I cannot solve the following question:
Given $G$ a finite group, $A$ an abelian subgroup of $G$ and $\rho:G\to \mathrm{GL}(V)$ an irreducible linear representation of G, where $V$ is a vector space over an algebraically closed field $\mathbb{K}$, we want to show that $\dim(V)\leq \frac{|G|}{|A|}$ where $|\cdot|$ denotes the cardinal of the underlying set.
I already showed that $\dim(V)\leq |G|$ by considering the representation morphism between the regular representation $\rho_r$ and $\rho$: $$ u_v:f\in \mathbb{K}^G\mapsto \sum_{g\in G}f(g) \rho(g)(v)\in V $$ where $v\in V\backslash\{0\}$, and that the endomorphism $\{\rho(h):h\in A\}$ are simultaneously diagonalizable.
My idea would be to find a specific $v$ for the $u_v$ morphism above, such that $\mathrm{Span}(v)$ and $\rho(g)(\mathrm{Span}(v))$ for $g\in G$ remains stable under the action of $A$, but I could not find such a $v$.
Any hint or help welcome !
I just figured out. In fact it was almost done : let $v\in V$ be one of the vectors of a basis in which $\{\rho(h) : h\in A\}$ are diagonalizable. Let also partition $G=\bigsqcup\limits_{i=1}^{\frac{|G|}{|A|}}g_iA$ into left classes. Then $\forall 1\leq i\leq \frac{|G|}{|A|},\, \forall g\in g_iA,\exists a\in A,\, g=g_ia$. Thus $\rho(g)(v)=\rho(g_ia)(v)=\rho(g_i)(\rho(a)(v))=\rho(g_i)(\lambda_a v)=\lambda_a \rho(g_i)(v)$. This shows that $\forall 1\leq i\leq \frac{|G|}{|A|},\, u_v(\{f\in \mathbb{K}^G:\forall g\not\in g_iA,\, f(g)=0\})=\mathrm{Span}(\rho(g_i)(v))$. As every $g\in G$ belongs to exactly one $g_i A$, this means that $u_v(\mathbb{K}^G)=\mathrm{Span}((\rho(g_i)(v))_{1\leq i\leq \frac{|G|}{|A|}})$. As $u_v(\mathbb{K}^G)$ is $G$-stable and $v\neq 0$, we conclude that $V=\mathrm{Span}((\rho(g_i)(v))_{1\leq i\leq \frac{|G|}{|A|}})$ by irreducibility.