Prove that irreducible character of subgroup extends to $[G:H]$ irreducible characters of abelian group

Question

Here we're working with finite-dimensional complex representations.

Let $H \le G$ be finite abelian groups and let $\varphi$ be an irreducible character of $H$ i.e. associated to an irreducible representation $\theta: H \to GL(V)$. Show that there are exactly $[G:H]$ irreducible representations $\rho: G \to GL(W)$ with character $\psi$ such that $\psi\big\vert_H = \varphi.$

My attempt: use the induced representation to get irreps of $G$ that are different and whose restriction to $H$ matches $\theta$. Prove that in irreducible representation of $G$ satisfying the hypotheses will appear in the decomposition of the induced representation.

We know that $GL(V) \cong \mathbb{C}^\times$ since $H$ is abelian. We can naturally consider its induced representation $\widetilde{\theta}:G \to GL(V_0)$ and we know that $\dim V_0 = [G:H]\dim V = [G:H]$. Again, since $G$ is abelian, $\widetilde{\theta}$ breaks down in $[G:H]$ irreducible (possibly isomorphic?) representations of $G$. Since $\widetilde{\theta}\big\vert_H = \theta$, it really seems that each factor of $\widetilde{\theta}$ should be $\theta$ or the trivial representation when restricted to $H$, but I couldn't prove that. Also, I couldn't prove that they are non-isomorphic to prove that there are at least $[G:H]$ representations.

Futhermore, I have to prove that there are no more than $[G:H]$, and my attempt was to show that since the restriction of the irreducible character of $G$ to $H$ is $\theta$, the associated representations must be equal - this is true. But I couldn't prove that this implies the irreducible representation of $G$ will be a factor of the induced representation.

This is homework, so if possible please do not give the full answer. Any hints are appreciated!

Answer 1

Let $\varphi\in\text{Irr}(H),\psi\in\text{Irr}(G)$, then by Frobenius reciprocity:

$$[\psi,\varphi^G]_G=[\psi|_H,\varphi]_H=\begin{cases}1&\text{if }[\psi,\varphi^G]_G>0,\\ 0&\text{if } [\psi,\varphi^G]_G=0.\end{cases}$$ This is becasue $\psi(1)=1$ ($G$ is abelian which implies that every irreducible character of $G$ is linear, and the same holds for $H$.)

And we know that every irreducible component $\psi$ enters $\varphi^G$ exactly once and only when $\psi|_H=\varphi$.

Conversely, every $\psi\in\text{Irr}(G)$ satisfying $\psi|_H=\varphi$ is an irreducible component of $\varphi^G$.

Finally, since $\varphi^G(1)=|G|/|H|$, we conclude that there are exactly $|G: H|$ irreducible representations of $G$ with character $\psi$ such that $\psi|_H=\varphi$.