Induced irreducible character of degree $2$ of $S_4/V_4$ to $S_5$.

Question

Fix the field to $\mathbb{C}$.

View $S_4$ as the subgroup of $S_5$ fixing $5$. Let $\chi_2$ be the irreducible degree $2$ character of $S_3$. View $\chi_2$ as an irreducible character of $S_4$ via $S_4/V_4 \cong S_3$, where $V_4$ is the normal Klein $4$-subgroup of $S_4$. Decompose the induced character $\text{Ind}^{S_5}_{S_4}\chi_2$ into irreducibles.

So if we have an isomorphism of groups, we can "pull back" via the isomorphism, a representation of one group, to a representation of the other. So let $\varphi:S_3 \overset{\sim}{\to} S_4/V_4$, and let $\chi_2:S_3 \to \mathbb{C}$. Let $\rho:S_3 \to \text{GL}(2,\mathbb{C})$ be the representation associated with $\chi_2$. Then we see that $\varphi^{*}\rho := \rho \circ \varphi:S_4/V_4 \to \text{GL}(2,\mathbb{C})$ is the irreducible degree $2$ representation of $S_4/V_4$ (it will have one, since I claim that isomorphic groups have isomorphic character tables; the converse does not hold, that is, two groups can have the same character table without being isomorphic, e.g. $D_8$ and $Q_8$).

We know that characters of $S_4/V_4$ are in correspondence with characters of $S_4$ with kernel containing $V_4$. Note that $V_4$ is abelian, so in particular, $V_4$ will be contained in any degree $1$ character of $S_4$ (so $\mathbb{1}_{S_4},\text{sgn}$).

So I believe the main problem is to find how the elements in $S_3$ get sent under $\varphi$ to $S_4/V_4$, since this will how $\chi_2$ behaves as a character of $S_4$. If one knows the character table of $S_4$ by heart (otherwise, one could perhaps derive it), then I believe one does not have to worry much about $S_4/V_4 \cong S_3$, am I correct?

We see from the accepted answer here Explicit isomorphism $S_4/V_4$ and $S_3$ that the explicit isomorphism from $S_4/V_4$ to $S_3$ is gotten by letting $A = \{a,b,c\}$ get acted on by $S_4$, where $a = (12)(34)$,$b = (13)(24)$ and $c = (14)(23)$. This induces a permutation representation $\sigma: S_4 \to S_{A} = S_{3}$.

One finds that $\chi_2(1V_4) = \tilde{\chi}_2(1) = 2$, that $\chi_2(12V_4) = \tilde{\chi}_2(12) = 0,\chi_2(123V_4) = 2,\chi_2((12)(34)V_4) = 1$ and $\chi_2((1234V_4)) = 0$. Here, $\tilde{\chi}_2$ is the character of the representation $\tilde{\rho}:S_4 \to \text{GL}(2,\mathbb{C})$ associated with $\rho:S_4/V_4 \to \text{GL}(2,\mathbb{C})$.

We then get $\text{Ind}^{S_5}_{S_4}\tilde{\chi}_2(g) := \frac{1}{24}\sum\limits_{x \in S_5} \tilde{\chi}_2^{0}(xgx^{-1})$ for $g \in S_5$, where

$$\tilde{\chi}_2^{0}(g) := \begin{cases} \tilde{\chi_2}(g), \ \text{if} \ g \in S_4\\ 0, \ \text{if}\ g \not \in S_4 \end{cases}.$$

We see directly that $\text{Ind}^{S_5}_{S_4}\tilde{\chi_2}(g) = 0$ if $g$ is a $5$-cycle, since $5$-cycles are only conjugate to other $5$-cycles, and there are none in $S_4$.

Recall that $\text{Ind}^{S_5}_{S_4}\tilde{\chi}_2$ is a character, so it is enough to compute the value once for each conjugacy class. We know that $\text{Ind}^{S_5}_{S_4}\tilde{\chi}_2(1) = [S_5:S_4] \chi_2(1) = 5 \cdot 2 = 10$.

If we look at $(12)$, a possible issue arises, which is; notice that $(15)(12)(15) = (25)$ and $(25)(12)(25) = (15)$, so that the induced character will be $0$ on these elements. But! This issue does not arise in this particular case, since regardless, $\tilde{\chi}^{0}_2(g) \equiv 0$ on $2$-cycles.

For $(12)(34)$, we see that for the $24$ elements in $S_4$ contained in $S_5$, $\tilde{\chi}^{0}_2(g) = 2$. We know that it will be $0$ on any elements taking either $1,2,3,4$ to $5$, so we need to count the elements where this does not happen. In this case, since generally $x(12)(34)x^{-1} = (x(1)x(2))(x(3)x(4))$, we see that any element in $S_5 \setminus{S_4}$ will permute $(12)(34)$ to something not in $S_4$, I claim. So $\text{Ind}^{S_5}_{S_4}\tilde{\chi}((12)(34)) = \frac{2 \cdot 24}{24} = 2$.

