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

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!

• Us the Frobenius Reciprocity Theorem to show that the irreducible components of the induced representation reduce to $\varphi$. Jun 28, 2021 at 7:58

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 $$\psi|_H=\varphi$$.