# Computing the characters of $\mathbf{1}_{D_n} \uparrow^{S_n}_{D_n}$

How can I compute the characters of the induced representation $\mathbf{1}_{D_n} \uparrow^{S_n}_{D_n}$?

Here, $S_n$ is the symmetric group over $n$ symbols and $D_n$ is the dihedral group of order $2 n$.

I can compute the character table of the symmetric group $S_n$ using the scheme mentioned in this blog article which uses the Murnaghan-Nakayama rule.

Is there a similar approach for computing the characters of $\mathbf{1}_{D_n} \uparrow^{S_n}_{D_n}$?

I understand that I can compute the transversal of $D_n$ in $S_n$ and use it to compute the matrix representation of each element in $S_n$ as shown in the section 1.12 of The Symmetric Group : Representations, Combinatorial Algorithms, and Symmetric Functions by Sagan. Then I can compute the characters by taking the traces. But this technique is tedious.

Is there any shorter procedure where I can use the Murnaghan-Nakayama rule to compute the characters of $\mathbf{1}_{D_n} \uparrow^{S_n}_{D_n}$?

Motivation

I am trying to detect the hidden subgroup $D_n$ of order $2 n$ within $S_n$ using Normal Subgroup Reconstruction and Quantum Computation Using Group Representations as reference. This boils down to computing the Fourier transform of the following indicator function of a left coset of $D_n$ in $S_n$ which is assumed.

$$f (g) = \begin{cases} \frac{1}{\sqrt{|D_n|}} & \quad \text{ if } g \in c D_n\\ 0 & \quad \text{ otherwise}\\ \end{cases}$$ for some $c \in S_n$.

If the quantum Fourier transform is $\hat{f}$, we seek to measure the labels of irreps, $\rho$. According to the Lemma 1 in the paper, the probability of measuring $\rho$ is $\frac{|D_n|}{|S_n|} d_\rho \langle \chi_\rho, \chi_{\mathbf{1}_H}\rangle_{D_n}$.

Using a special case of Frobenius reciprocity as shown in the Lemma 2 in the paper, we can say that, $$\langle {\chi \uparrow^{S_n}_{D_n}}_{\mathbf{ 1}_{D_n}} , \chi_\rho \rangle_{S_n} = \langle \chi_{\mathbf{ 1}_{D_n}}, \chi_\rho \downarrow^{S_n}_{D_n} \rangle_{D_n}$$

That's why I need to compute $\langle {\chi \uparrow^{S_n}_{D_n}}_{\mathbf{ 1}_{D_n}} , \chi_\rho \rangle_{S_n}$. To do that I need to determine the character table of $\mathbf{1}_{D_n} \uparrow^{S_n}_{D_n}$.

• So if that notation means "an induced representation," you're just looking for its (singular) character, and not character table right? – pjs36 May 25 '16 at 16:08
• A "character table" is something associated to a group. What you're asking for the character table of is a representation (usually reducible) of $S_n$; it doesn't have a character table. The closest you could ask for is its decomposition into reducible characters (which, in general, is going to be fairly large; the order of $S_n$ grows as $n!$, while the order of $D_n$ grows as $2n$). – PL. May 25 '16 at 18:02
• Or it occurs to me that perhaps you could want to decompose your induced character into irreducibles and hope that this gives enough of the characters of $S_n$ to fill in the rest of the table? If so, I don't think that's a very good approach due to the fact that the induced character will in general be of huge length. Young tableaux methods really are the best way to deal with representations of $S_n$. – PL. May 25 '16 at 18:06
• Your edit does not address that the thing does not have a character table but a decomposition. – Tobias Kildetoft May 25 '16 at 19:41
• You appended your motivation (which I don't understand, btw) but the problem is that we don't understand what you are trying to do, independently of why you want to do it. As people have told you, representations do not have «character tables» so your question does not make sense at the moment. – Mariano Suárez-Álvarez May 25 '16 at 19:55