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Let $D_n$ be the dihedral group with order $2 n$. 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:

  1. The trivial representation, sending all group elements to the $1 \times 1$ matrix $\begin{pmatrix}1\end{pmatrix}$.
  2. The representation, sending all elements in $\langle x \rangle$ to $\begin{pmatrix}1\end{pmatrix}$ and all elements outside $\langle x \rangle$ to $\begin{pmatrix}-1\end{pmatrix}$.
  3. The representation, sending all elements in $\langle x^2, y \rangle$ to $\begin{pmatrix}1\end{pmatrix}$ and $x$ to $\begin{pmatrix}-1\end{pmatrix}$.
  4. 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:

  1. The trivial representation, sending all group elements to the $1 \times 1$ matrix $\begin{pmatrix}1\end{pmatrix}$.
  2. 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}

Now following the Frucht's theorem, any product (direct or wreath) between $D_n$ and any Finite group is an automorphism group of a graph. I am interested about the representation theory of this automorphism group.

Can the irreps of this product, $D_n \times G$, where $G$ is any finite group, have similar matrices as shown above where the traces will be zero for some of the group elements?

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    $\begingroup$ Perhaps $\S8.4$ of this PDF will help for direct products. It's about characters, not representations, but you seem primarily interested in traces. $\endgroup$
    – pjs36
    Jun 19, 2016 at 3:03
  • $\begingroup$ @pjs36, I have checked out $\S8.4$ of the pdf. I would also quote the following from mathworld.wolfram.com/ExternalTensorProduct.html. "When V and W are irreducible representations of G and H respectively, then so is the external tensor product. In fact, all irreducible representations of G×H arise as external direct products of irreducible representations." $\endgroup$ Jun 19, 2016 at 3:41
  • $\begingroup$ @pjs36, it means, the irreps of $D_n \times G$ will have zero traces when they are the external tensor product between the irreps of $D_n$ with zero traces and the irreps of $G$. $\endgroup$ Jun 19, 2016 at 3:46

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In general, suppose that $F$ is a field and $G$ is a group that can be expressed as a direct product $G=H\times K$. Let $\rho$ and $\sigma$ be representations of $H$ and $K$ over $F$, respectively. Then a corresponding representation of $G$ over $F$ may be constructed from $\rho$ and $\sigma$ by using tensor products.

Suppose that $\rho$ and $\sigma$ arise from an $FH-$module $M$ and an $FK-$module $N$, respectively. Form the tensor product $T=M\otimes_{F}N$ and make $T$ into a right $FG-$module by the rule $(a\otimes b)(x,y)=(ax)\otimes (by)$, where $a\in M$, $b\in N$, $x\in H$ and $y\in K$. Then $T$ affords an $F-$representation $\rho\sharp \sigma$ called the Kronecker (or outer tensor) product of $\rho$ and $\sigma$. The degree of $\rho\sharp \sigma$ equals the product of the degrees of $\rho$ and $\sigma$.

It is easy to show that if $\rho$ has character $\chi$ and $\sigma$ has character $\psi$, the character $\phi$ of $\rho\sharp \sigma$ is given by $(x, y)\phi=(x)\chi(y)\psi$.
There is a theorem in representation theory which state that, if $F$ is an algebraically closed field, $G$ is finite, the characteristic of $F$ does not divide the order of $G$ and $\{\rho_1,\ldots,\rho_h\}$ and $\{\sigma_1,\ldots,\sigma_k\}$ are complete sets of inequivalent irreducible $F-$representations of $H$ and $K$, then the $\rho_i\sharp \sigma_r$, for $i=1,\ldots h$ and $r = 1,\ldots,k$ , form a complete set of inequivalent irreducible $F-$representations of $G$.

With the discussion above, a complete set of inequivalent irreducible $\mathbb{C}-$characters of $D_n\times G$, where $G$ is a finite group, can be constructed by the rule $(x, y)\phi_j=(x)\chi_i(y)\psi_r$, where $i=1,\ldots h$, $r = 1,\ldots,k$, $x\in D_n$, $y\in G$, and $\{\chi_1,\ldots,\chi_h\}$ and $\{\psi_1,\ldots,\psi_k\}$ are complete sets of inequivalent irreducible $\mathbb{C}-$characters of $D_n$ and $G$.

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  • $\begingroup$ could you please give me the reference for the theorem? I will also try to find it in Fulton and Harris. $\endgroup$ Jun 19, 2016 at 14:08
  • $\begingroup$ I just found the theorem in Section 6.4 of Group Theory: Application to the Physics of Condensed Matter by Mildred S. Dresselhaus, Gene Dresselhaus, Ado Jorio. $\endgroup$ Jun 19, 2016 at 14:28
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    $\begingroup$ For more details you can read the chapter 8 of the book "A Course in the Theory of Groups" by Derek J.S. Robinson. $\endgroup$ Jun 19, 2016 at 18:29

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