Show that if $H$ is an abelian subgroup of order $p$ of finite group $G$ of order $n$, then every irreducible representation is of dimension $\leq n/p$.
I'm really confused of how an order of a group can define the dimension of a representation. Anyone care to explain or give me some hints? Thank you in advance!\
This is a problem from Yvette Kosmann-Schwarzbach's book called Groups and Symmetries from Chapter two ''Representations of Finite Groups''
You do not need the notion of induced representations to prove this statement. The biggest idea needed is the orthogonality of characters. Here is my proof:
Let $\Gamma$ be an irreducible $\mathbb{C}$-representation of $G$ into a finite dimensional vector space $V$. Let $\chi$ be the character of this representation. I claim that $\chi(1) \leq [\chi_H,\chi_H]$ where $\chi_H$ is the restriction of $\chi$ from $G$ to the abelian subgroup $H$ and $[\cdot,\cdot]$ is the inner product. We can prove this directly using the orthogonality relations of characters. Let $\chi_H = \sum_{i} n_i \psi_i$ where each $\psi_i$ is an irreducible $H$-character and each $n_i \in \mathbb{Z}^+$ (Note that restricting irreducible characters can yield reducible characters.). We have that for $x \in H$,
$$ \chi(x)\bar{\chi}(x) = \sum_{i} n_{i}^{2}\psi_i (x) \bar{\psi_i}(x) + \sum_{i \neq j} n_i n_j \psi_i (x) \bar{\psi_j}(x) $$ If we sum the rightmost expression over all $x \in H$, the sum on the right is zero by the Orthogonality Relations, so we have that
$$ \frac{1}{|H|}\sum_{x \in H} |\chi(x)|^2 = \frac{1}{|H|}\sum_{x \in H} \sum_{i} n_{i}^{2}\psi_i (x) \bar{\psi_i}(x) \geq \frac{1}{|H|}\sum_{x \in H} \sum_{i} n_{i}\psi_i (x) \bar{\psi_i}(x) = \sum_{i} n_i = \chi(1) $$
This implies the desired inequality that $\chi(1) \leq [\chi_H,\chi_H]$. We may then use this inequality to obtain that
$$ |H|\chi(1) \leq \sum_{x \in H} |\chi(x)|^2 \leq \sum_{x \in G} |\chi(x)|^2 = |G|[\chi,\chi] = |G| $$ Where $[\chi,\chi] = 1$ since we assume $\chi$ is obtained from an irreducible representation. Now, $\chi(1) \leq |G:H| = \frac{n}{p}$ like you wanted to show.
This proof was outlined as Problem 2.9 in the Isaacs' book on character theory, and that's where this came from. If I used any confusing notation, let me know and I will clarify.