $G=\mathbb Z/(p-1)^2\mathbb Z, H=\mathbb Z/(p-1)\mathbb Z,F=\mathbb F_p,F^\times=\langle g\rangle$ and $H\to F^\times,1\mapsto g$ lead to contradiction.

