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I have come across the following statement in various sources without any proof. Apparently they say the proof is trivial. However, I don't see the triviality in this case:

$\rho_1,\cdots , \rho_r$ are isomorphism classes of irreducible repreentations of finite group G and $\rho$ is any representation of G. We also have $\chi_i$ and $\chi$ as the characters of $\rho_i$ and $\rho$ and $n_1=<\chi, \chi_i>$

Theorem: Two representations $\rho$ and $\rho'$ of a finite group are equivalent if and only if their characters are equal.

I was able to prove (not rigorously) that the $\chi = n_1\chi_1+\cdots n_r\chi_r$ but I dont see how this helps in our proof. Maybe the forward direction is trivial. That is if two representations are equivalent then their matrix representation will be equivalent too (sa,e matricies), so they will have the same trace. I don't know how to prove the reverse direction.

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    $\begingroup$ This is definitely not a trivial theorem, and it fails in the positive-characteristic case. The proof in characteristic $0$ should be in many texts, however. See Corollary 3.9 in Peter Webb's Representation Theory ( math.umn.edu/~webb/RepBook ) for a proof. $\endgroup$ – darij grinberg Dec 16 '13 at 3:47
  • $\begingroup$ It also fails for infinite groups. So there are various signs that this is nontrivial. $\endgroup$ – Qiaochu Yuan Dec 16 '13 at 9:42
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It follows from the orthogonality relations for characters. Note that if $\chi = \sum n_i \chi_i,$ then $n_i = \langle \chi, \chi_i \rangle.$ So, if two reps $\rho, \sigma$ have the same characters, they have the same set of $n_i.$ If that is true, you are saying that both $\rho$ and $\sigma$ are equivalent to the same representation. Thus they are eqiuivalent.

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The following are standard theorems which should be in any introductory text on representation theory:

Theorem 1: Every finite-dimensional representation of a finite group is a direct sum of irreducible representations.

Theorem 2: Let $\chi_1,\ldots,\chi_r$ be the characters of the irreps of $G$. These form an orthonormal basis for the space of class functions on $G$.

By Theorem 1, we know that $$\rho=\rho_1^{\oplus n_1}\oplus \cdots\oplus \rho_r^{\oplus n_r}$$ $$\rho'=\rho_1^{\oplus m_1}\oplus \cdots\oplus \rho_r^{\oplus m_r}$$ for some $n_i,m_i$ and thus $$\chi=n_1\chi_1+\cdots+n_r\chi_r$$ $$\chi'=m_1\chi_1+\cdots+m_r\chi_r$$ where $\chi$ and $\chi'$ are the characters of $\rho$ and $\rho'$ respectively. It is easy to show that $\rho$ and $\rho'$ are isomorphic iff $n_i=m_i$ for all $i$. But by Theorem 2, $n_i=\langle \chi_i,\chi\rangle$ and $m_i=\langle \chi_i,\chi'\rangle$, so if $\chi=\chi'$ these are equal for all $i$. Conversely, since the $\chi_i$ form a basis, if these are equal for all $i$ then $\chi=\chi'$.

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Last semester I took a course on group theory, which included a bit of representation theory. My teacher made the following statement ''The characters of 2 equivalent representations are the same'', and then he proved it briefly (I include the proof just in case you haven't done this yet). He then said "two representations are equivalent iff their characters are equal" but didn't prove both sides. Maybe it is quite direct if you reverse this proof: \begin{equation} {D}^{A} = {D}^{B} \iff \forall g \in G, \exists S: {D}^{A}(g) ={S}^{-1}{D}^{B}(g)S. \end{equation} (A and B are just labels of the representations.) Then:

\begin{equation} \forall g \in G: Tr({D}^{A}(g)) =Tr({S}^{-1}{D}^{B}(g)S). \end{equation} Since the trace has the cyclical property: \begin{equation} Tr({D}^{A}(g)) =Tr(S{S}^{-1}{D}^{B}(g)) \end{equation}

\begin{equation} Tr({D}^{A}(g)) =Tr({D}^{B}(g)), \end{equation} then, clearly:

\begin{equation} {\chi}^{A}(g)={\chi}^{B}(g). \end{equation} Hope this helps.

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  • $\begingroup$ But clearly one cannot just reverse the proof, as the reverse direction needs much more to be true. $\endgroup$ – Tobias Kildetoft Jul 30 '17 at 17:17

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