I am using Linear Representations of Finite Groups by Jean-Pierre Serre as my reference for this discussion. I would like to use the ideas from the book to determine the character of an irrep of the product of two dihedral groups, $D_4 \times D_8$.
Section 3.2 introduces Product of two groups. It is defined as follows.
Let $G_1$ and $G_2$ be two groups, and let $G_1 \times G_2$ be their products, that is, the set of pairs $(s_1, s_2)$, with $s_1 \in G_1$ and $s_2 \in G_2$.
So, if $(s_1, s_2), (t_1, t_2) \in G_1 \times G_2$, $(s_1, s_2) \cdot (t_1, t_2) = (s_1 t_1, s_2 t_2) \in G_1 \times G_2$.
Let's take our example $D_4 \times D_8$ in consideration. Here, $D_n$ is the dihedral group of order $2 n$. According to Serre, the order of $D_4 \times D_8$ would be $8 \cdot 16 = 128$.
The irreps of a dihedral group $D_n$ of order $2 n$ is given below.
The total number of irreducible representations for $D_n$ is as follows. When $n$ is even, the total number is $\frac{n-2}{2} + 4 = \frac{n}{2} + 3$. When $n$ is odd, it is $\frac{n-1}{2} + 2 = \frac{n+3}{2}$. The one dimensional representations are:
When $n$ is even:
- The trivial representation, sending all group elements to the $1 \times 1$ matrix $\begin{pmatrix}1\end{pmatrix}$.
- The representation, sending all elements in $\langle x \rangle$ to $\begin{pmatrix}1\end{pmatrix}$ and all elements outside $\langle x \rangle$ to $\begin{pmatrix}-1\end{pmatrix}$.
- The representation, sending all elements in $\langle x^2, y \rangle$ to $\begin{pmatrix}1\end{pmatrix}$ and $x$ to $\begin{pmatrix}-1\end{pmatrix}$.
- The representation, sending all elements in $\langle x^2, x y \rangle$ to $\begin{pmatrix}1\end{pmatrix}$ and $x$ to $\begin{pmatrix}-1\end{pmatrix}$.
When $n$ is odd:
- The trivial representation, sending all group elements to the $1 \times 1$ matrix $\begin{pmatrix}1\end{pmatrix}$.
The representation, sending all elements in $\langle x \rangle$ to $\begin{pmatrix}1\end{pmatrix}$ and all elements outside $\langle x \rangle$ to $\begin{pmatrix}-1\end{pmatrix}$.
The $k$-th two dimensional irreducible representations for the general group elements.
\begin{align} x \mapsto \begin{pmatrix} e^{\frac{2 \pi i k}{n}}&0\\ 0&e^{-\frac{2 \pi i k}{n}} \end{pmatrix} \nonumber\\ x^l \mapsto \begin{pmatrix} e^{\frac{2 \pi i k l}{n}}&0\\ 0&e^{-\frac{2 \pi i k l}{n}} \end{pmatrix} \nonumber\\ y \mapsto \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} \nonumber\\ x^l y \mapsto \begin{pmatrix} 0&e^{\frac{2 \pi i k l}{n}}\\ e^{-\frac{2 \pi i k l}{n}}&0 \end{pmatrix} \end{align}
Let's take the case of $x \mapsto \begin{pmatrix} e^{\frac{2 \pi i k}{n}}&0\\ 0&e^{-\frac{2 \pi i k}{n}} \end{pmatrix}$ for both $D_4$ and $D_8$. Let the $x$ be $x_4$ for $D_4$, and $x_8$ for $D_8$. So,
$$x_4 \mapsto \begin{pmatrix} e^{\frac{2 \pi i k}{4}}&0\\ 0&e^{-\frac{2 \pi i k}{4}} \end{pmatrix}$$
and
$$x_8 \mapsto \begin{pmatrix} e^{\frac{2 \pi i k}{8}}&0\\ 0&e^{-\frac{2 \pi i k}{8}} \end{pmatrix}$$.