Similar to the case $(12)$, we see that $\text{Ind}^{S_5}_{S_4}\tilde{\chi}_2((1234)) = 0$.

Things get a bit trickier for $(123)$. We get $24$ elements with each value $-1$ in the sum of the induced character. Any $\tau \in S_5 \setminus{S_4}$ which permutes $1,2,3$ to $5$ gives us a $0$-term. How many such elements are there, one might ask. For $2$-cycles, there are $3$ such elements ($(15),(25),35)$). For $3$-cycles, we fix $5$, and then we can choose $\binom{3}{2} \cdot 2! = 6$ such elements, but we can have even more, for example $(543)$. It seems like things get tricky to work out here. I'll post this as a partial idea, and if someone has suggestions for how to work it efficiently, I'd be happy.

When one has computed the character table for the induced character, assuming one has access to $S_5$:s character table, one has to decompose it, which one can do explicitly by computing the hermitian inner product on the set of class functions from $S_5$ to $\mathbb{C}$ by coupling the induced character with the irreducible characters, or one can in the optimistic case just inspect the character table carefully.

Remark: After thinking about it, I am not sure one has to do what I did in the beginning, when pulling back to $S_4/V_4$ (I left out parts of the details there), atleast if one knows the unique irreducible degree $2$ character of $S_4$ by heart.

Answer 1

To make the calculation easier I include the relevant row from the character table of $S_4$ together with conjugacy class data: $$ \begin{array}{c|c|c|c|c|c} g&1&(12)&(12)(34)&(123)&(1234)\\ \hline |[g]|&1&6&3&8&6\\ \hline \chi_2&2&0&2&-1&0\\ \end{array} $$ Every element of $V_4$ is in the kernel, and the images of 2-cycles and 4-cycles in the quotient group $S_4/V_4\simeq S_3$ are 2-cycles (hence with a single fixed point and character value zero), but the 3-cycles are mapped to 3-cycles (hence no fixed points, and character value $-1$).

You seem to be using the following formula for the induced character: $$ \mathrm{Ind}_H^G(\chi)(x)=\frac1{|H|}\sum_{g\in G}\chi(gxg^{-1}),\qquad(*) $$ where $x\in G$ is arbitrary, and we adopted the convention that $\chi(y)=0$ whenever $y\in G\setminus H$.

The full character table of $S_5$ looks like $$ \begin{array}{c|c|c|c|c|c|c|c} g&1&(12)&(123)&(12)(34)&(1234)&(123)(45)&(12345)\\ \hline c_i&1&10&20&15&30&20&24\\ \hline \psi_1&1&1&1&1&1&1&1\\ \psi_2&1&-1&1&1&-1&-1&1\\ \psi_3&4&2&1&0&0&-1&-1\\ \psi_4&4&-2&1&0&0&1&-1\\ \psi_5&5&1&-1&1&-1&1&0\\ \psi_6&5&-1&-1&1&1&-1&0\\ \psi_7&?&?&?&?&?&?&? \end{array} $$ (copy/pasted from a source that left the calculation of the values of the last 6-dimensional character as an exercise).

The chief problem to the OP seems to be calculate the values of the induced character $\theta=\mathrm{Ind}_{S_4}^{S_5}(\chi_2).$ He already calculated $\theta(1)=10$.

The conjugacy classes can be equated with cycle patterns. It immediately follows that $\theta(x), x\in S_5$, vanishes unless $x=1$, $x$ is a product of two disjoint 2-cycles, or $x$ is a 3-cycle.

We saw that there are $20$ 3-cycles in $S_5$, so each with a centralizer of size $6$. Let $x\in S_5$ be a 3-cycle. There are $8$ 3-cycles in the subgroup $S_4$, so $gxg^{-1}$ is an element of $S_4$ for $8\cdot6=48$ choices of $g\in S_5$ — each of those eight occuring exactly six times (by orbit-stabilizer). Therefore the formula $(*)$ gives $$\theta(x)=\frac{48}{|S_4|}\chi_2((123))=-2.$$ Similarly, if $y\in S_5$ is a product of two disjoint 2-cycles, its centralizer has size $120/15=8$. Consequently each of the three permutations of this type $\in S_4$ occurs as $gyg^{-1}$ eight times, and $gyg^{-1}\in S_4$ for $24$ choices of $g$ altogether. Hence, by $(*)$, $$ \theta(y)=\frac{24}{24}\chi_2((12)(34))=2. $$ The rest is a straight forward calculation. $\langle\theta,\theta\rangle=2$, so it is a sum of two irreducible characters of $S_5$. Consulting the table quickly shows that $\theta=\psi_5+\psi_6$ is the sum of the two 5-dimensional irreducible characters of $S_5$.