# Character table of direct group product

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.

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).
• I thought this only held when you wanted representations of the tensor product of the spaces $V_1$ and $V_2$ Oct 30, 2019 at 3:42
• 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? Oct 30, 2019 at 3:46
• 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. Oct 30, 2019 at 3:48