Now, according to Serre, the linear representation of the element $(x_4, x_8) \in D_4 \times D_8$ is $\begin{pmatrix} e^{\frac{2 \pi i k}{4}}&0\\ 0&e^{-\frac{2 \pi i k}{4}} \end{pmatrix} \otimes \begin{pmatrix} e^{\frac{2 \pi i k}{8}}&0\\ 0&e^{-\frac{2 \pi i k}{8}} \end{pmatrix}$.
According to Theorem $10$ from the same section of Serre, as both $\begin{pmatrix} e^{\frac{2 \pi i k}{4}}&0\\ 0&e^{-\frac{2 \pi i k}{4}} \end{pmatrix}$ and $\begin{pmatrix} e^{\frac{2 \pi i k}{8}}&0\\ 0&e^{-\frac{2 \pi i k}{8}} \end{pmatrix}$ are irreducible for $D_4$ and $D_8$ respectively, $\begin{pmatrix} e^{\frac{2 \pi i k}{4}}&0\\ 0&e^{-\frac{2 \pi i k}{4}} \end{pmatrix} \otimes \begin{pmatrix} e^{\frac{2 \pi i k}{8}}&0\\ 0&e^{-\frac{2 \pi i k}{8}} \end{pmatrix}$ is irreducible for $D_4 \times D_8$.
Now, we will simplify it.
$$\begin{pmatrix} e^{\frac{2 \pi i k}{4}}&0\\ 0&e^{-\frac{2 \pi i k}{4}} \end{pmatrix} \otimes \begin{pmatrix} e^{\frac{2 \pi i k}{8}}&0\\ 0&e^{-\frac{2 \pi i k}{8}} \end{pmatrix}\\ = \left( \begin{array}{cccc} e^{\frac{3 i k \pi }{4}} & 0 & 0 & 0 \\ 0 & e^{\frac{i k \pi }{4}} & 0 & 0 \\ 0 & 0 & e^{-\frac{1}{4} i k \pi } & 0 \\ 0 & 0 & 0 & e^{\frac{1}{4} (-3) i k \pi } \\ \end{array} \right) $$
Serre also gave a relation for the character of product groups. It is given below.
$$ \chi(x_4 \cdot x_8) = \chi(x_4) \cdot \chi(x_8) $$
I would like to test this relation for our example.
Right hand side: $$ \chi(x_4) \cdot \chi(x_8)\\ = Tr(\begin{pmatrix} e^{\frac{2 \pi i k}{4}}&0\\ 0&e^{-\frac{2 \pi i k}{4}} \end{pmatrix}) \cdot Tr(\begin{pmatrix} e^{\frac{2 \pi i k}{8}}&0\\ 0&e^{-\frac{2 \pi i k}{8}} \end{pmatrix}) \\ = (e^{\frac{2 \pi i k}{4}} + e^{-\frac{2 \pi i k}{4}}) \cdot (e^{\frac{2 \pi i k}{8}} + e^{-\frac{2 \pi i k}{8}})\\ = \left(e^{-\frac{1}{4} i \pi k}+e^{\frac{i \pi k}{4}}\right) \left(e^{-\frac{1}{2} i \pi k}+e^{\frac{i \pi k}{2}}\right)\\ = e^{-\frac{1}{4} i \pi k}+e^{\frac{i \pi k}{4}}+e^{\frac{1}{4} (-3) i \pi k}+e^{\frac{3 i \pi k}{4}}\\ = Tr(\left( \begin{array}{cccc} e^{\frac{3 i k \pi }{4}} & 0 & 0 & 0 \\ 0 & e^{\frac{i k \pi }{4}} & 0 & 0 \\ 0 & 0 & e^{-\frac{1}{4} i k \pi } & 0 \\ 0 & 0 & 0 & e^{\frac{1}{4} (-3) i k \pi } \\ \end{array}\right))\\ \text{ = Left hand side} $$
So, it satisfies Serre's relation.
So, that's how I can determine the characters of a product group from those of the multiplicand groups.
Right?