Irreducible Representations of (Sₖ × Sₖ) ⋊ ℤ₂

Irreducible Representations of $(S_k\times S_k)⋊{\mathbb{Z}}_2$

Notes — 29 September 2025 I stayed up late. -Sam

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Let $W=(S_k\times S_k)⋊{\mathbb{Z}}_2$ with ${\mathbb{Z}}_2=⟨\tau ⟩$ where $\tau$ is the nontrivial element of ${\mathbb{Z}}_2$ that swaps the two factors of $S_k\times S_k$, and $k\in {\mathbb{Z}}_{>0}$. Let $N:=S_k\times S_k$ and build the quotient group $Q:=W/N\cong {\mathbb{Z}}_2=\{1,\tau \}$ (which is normally just a coset but since $N◃W$ this forms a quotient group).

The goal here is to classify $\hat{W}$, the irreps of $W$, using Clifford Theory. To do this, we follow the following steps:

1. Understand $\hat{N}$ and how $Q$ acts on it by conjugation.
2. For each $Q$-orbit $\mathcal{O}\subset N$, pick a representative $\pi \in \mathcal{O}$.
3. Let $I:={\text{Stab}}_W(\pi )$ be the inertia subgroup (this seems to be a historical term indicating the stabilizer under conjugation).
4. Irreps of $W$ lying over $\mathcal{O}$ are then obtained by (a) extending $\pi$ to $I$ (in principle is not always possible, but here it is), then (b) inducing to $W$.

Remark 1

The statement “extending $\pi$ to $I$” should be clarified a bit. We start with an irrep $\pi$ of $N$ and we know, by construction, that $N\le I\le W$. But here, $I$ is either $N$ or $W$ depending on whether we look at the diagonal or off-diagonal case. An extension of $\pi$ to $I$ really means to find a representation $\tilde{\pi }:I\to \text{GL}(V)$ such that

${\text{Res}}_N^I\tilde{\pi }\cong \pi ,$

where we will define what this ${\text{Res}}_N^I$ thing means shortly.

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First, let’s look at the irreps of $N=S_k\times S_k$ and the ${\mathbb{Z}}_2$ action on these objects. We have the following theorem from representation theory:

Theorem 1 (Fulton & Harris Exercise 2.36, lol)

The irreps of a direct product group $G_1\times G_2$ are precisely the outer tensor products $\rho ⊠\sigma$, where $\rho$ and $\sigma$ are irreps of $G_1$ and $G_2$, respectively.

Remark 2

The external tensor product has an underlying vector space associated with it and equipped with the action defined as

$(\rho ⊠\sigma )(g_1,g_2)=\rho (g_1)\otimes \rho (g_2).$

Why is this different from the “internal” tensor product? The internal tensor product takes the representations from the same group, say $\rho ,\sigma \in \hat{G}$, and the form of the tensor product $\rho \otimes \sigma$ acts on a single object $g\in G$ as

$(\rho \otimes \sigma )(g)=\rho (g)\otimes \sigma (g),$

which is again a representation of $G$. This means that the internal tensor product acts diagonally on the tensor product space.

Now, the ${\mathbb{Z}}_2$ action on $S_k\times S_k$ by conjugation with $\tau$ swaps the two factors, i.e.,

$(\rho ⊠\sigma {)}^{\tau }\cong \sigma ⊠\rho .$

This means that the orbits $\mathcal{O}\subset N$ come in two flavors:

• Off-diagonal elements: if $\rho \ne \sigma$, then the orbit size is 2: $\{\rho \otimes \sigma ,\sigma \otimes \rho \}$.
• Diagonal elements: if $\rho =\sigma$, then the orbit size is 1: $\{\rho \otimes \rho \}$.

