Order of groups and dimensions of representations 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''
 A: 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.
