Degrees of irreducible representations of direct product are pairwise products of degrees of irreducible representations of direct factors

Over any field
Suppose $$G_1$$ and $$G_2$$ are finite groups. Suppose $$K$$ is a field. Suppose $$d_{11},d_{12},\dots,d_{1r}$$ are the degrees of irreducible representations (possibly with repetitions) of $$G_1$$ over $$K$$ and $$d_{21},d_{22},\dots,d_{2s}$$ are the degrees of irreducible representations of $$G_2$$.

Then, the degrees of irreducible representations of $$G_1 \times G_2$$ over $$K$$ are:

$$d_{11}d_{21}, d_{11}d_{22}, \dots,d_{11}d_{2s},d_{12}d_{21},d_{12}d_{22},\dots,d_{12}d_{2s},d_{1r}d_{21},\dots,d_{1r}d_{2s}$$

Over a splitting field
If we take $$K$$ to be a splitting field for both $$G_1$$ and $$G_2$$, then the above relates the degrees of irreducible representations for $$G_1,G_2,G_1 \times G_2$$ over a splitting field. For instance, we could take $$K$$ to be $$\mathbb{C}$$ or $$\overline{\mathbb{Q}}$$. Note that degrees of irreducible representations are the same for all splitting fields (as long as the characteristic does not divide the group order), so this relates universal lists of numbers for $$G_1,G_2,G_1 \times G_2$$.

Related facts

 * Conjugacy class sizes of direct product are pairwise products of conjugacy class sizes of direct factors
 * Number of conjugacy classes in a direct product is the product of the number of conjugacy classes in each factor

Facts used

 * 1) uses::Tensor product establishes bijection between irreducible representations of direct factors and direct product