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\begin{document}
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\fancyhead[EL]{\nouppercase\leftmark}
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\title{Introduction to Representation theory}
\maketitle
\begin{corollary}
Let $h$ be a linear mapping of $V_1$ into $V_2$. Let
\[
h^0 = \frac{1}{g} \sum_{t\in G} (\rho_t^2)^{-1} h \rho_t^1
\]
Then
\begin{enumerate}[label=(\roman*)]
\item If $\rho^1$ and $\rho^2$ are not isomorphic then we have $h^0=0$
\item If $V_1 = V_2$ and $\rho^1 = \rho^2$, $h^0$ is a scalar multiple of the identity, the scalar being $\frac{1}{n} Tr(h)$ where $n = \dim(V_1)$
\end{enumerate}
\end{corollary}
\begin{proof}
\begin{enumerate}[label=(\roman*)]
\item For any $s$, we have
\[(\rho_{s}^2)^{-1} h^0 \rho_s^1 = \frac{1}{g} \sum_{t \in G}(\rho_{s}^2)^{-1}(\rho_{t}^2)^{-1} h \rho_t^1 \rho_s^1 = \frac{1}{g} \sum_{t \in G}(\rho_{ts}^2)^{-1}h \rho_{ts}^1 = h^0
\]
Applying part $(i)$ of Schur's lemma to $f=h^0$, we see that in case $(i)$ that $h^0 = 0$.
\item Now by part $(ii)$ of Schur's lemma we have $h^0 = \lambda I$. Let us take the trace of both sides
\[
Tr(h^0) = \frac{1}{g} \sum_{t \in G} Tr((\rho_{t}^1)^{-1}h \rho_{t}^1)=Tr(h)
\]
We also know that $Tr(\lambda I) = n \cdot \lambda$ hence $\lambda = \frac{1}{n} Tr(h)$
\end{enumerate}
\end{proof}
\end{document}
