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Each irreducible representation of $G_1\times G_2$ is isomorphic to a representation $\rho_1\otimes\rho_2$, where $\rho_i$ is an irreducible representation of $G_i$ for $i=1,2.$

So, this is the theorem $10$ in Jean-Pierre Serre book "Linear Representations of Finite Groups".

I understood half of the proof, but there is a line which he says: "it suffices to show that each class function $f$ on $G_1\times G_2$, which is orthogonal to the characters on the form of $\chi_1(s_1)\chi_2(s_2)$, is zero".

I don't get why showing that each class function is zero proves that each irreducible representation of $G$ is of the form $\rho_1\otimes\rho_2$, my guess is that this idea has some relation with the theorem 6 of the book that says: "The characters $\chi_1,...,\chi_h$ form an orthonormal basis of $H$", but i can't fully comprehend this.

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    $\begingroup$ If the given characters did not form a basis, then there would be some non-zero element in the orthogonal complement of their span. He shows that there is not. $\endgroup$ Mar 30 '20 at 17:08
  • $\begingroup$ And why this proves that each irreducible representation is of the form i cited above? $\endgroup$
    – Vityôk
    Mar 30 '20 at 17:15
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    $\begingroup$ Because the characters mentioned there are precisely the characters of the representations claimed to form a basis. $\endgroup$ Mar 30 '20 at 17:17
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Here you have and alternative (and I think more understandable and full detailed) proof of the result you want to prove.

$\textbf{Proposition}$ Let $\phi^1:G_1\to \text{GL}(V)$ and $\phi^2:G\to \text{GL}(V)$ two group representations of $G_1$ and $G_2$ with dimensions $d_1$ and $d_2$, respectively, over the same vectorial space $V$. Denoting as $D^1(g^1_i)$ y $D^2(g_j^2)$ the matrices of the representations $\phi^1$ and $\phi^2$, then $D^1(g^1_i) \otimes D^2(g_j^2)$ is a representation with dimension $d_1d_2$ of $G_1\times G_2$.

$\textit{Proof}.$ By the properties of the Kronecker product, it followes that \begin{equation*} \begin{split} [D^1(g_i^1)\otimes D^2(g_j^2)][D^1(g_k^1)\otimes D^2(g_l^2)] & =[D^1(g_i^1)D^1(g_k^1)]\otimes [D^2(g_j^2)D^2(g_l^2)]\\ &=D^1(g_i^1g_k^1)\otimes D^2(g_j^2g_l^2). \end{split} \end{equation*}

$\textbf{Theorem}$ With the same notation, if the representations $\phi^1$ and $\phi^2$ are irreducible then $D^1(g^1_i) \otimes D^2(g_j^2)$ of hte group $G_1\times G_2$ is also irreducible. Moreover, all the irreducible representations of $G_1\times G_2$ are the direct product of an irreducible representation of $G_1$ times other of $G_2$.

$\textit{Proof}$ It is known that the irreducibility of $\phi^1$ and $\phi^2$ is equivalent to the condition \begin{equation*} \sum_{g\in G_1} \chi^1(g)^*\chi^1(g)=|G_1| \quad \text{y} \quad \sum_{h\in G_2} \chi^2(h)^*\chi^2(h)=|G_2|, \end{equation*} where $\chi^1$ and $\chi^2$ are the characters of $\phi^1$ y $\phi^2$, respectively. Then, \begin{equation*} \begin{split} |G_1\times G_2|& =|G_1||G_2|=\left(\sum_{g\in G_1} \chi^1(g)^*\chi^1(g)\right)\left(\sum_{h\in G_2} \chi^2(h)^*\chi^2(h)\right)\\ &=\sum_{g\in G_1}\sum_{h\in G_2} \left(\chi^1(g)\chi^2(h)\right)\left(\chi^1(g)^*\chi^2(h)^*\right). \end{split} \end{equation*}

On the other hand, using that the characters of the Kronecker product of the two representations is the product of the characters of the two representations, then \begin{equation*} |G_1\times G_2|=\sum_{g'\in G_1\times G_2} \chi(g')\chi(g')^*, \end{equation*} where $\chi$ are the characters of the representation on the direct product $G_1\times G_2$. Thus, we infer that the representation $D^1(g^1_i) \otimes D^2(g_j^2)$ of the group $G_1\times G_2$ is also irreducible.

For the second part, let $m_1$ and $m_2$ the number of irreducible representations of $G_1$ and $G_2$, respectively, whose dimensions are denoted as $d_{i,1}$ and $d_{j,2}$. Hence, \begin{equation*} |G_1|=\sum_{i=1}^{m_1} d_{i,1}^2 \quad \text{y} \quad |G_2|=\sum_{j=1}^{m_2} d_{j,2}^2. \end{equation*} The irreducible representations of $G_1\times G_2$ that are obtained as the Kronecker product of irreducible representations of $G_1$ and $G_2$ will have dimension $d_{i,1}d_{j,2}$. Therefore, the sum of the squares of the dimensions $d_k$ of the irreducible representations $G_1\times G_2$ is \begin{equation*} \sum_{k=1}^{m_1m_2}d_k^2= \sum_{i=1}^{m_1} \sum_{j=1}^{m_2} d_{i,1}^2d_{j,2}^2 = |G_1||G_2|=|G_1\times G_2|. \end{equation*} We conclude that the Kronecker product of the irreducible representations of $G_1$ and $G_2$ determine all the irreducible representacions of $G_1\times G_2$.

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