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Definitions: Restriction and Induction

Definition 1: Restriction ${\text{Res}}_H^G$

Let $G$ be a group and $H\le G$ a subgroup. If $\rho :G\to \text{GL}(V)$ is a representation of $G$, then the restriction of $\rho$ to $H$ is the same linear action but only considering elements of $H$:

${\text{Res}}_H^G\rho :H\to \text{GL}(V),$ $h\mapsto \rho (h).$

So, in simpler terms, you literally just “forget” about the action on the whole group and only use the subgroup’s elements. Mathematicians are just trying to confuse us to keep their jobs!

Example 1

Let $G=S_3$ and $H={\mathbb{Z}}_2$. We have the standard 2D representation $\rho$ on $S_3$ which is a permutation action on ${\mathbb{R}}^3$ modulo some other stuff. We can write ${\mathbb{Z}}_2=⟨(12)⟩$ and then we can see that restricting $\rho$ to $H$ yields two objects: the identity acting as the identity, and $(12)$ acting by swapping coordinates, which is just a reflection. This is already the usual (reflection) representation of ${\mathbb{Z}}_2$, so ${\text{Res}}_{{\mathbb{Z}}_2}^{S_3}\rho$ is the representation of ${\mathbb{Z}}_2$.

Let’s make this even more concrete. The element $(12)$ acts on ${\mathbb{R}}^3$ with basis $\{e_1,e_2,e_3\}$ by

$(12):\{\begin{array}{l}{e}_1\mapsto e_2,\\ e_2\mapsto e_1,\\ e_3\mapsto e_3.\end{array}$

So as a matrix representation in this basis, we have

$\rho (12)=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right].$

Next, we define the standard 2-dimensional representation. The permutation representation is 3-dimensional, but it contains the trivial 1-dimensional subrepresentation which is spanned by the line $e_1+e_2+e_3$ (i.e., this line doesn’t change under $\rho$). The standard representation is the 2-dimensional quotient $\{(x,y,z)\mid x+y+z=0\}$.

Now, we need to select some basis vectors for this subspace. We need two of them and they need to satisfy the quotient, so we can select

$v_1=\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]\text{and}{v}_2=\left[\begin{array}{c}1\\ 1\\ -2\end{array}\right].$

The action of $(12)$ on this new basis just swaps the first two entries, so we find

$v_1=\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]\stackrel{(12)}{\to }\left[\begin{array}{c}-1\\ 1\\ 0\end{array}\right]=-v_1,$

$v_2=\left[\begin{array}{c}1\\ 1\\ -2\end{array}\right]\stackrel{(12)}{\to }\left[\begin{array}{c}1\\ 1\\ -2\end{array}\right]=v_2,$

so in this basis, we can see that

${\rho }_{\text{res}}(12)=\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]\text{and}{\rho }_{\text{res}}(1)=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],$

with the subscript on $\rho$ indicating that this is the restriction to ${\mathbb{Z}}_2$. Now, restricting to ${\mathbb{Z}}_2$, the subgroup acts by the identity and transposition, which is exactly the reflection representation of ${\mathbb{Z}}_2$.

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Definition 2: Induction ${\text{Ind}}_H^G$

Let $G$ be a group and $H\le G$ a subgroup. Suppose $\sigma :H\to \text{GL}(V)$ is a representation of $H$. The induced representation ${\text{Ind}}_H^G\sigma$ is a representation of $G$ built on a larger vector space and is defined as

${\text{Ind}}_H^G(\sigma )=\{f:G\to V\mid h\in H,g\in G,f(hg)=\sigma (h)f(g)\}$

with the dimension of this representation being $[G:H]\cdot \dim V$.

Example 2

Let’s take the same example as before. Let $\sigma$ be the trivial representation of ${\mathbb{Z}}_2$ where every $h\in H$ acts as 1. A quick calculation gives us

$\dim {\text{Ind}}_{{\mathbb{Z}}_2}^{S_3}\sigma =[S_3:{\mathbb{Z}}_2]=3,$

so we get a 3-dimensional representation of $S_3$. This induced representation is just the permutation representation of $S_3$ acting on the three cosets of ${\mathbb{Z}}_2$ which are $\{{\mathbb{Z}}_2,(13){\mathbb{Z}}_2,(23){\mathbb{Z}}_2\}$, and $S_3$ permutes them. This means that ${\text{Ind}}_{{\mathbb{Z}}_2}^{S_3}1$ recovers the 3-dimensional permutation representation of $S_3$.

