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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?

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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 $G_1 = G_2 = \mathbb{Z} / 2 \mathbb{Z}$.

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