Group Algebra of Direct Product is Tensor Product of Group Algebras

Question

Given groups $G$ and $H$, I want to show that $R[G \times H] \cong R[G] \otimes R[H]$. I'm not sure why, but I'm struggling with it. I'd like to map the basis elements from one to the other in the obvious way, that is $$ F:R[G] \otimes R[H] \rightarrow R[G \times H]$$ $$ g \otimes h \mapsto (g,h) $$ but I'm having trouble with the details. Clearly this is invertible, so I just need to show it respects the algebra structure. I've got that $$ F((g\otimes h) \cdot (g'\otimes h')) = F((gg') \otimes (hh')) = (gg',hh') = (g,h)(g',h') = F(g \otimes h) F(g' \otimes h) $$ and that $$ F(r(g \otimes h)) = r(g,h) = rF(g \otimes h), $$ so how do I go about showing that $$ F( (g\otimes h) + (g' \otimes h')) = F( g\otimes h) + F(g' \otimes h'). $$ Clearly, $g \otimes h + g' \otimes h' \neq (g+g') \otimes (h + h')$ in general so I feel like I have to be missing something silly or I'm going about this completely wrong? Any help is appreciated!

Answer 1

It might be easier to appeal to the universal property of the tensor product and the fact that universal objects are defined uniquely up to (unique) isomorphism.

So, we want to show that a given algebra homomorphism $\varphi\colon R[G]\times R[H]\to A$ uniquely induces an algebra homomorphism $\overline\varphi\colon R[G\times H]\to A$ such that $\varphi=\overline\varphi\circ\iota$ where $\iota\colon R[G]\times R[H]\to R[G\times H]$ is the obvious embedding. But this is clear as $\varphi$ is uniquely determined by its values on $g\in G$ and $h\in H$ and hence we can easily define $\overline\varphi$ (uniquely) on the pairs $(g,h)$ which form a basis for $R[G\times H]$.

Answer 2

One other way to see it is to consider the map $\varphi : R[G \times H] \to R[G] \otimes_R R[H]$ that sends $$\sum_{(g, h) \in G \times H} a_{(g, h)} (g, h) \mapsto \sum_{(g, h) \in G \times H} a_{(g, h)} (g \otimes_R h).$$ Understanding the sum on the right-hand side as a double sum $$\sum_{h \in H} \sum_{g \in G} b_g c_h (g \otimes_R h) = \biggl(\sum_{g \in G} b_g g \biggr) \otimes_R \biggl(\sum_{h \in H} c_h h \biggr)$$ with $b_g c_h = a_{(g, h)},$ it follows that $\varphi$ is surjective, as every element of $R[G] \otimes_R R[H]$ is a finite $R$-linear combination of elementary tensors. Of course, $\varphi$ is injective, so it is an isomorphism.