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Consider a finite group $G$ and assume it decomposes as

$$G\cong\displaystyle\bigoplus_{k=1}^n G_k.$$

Say that I know the character table for all of $G_k$. Can I construct the character table for $G$?

We consider complex irreducible representations, i.e. group homomorphisms $\pi : G\to\operatorname{End}(\mathbb C^n).$

In particular, the case $n=2$ is of interest, since I have a situation like this.

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1 Answer 1

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Yes. The complex irreducible representations of $G_1\times G_2$ are tensor products $(V_1\otimes V_2,\rho_1\otimes\rho_2)$ of complex irreducible representations $(V_1,\rho_1)$ of $G_1$ and $(V_2,\rho_2)$ of $G_2$. The character associated to a tensor product is just a product of characters, $\chi_{\small V_1\otimes V_2}(g_1,g_2)=\chi_{V_1}(g_1)\chi_{V_2}(g_2)$. The conjugacy classes of a direct product $G_1\times G_2$ are Cartesian products $K_1\times K_2$ of conjugacy classes $K_1$ of $G_1$ and $K_2$ of $G_2$. Therefore, the character table of $G_1\times G_2$ is just the "Kronecker product" of the tables for the groups $G_1$ and $G_2$ (viewing the tables as matrices).

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  • $\begingroup$ I thought this only held when you wanted representations of the tensor product of the spaces $V_1$ and $V_2$ $\endgroup$ Oct 30, 2019 at 3:42
  • $\begingroup$ @MarkusKlyver What does the word "this" in your sentence refer to? The tensor products of irreps give all irreps of a direct product, which is what you wanted. $\endgroup$
    – runway44
    Oct 30, 2019 at 3:43
  • $\begingroup$ Consider my case. I have $\mathbb Z/2\mathbb Z \oplus S_3$. The cyclic group of order 2 has 2 irreducible representations and $S_3$ has 3. Despite this, their product, the symmetries of a regular hexagon, only has 5 irreducible representations. What's wrong here? $\endgroup$ Oct 30, 2019 at 3:46
  • $\begingroup$ Your statement that $C_2\times S_3$ has only 5 irreps must be wrong. Its 6 irreps are the tensor product of the 2 from $C_2$ and the 3 from $S_3$. I expect most (mathematical) textbooks covering complex representations of finite groups state and prove this fact about irreps of direct products. $\endgroup$
    – runway44
    Oct 30, 2019 at 3:48
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    $\begingroup$ @MarkusKlyver That article clearly shows there are 6 irreps, indexed 0-5. :-) $\endgroup$
    – runway44
    Oct 30, 2019 at 3:51

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