Let’s treat this more concretely like we did with the restriction. We can write $\sigma =1_{{\mathbb{Z}}_2}$ to be the trivial representation of ${\mathbb{Z}}_2$. We have three cosets from the calculation $[S_3:{\mathbb{Z}}_2]=3$, so we need to select three coset representatives. Let’s select the representatives to be

$r_1=(1),r_2=(13),\text{and}{r}_3=(23).$

This gives us the left cosets we were looking for. Now, we will use the basis $\{e_1,e_2,e_3\}$ where $e_i$ indicates the coset $r_i{\mathbb{Z}}_2$. For any $g\in S_3$ the induced action is just

$g\cdot e_i=e_j\text{where}g{r}_i{\mathbb{Z}}_2=r_j{\mathbb{Z}}_2,$

so each $g$ acts by a $3\times 3$ permutation matrix. If we write out each of these cosets, we find

$r_1{\mathbb{Z}}_2=\{(1),(12)\},r_2{\mathbb{Z}}_2=\{(13),(123)\},\text{and}{r}_3{\mathbb{Z}}_2=\{(23),(132)\},$

which is clearly everything in $S_3$. Let’s now consider the action of $(12)$ by computing the images of the cosets under left multiplication of $(12)$. This gives us

$(12)\cdot r_1{\mathbb{Z}}_2=(12){\mathbb{Z}}_2\cong {\mathbb{Z}}_2⟹e_1\mapsto e_1,$ $(12)\cdot r_2{\mathbb{Z}}_2=(12)(13){\mathbb{Z}}_2\cong (132){\mathbb{Z}}_2\cong (23){\mathbb{Z}}_2⟹e_2\mapsto e_3,$ $(12)\cdot r_3{\mathbb{Z}}_2=(12)(23){\mathbb{Z}}_2\cong (13){\mathbb{Z}}_2⟹e_3\mapsto e_2$

so we can see that if the action of $(12)$ on these cosets takes the form of

${\rho }_{\text{ind}}(12)=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right].$

Similarly, if we calculate the action of $(23)$ we find

${\rho }_{\text{ind}}(23)=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]$

and that $\rho (123)=\rho (12)\rho (23)$. This feels less satisfying than the previous example, but we can immediately see that on the vector ${\left[\begin{array}{ccc}1& 1& 1\end{array}\right]}^T$ these matrices keep this vector fixed, and that in the basis

$\left\{\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right],\left[\begin{array}{c}1\\ 1\\ -2\end{array}\right]\right\},$

the two generator matrices block-diagonalize into the identity and the $2\times 2$ block that acts as the standard reflection/rotation action of $S_3$, being isomorphic to the representation we calculated in the restriction example.

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Mackey’s Formula

We also have another very important formula which is Mackey’s formula:

Theorem 2 (Mackey's Formula)

Let $G$ be a finite group and $H,K\le G$ subgroups. Let $\sigma$ be a representation of $H$. Then the restriction to $K$ of the induced representation from $H$ to $G$ decomposes as

