Number of characters II

Question

This is similar/related to this topic. I want to prove the following.

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 $k=[G:H]$ homorphisms $\chi':G\to F^\times$ such that $\left.\chi'\right|_H=\chi.$

I think the proof is not hard (I guess), but there is one minor obstacle.

So once again (partial copy-paste from the previous topic).

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 $\bigl(\tilde\chi(a)\bigr)^m=\tilde\chi\left(a^m\right)=\chi\left(a^m\right).$ Denote $\zeta,\zeta^2,\ldots,\zeta^n=1$ all $n$-th roots of unity in $F.$ Then, $\chi\left(a^m\right)$ is a $(n/k)$-th root of unity, that is $\chi\left(a^m\right)=\zeta^{in/k}$ for some $i\in\{1,\ldots,k\}.$ Subsequently, $\tilde\chi(a)$ is an $m$-th root of $\chi\left(a^m\right)$ meaning $\tilde\chi(a)=\zeta^{ijn/(mk)}$.

And here's the problem. In general, $1<k\le n$ and $m\mid k\mid n$ is not strong enough to imply the relation $mk\mid n$ which I obviously need...

Answer 1

I don't think one needs to make these calculations. One needs this

Fact Let $A$ be a finite abelian group and $F$ a field where $x^{|A|}-1$ splits into $|A|$ distinct linear factors. Then $A$ has $|A|$ $F$-characters, which form a group $\hat{A}$ under pointwise multiplication.

In the notation of the Proposition let $\hat{G}$ be the character group of $G$ and let $$ \text{Fix}(H):=\{ \phi\in\hat{G}\ :\ \phi|_H=\text{id}\}. $$ Clearly $\text{Fix}(H)$ is a subgroup of $\hat{G}$, and each of its elements is effectively a homomorphism $G/H\to F^{*}$, and every such homomorphism gives an element of $\text{Fix}(H)$. That is, $\text{Fix}(H)$ is isomorphic to the character group of $G/H$. We therefore have that $\text{Fix}(H)$ has order $|G:H|$ and index $|H|$.

Now consider the cosets of $\text{Fix}(H)$; it is clear that two elements of $\hat{G}$ are in the same coset of $\text{Fix}(H)$ if and only if they restrict to the same character of $H$. As there are $|H|$ of these cosets, every one of the $|H|$ characters of $H$ must arise; as the number of elements in a coset is $|G:H|$, each character of $H$ has $|G:H|$ extensions to $G$.

Answer 2

Let $G = G_1 \times G_2$ then a character $\chi$ is defined as the product of two characters $\chi_1$ and $\chi_2$ each defined on the subgroups $G_1$ and $G_2$. It is not hard to prove that the proposition is proved if it is proved for each of the factors $G_i$. So it is reasonale to prove the proposition where the order of $G$ is of order of $p^q$ where $p$ is a prime. The problem you mention at the end of your question is now solved because all of the numbers involved $m, k, n$ are all powers of the prime $p$.