Simple Q about direct product representation of a group

[AxiomOfChoice]

(At least, I think it's simple.)

Disclaimer: I'm approaching this subject from the vantage point of a chemist, so be careful with how much lingo/jargon/rigor you lay on me

The claim is that if you have two representations of a group, $\Gamma_1$ and $\Gamma_2$, with bases $\{ f_i \}$ and $\{ g_j \}$, then the set of products $\{f_i g_j\}$ is a basis for the direct product representation. Fine. But in the process of showing this, the following assertion is made: If $\hat R$ is an element of the group, then

$$\hat R(f_i g_j) = \hat R(f_i) \hat R(g_j).$$

...huh? How does that work? How do I know that if I apply a group operation to a product of basis elements, then it's just the product of the operation applied to the individual basis elements? I missed something there, because (a priori) that move is just as suspect as saying that $x(yz) = (xy)(xz) = x^2yz$ for $x,y,z\in \mathbb R$.

 

[lavinia]

(At least, I think it's simple.)

Disclaimer: I'm approaching this subject from the vantage point of a chemist, so be careful with how much lingo/jargon/rigor you lay on me

The claim is that if you have two representations of a group, $\Gamma_1$ and $\Gamma_2$, with bases $\{ f_i \}$ and $\{ g_j \}$, then the set of products $\{f_i g_j\}$ is a basis for the direct product representation. Fine. But in the process of showing this, the following assertion is made: If $\hat R$ is an element of the group, then

$$\hat R(f_i g_j) = \hat R(f_i) \hat R(g_j).$$

...huh? How does that work? How do I know that if I apply a group operation to a product of basis elements, then it's just the product of the operation applied to the individual basis elements? I missed something there, because (a priori) that move is just as suspect as saying that $x(yz) = (xy)(xz) = x^2yz$ for $x,y,z\in \mathbb R$.

If each representation is thought of a a set of matrices then the product representation is just the matrices of each representation joined into a diagonal block. Multiplication preserves these blocks.