Estimate the number of conjugacy classes in a group Let $\rho: G \to GL(V)$ be a finite dimensional complex irreducible representation of $G$. Let $H$ be an abelian subgroup of $G$. I want to show that the number of irreducible characters of $G$ is greater than or equal to $\frac{\lvert H \rvert^2}{\lvert G \rvert}$.
Here's what I did: Suppose $W \neq 0$ be a subspace of $V$ that is invariant under the $H$-action, i.e., $\rho(h)W \subseteq W$ for all $h \in H$. Since $H$ is abelian, we can assume $\text{dim}_{\mathbb{C}}(W) = 1$. Consider the subspace $W_0 = \sum_{g \in G}\rho(g) W$. Since $W_0$ is $G$-invariant and $\rho$ is irreducible, it follows that $W_0 = V$. In addition, we note $\rho(g_1)W = \rho(g_2)W$ if $g_1H = g_2H$. Hence, $W_0 = \sum_{i = 1}^{k}\rho(g_i)W$, where $g_1,\dots,g_k$ are coset representatives of $G/H$.
Therefore, $\text{dim}_{\mathbb{C}}(V) \leqslant \text{dim}_{\mathbb{C}}(W_0) \leqslant k = \frac{\lvert G \rvert}{\lvert H \rvert}$. To prove the statement, it suffices to show that the number of conjugacy classes in $G$ is greater than or equal to $\frac{\lvert H \rvert}{\text{dim}_{\mathbb{C}}(V)}$. But I don't know how to do the last part.
 A: Let me provide you with a character theoretic proof. All characters below are complex, and groups finite. We start with a small lemma.
Lemma Let $A$ be an abelian subgroup of the group $G$ and $\chi \in Irr(G)$. Then $\chi(1) \leq |G:A|$.
Proof We can write for the restriction of $\chi$ to $A$
$$\chi_A=\sum_{\lambda \in Irr(A)}a_{\lambda}\lambda$$
where the $a_{\lambda}$ are non-negative integers and all the $\lambda$'s are linear. Hence
$$[\chi_A,\chi_A]=\sum_{\lambda \in Irr(A)}a_{\lambda}^2 \geq \sum_{\lambda \in Irr(A)}a_{\lambda}=\chi(1).$$
On the other hand,
$$[\chi_A,\chi_A]=\frac{1}{|A|}\sum_{a \in A}|\chi(a)|^2 \leq |G:A|(\frac{1}{|G|}\sum_{g \in G}|\chi(g)|^2)=|G:A|[\chi,\chi]=|G:A|.$$
We conclude that $\chi(1) \leq |G:A|$ as required. $\square$
Let $k(G)$ denote the number of conjugacy classes of $G$, and $A$ an abelian subgroup of $G$. Note that $k(G)=\#Irr(G)$. Now by the Frobenius-Molien formula and the lemma, $$|G|=\sum_{\chi \in Irr(G)}\chi(1)^2 \leq k(G)|G:A|^2.$$
from which it follows that $$k(G) \geq \frac{|A|^2}{|G|}$$
as wanted.
