# direct sum of representation of product groups

Given two finite groups $$G_1$$ and $$G_2$$, and some representations $$\rho_1: G_1 \to V_1$$ and $$\rho_2: G_2 \to V_2$$, it seems the standard way to create a representation for $$G_1 \times G_2$$ is to use the tensor product $$\rho_1(g_1) \otimes \rho_2(g_2) \quad g_1,g_2 \in G_1,G_2.$$ It seems to me that one could also use the direct sum $$\rho_1(g_1) \oplus \rho_2(g_2)$$, because the blocks in the matrix form of the representation do not interact and one gets the desired effect. Given that this representation could have a lower dimension than using tensor product, why is it not used?

## 1 Answer

When $$V_1$$ and $$V_2$$ are representations of $$G_1$$ and $$G_2$$ respectively, I'll use $$V_1 \boxtimes V_2$$ to mean the representation of $$G_1 \times G_2$$ with underlying vector space $$V_1 \otimes_{\mathbb{C}} V_2$$, and $$V_1 \boxplus V_2$$ to mean the representation of $$G_1 \times G_2$$ with underlying vector space $$V_1 \oplus V_2$$.

If $$V$$ is an irreducible representation of $$G_1 \times G_2$$, then $$V$$ is isomorphic to $$V_1 \boxtimes V_2$$ for some irreducible representations $$V_1$$ and $$V_2$$ of $$G_1$$ and $$G_2$$ respectively. This means that if we know the representations of $$G_1$$ and $$G_2$$, then using the $$\boxtimes$$ construction we can get to all the (irreducible) representations of $$G_1 \times G_2$$. Conversely, the $$\boxtimes$$ product of two irreducible representations always produces an irreducible representation of $$G_1 \times G_2$$.

On the other hand, $$V_1 \boxplus V_2$$ is always reducible as a $$G_1 \times G_2$$ representation, since both vector subspaces $$V_1$$ and $$V_2$$ are stable under the $$G_1 \times G_2$$ action. On the $$V_1$$ subspace, really only the $$G_1$$ part of the group acts, and the $$G_2$$ part acts trivially, and similarly for the $$V_2$$ subspace. We cannot produce all irreducible representations of $$G_1 \times G_2$$ using this construction, which can already be seen in the example $$G_1 = G_2 = \mathbb{Z} / 2 \mathbb{Z}$$.