Dimension of an irreducible representation and the index of the group's center

Question

Let $(\pi,V_\pi)$ be an irreducible representation of finite group $G$, over algebraically closed field $F$ s.t. $char(F)$ is coprime to $|G|$ (for example $\mathbb{C}$).

Prove that $dim(\pi)^2 \le [G:Z(G)]$ where $Z(G)$ is the group's center.

My thoughts so far: let us recall that there exists a character of G (i.e. 1-dim representation) of the center: $\omega _\pi : Z(G) \to F^*$ such that for every $z\in Z(G): \pi(z) = \omega_\pi (z) Id_{V_\pi}$. also we know $dim(Ind_{Z(G)}^G (\omega _\pi )) = [G:Z(G)]$, i'm not sure how to continue or even if my intuition so far is correct.

thanks ahead

Answer 1

First, note that $\pi|_{Z(G)}=\dim(\pi)\omega_\pi$ and $\pi$ is a summand of $\mathrm{Ind}_{Z(G)}^G\omega_\pi$. The inequality now follows since $\dim\mathrm{Ind}_{Z(G)}^G\omega_\pi=[G:Z(G)]$.

EDIT: To see that $\pi$ is a direct summand of $\mathrm{Ind}_{Z(G)}^G\omega_\pi$, use Frobenius reciprocity: $$\langle\mathrm{Ind}_{Z(G)}^G\omega_\pi,\pi\rangle_G=\langle\omega_\pi,\mathrm{Res}^G_{Z(G)}\pi\rangle_{Z(G)}=\langle \omega_\pi,\dim(\pi)\omega_\pi\rangle_{Z(G)}=\dim\pi.$$ In fact, this means that $\pi$ appears with multiplicity $\dim(\pi)$ in $\mathrm{Ind}_{Z(G)}^G\omega_\pi$.

Answer 2

I'll try to write a proof without using induced representations. Please let me know if it is correct.

Let $n=dim(\pi)$. The following lemma has an easy proof.

Lemma. For $g\in Z(G)$ it is $|\chi(g)|=n$ (where $\chi$ is the character of $\pi$).

Now, since $\pi$ is irreducible we have that $\langle \chi,\chi\rangle =1$. So

$$1=\langle \chi,\chi\rangle=\dfrac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\chi(g)}=\dfrac{1}{|G|}\sum_{g\in G-Z(G)}\chi(g)\overline{\chi(g)}+\dfrac{1}{|G|}\sum_{g\in Z(G)}\chi(g)\overline{\chi(g)}$$

Hence $$|G|=|Z(G)|n^2+\sum_{g\in G-Z(G)}\chi(g)\overline{\chi(g)}\ge |Z(G)|n^2\Rightarrow n^2\leq |G:Z(G)|$$