How to show that $\sum_{\phi \in \hat{G}} d_{\phi}^2 \leq \vert G \vert$ for a finite group $G$? I am trying to prove that $\vert \hat{G}\vert \leq \sum_{\phi \in \hat{G}} d_{\phi}^2 \leq \vert G  \vert$ for a finite group $G$ where $\hat{G}=\{\text{inequivalent irreducible unitary representation } \phi \text{ of } G\}$. I am able to prove that $\vert \hat{G}\vert \leq \vert G \vert$. By considering the map $\alpha : \hat{G} \longrightarrow L^{2}(G)$ given by $\alpha(\phi)=\text{Trace}(\phi)$ which is well defined as trace is well defined (since similar matrices have same trace) and it is also injective by the orhogonality of the character. Let $A_{\phi}=\{\phi_{ij} : 1 \leq i,j\leq d_{\phi}\}$, and $\vert A_{\phi}\vert = d_{\phi}^2$. Now consider the map $\beta: \hat{G} \longrightarrow \cup_{\phi \in \hat{G}} A_{\phi} $ defined by $\beta(\phi)=\phi_{11}$, where $\cup_{\phi \in \hat{G}} A_{\phi}$ has finitely many elements. $\beta$ is injective by schur's orhogonality relations, but I am not able to show why $\beta$ is well defined? Is there any other map which will work? It might be a small thing but I am unable to get it. Any help in this regard would be appreciated!
 A: The proof I've seen (in "Erdos-Ko-Rado Theorems: Algebraic Approaches" by Godsil and Meagher) is to look at the left-regular representation $\Lambda:G\rightarrow GL(L(G))$, where $L(G)$ is the set of all complex functions on $G$. The action is, for $\phi\in L(G)$ and $g,x\in G$, that 
$$(\Lambda(g)\phi)(x)=\phi(g^{-1}x)$$
In other words, $\Lambda(\phi)=\phi'$, where $\phi'$ is a new function such that $\phi'(x)=\phi(g^{-1}x)$ (you can check that it's a representation).
Then we take functions $\delta_g\in L(G)$, defined as $\delta_g(g)=1$, $\delta_g(x)=0$ if $x\neq g$. These form a basis for $L(G)$, and we note that $(\Lambda(a)\delta_g)(x)=\delta_g(a^{-1}x)=\delta_{ag}$. Then if we define $\lambda(g)=Tr(\Lambda(g))$:
$$\lambda(a)=\sum_{g\in G}\langle \delta_g,\Lambda(a)\delta_g\rangle=\sum_{g\in G}\langle \delta_g,\delta_{ag}\rangle$$
And the terms in the sum are only non-zero if $ag=g$, so it's only non-zero if $a=e$. Thus we have that $\lambda(e)=\vert G\vert$, and $\lambda(x)=0$ for $x\neq e$. Using this, we can use the fact that the irreducible characters are an orthonormal basis of all characters and say that:
$$\lambda=\sum_{\phi\in\hat{G}}m_\phi \phi$$
And then to find the coefficients, use the inner product:
$$m_\phi=\langle \lambda,\phi\rangle=\frac{1}{\vert G\vert}\sum_{g\in G}\phi(g^{-1})\lambda(g)=\frac{\phi(e)}{\vert G\vert}$$
(where all but one of the terms in the sum disappeared since $\lambda(g)=0$ if $g\neq e$). This means that $\lambda=\frac{1}{\vert G\vert}\sum_{\phi\in\hat{G}}\phi(e)\phi$. Then since $\lambda(e)=1$:
$$1=\lambda(e)=\frac{1}{\vert G\vert}\sum_{\phi\in\hat{G}}\phi(e)\phi(e)$$
And $\phi(e)=d_\phi$, so that gives that $\sum_{\phi\in\hat{G}}d_\phi^2=\vert G\vert$.
Running with the method you were trying, I don't think $\beta$ is well-defined. For any irreducible representation, there is an equivalent one where I just swap any two vectors. By the orthogonality, we know that for any representation $\phi$, if $i\neq j$, then $\phi_{ii}\neq\phi_{jj}$. Then I can construct an equivalent representation $\phi'=M^{-1}\phi M$, where $M$ swaps $v_1$ and $v_j$, then $\beta(\phi)\neq\beta(\phi')$. But what stops you from simply choosing a representative of each equivalence class of irreducible representations and constructing a set $\hat{G}'$ of these? Then you can define $\beta':\hat{G}'\rightarrow \cup A_\phi$ in the same way and avoid problems of equivalence.
