I have to prove the following statement:
Let $\rho: G \to GL(V), \tau: G \to GL(W)$ be representations of a
finite group $G$. Then $\rho$ and $\tau$ are equivalent if and only if
$\chi_\rho = \chi_\tau$
I can use the following theorems, and nothing else:
If $(.|.): \mathbb{C}^G\times \mathbb{C}^G \to \mathbb{C}^G:
(\phi|\psi) \mapsto \frac{1}{|G|}\sum_{g \in G}
\phi(g)\overline{\psi(g)}$, then we have:If $\chi$ is a complex, irreducible character of a finite group $G$,
then $(\chi|\chi) = 1$If $\chi_\rho, \chi_\tau$ are irreducible, complex characters and
$\rho$ not equivalent to $\tau$, then $(\chi_\rho|\chi_\tau) = 0$
If $\rho = \bigoplus_{i=1}^n \rho_i $, and $\chi_\rho'$ is an
irreducible character of a certain representation $\tau$, then the
amount of irreducible components in the direct sum decomposition (as
written above) equivalent with $\rho'$, is given by
$(\chi_\rho|\chi_\rho')$
My attempt:
$\boxed{\Rightarrow}$ Let $\rho$ and $\tau$ be equivalent representations, then for $g \in G$ and for a certain isomorphism $f: V \to W$ we have
$$\rho_g = f \circ \tau_g \circ f^{-1}$$
and $Tr(\rho_g) = Tr([f \circ \tau_g \circ f^{-1}]) = Tr(\tau_g)$ by the cyclic property of trace. Hence, $\chi_\rho = \chi_\tau$
$\boxed{\Leftarrow}$ Here I'm stuck. Can someone hint me in the right direction here? I tried to use the second theorem I listed, but could not find a correct argument.
Best Answer
Note that
$$\langle \phi, \psi \rangle := \frac{1}{|G|}\sum_{g \in G} \phi(g)\overline{\psi(g)}$$
is an inner product on the vector space $\mathbb{C}^G$ and then, according to the second theorem, the characters of non-isomorphic irreducible representations are linearly independent.
Now, let $V_1, ..., V_k$ be an isomorphy list of all simple $G$-modules. Assume that $M = \bigoplus_{i=1}^k m_i V_i, W = \bigoplus_{i=1}^k w_i V_i$ (where the $m_i$'s and the $w_i$'s are the multiplicities, meaning $m_i V_i = V_i \oplus ... \oplus V_i$, $m_i$ times of addition) have the same characters. Moreover, the third theorem yields:
$$\chi_M = \sum_{i=1}^k m_i \cdot \chi_{V_i} = \sum_{i=1}^k w_i \cdot \chi_{W_i} = \chi_W.$$
Owing to the linear independence, we deduce $m_i = w_i$ for all $i$ and hence $M \simeq W$.