Let’s say we have two finite groups and , and we have (left) representations of each one: and . It turns out that the tensor product naturally carries a representation of the product group . Equivalently, it carries a representation of each of and , and these representations commute with each other. In our module notation, we write .
The action is simple enough. Any vector in can be written (not always uniquely) as a sum of vectors of the form . We let act on the first “tensorand”, let act on the second, and extend by linearity. That is:
and the action of either or on the sum of two tensors is the sum of their actions on each of the tensors.
Now, might the way we write the sum make a difference? No, because all the relations look like
where in the last equation is a complex constant. Now, we can check that the actions of and give equivalent results on either side of each equation. For instance, acting by in the first equation we see
just as we want. All the other relations are easy enough to check.
But do the actions of and commute with each other? Indeed, we calculate
So we really do have a representation of the product group.
We have similar “outer” tensor products for other combinations of left and right representations:
Now, let’s try to compute the character of this representation. If we write the representing homomorphisms and , then we get a representing homomorphism . And this is given by
Indeed, this is exactly the endomorphism of that applies to and applies to , just as we want. And we know that when expressed in matrix form, the tensor product of linear maps becomes the Kronecker product of matrices. We write the character of as , that of as , and that of their tensor product as , and calculate:
That is, the character of the tensor product representation is the product of the characters of the two representations.
Finally, if both and are irreducible, then the tensor product is as well. We calculate:
In particular, we find that . If and are both irreducible characters, then our character properties tell us that both of the inner products on the right are , and we conclude that the inner product on the left is as well, which means that is irreducible.