${\text{Res}}_K^G\left({\text{Ind}}_H^G\sigma \right)\cong {⨁}_{g\in K\backslash G/H}{\text{Ind}}_{K\cap gH{g}^{-1}}^K\left({\text{Res}}_{K\cap gH{g}^{-1}}^{gH{g}^{-1}}{\sigma }^g\right)$

where $K\backslash G/H$ is the double coset of $G$ with $K$ and $H$ defined as

$K\backslash G/H:=\{KgH\mid g\in G\}$

and ${\sigma }^g$ is the representation of $gH{g}^{-1}$ defined as

${\sigma }^g(x)=\sigma (g^{-1}xg)\text{where}x\in gH{g}^{-1}.$

Now, let’s plug in $K=H=N$ and $G=W$. It should be clear that $N\backslash W/N\cong G/N$, $sN{s}^{-1}=N$, and $N\cap sN{s}^{-1}=N$. Selecting $\sigma =\pi$ to be a representation of $N$, we see that everything boils down to

${\text{Res}}_N^W{\text{Ind}}_N^W\pi \cong {⨁}_{s=1,\tau }{\pi }^s\cong \pi \oplus {\pi }^{\tau }.$

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Off-diagonal case: $\rho \ne \sigma$

Select $\pi =\rho \otimes \sigma$ on $V_{\rho }\otimes V_{\sigma }$. Since the action $\tau$ doesn’t stabilize $\pi$, the inertia subgroup is just $I=N$. Let’s now consider ${\text{Ind}}_N^W\pi$. Since we know that $[W:N]=2$, by Mackey’s formula we have

${\text{Res}}_N^W{\text{Ind}}_N^W\pi \cong \pi \oplus {\pi }^{\tau }\cong (\rho ⊠\sigma )\oplus (\sigma ⊠\rho ).$

Next, we need to use Mackey’s irreducibility criterion.

Theorem 3 (Mackey's irreducibility criterion)

Let $G$ be a group and $H\le G$ a subgroup, and $\pi$ an irrep of $H$. Then ${\text{Ind}}_H^G\pi$ is irreducible if and only if $\pi$ is irreducible and for every $g\in G\setminus H$, $\pi$ and ${\text{Res}}_{H\cap gH{g}^{-1}}^H({\pi }^g)$ have no common irreducible constituents.

Applying this criterion to our situation with $G=W$ and $H=N$, we need to check this criterion for any $g\in W\setminus N$. But since $N$ has index 2 in $W$, every $g$ is in the coset $\tau N$, i.e., all of these $g$‘s give the same intersection. This means that we just need to check one representative, so we can select $\tau$.

Computing the intersection subgroup, we just have $N\cap N=N$. The conjugate representation is just swapping the factors like we saw, so ${\pi }^{\tau }=\sigma ⊠\rho$. Since we can see that $\pi$ and ${\pi }^{\tau }$ don’t share any irreducible constituents and therefore ${\text{Ind}}_N^W\pi$ is irreducible as a representation of $W$, which is what we needed to identify.

Calculating the dimension of this irrep, we get

$\dim {\text{Ind}}_N^W\pi =2\cdot d_{\rho }{d}_{\sigma }.$

Remark 3

Moreover, for each unordered pair $\{\rho ,\sigma \}$ with $\rho \ne \sigma$, we have one irrep of $W$ produced from this. Since the partitions of $k$ and the irreps of $S_k$ are in one-to-one correspondence with one another, we can see that there are

$\left(\genfrac{}{}{0ex}{}{p(k)}{2}\right)$

total unordered pairs, where $p(k)$ is the number of partitions of $k$.

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Diagonal case: $\rho =\sigma$

Let $\pi =\rho ⊠\rho$, which is fixed by $\tau$ so $I=W$. This means that we need to extend $\pi$ to $W$. To do this, let’s define an action of $\tau$ on $V_{\rho }\otimes V_{\rho }$ as $T$ such that $T(v\otimes w)=w\otimes v$. This operator clearly commutes with the diagonal action of $N$ and is therefore a well-defined extension of $\pi$ to $W$. This gives us two possible extensions. The first corresponds to just one action of $T$, and the second corresponds to $T^2$, which is clearly just the identity. This means that the eigenvalues of $T$ (and, ultimately, $\tau$) are $\pm 1$ when acting on the diagonal elements, so we can break our space into

$V_{\rho }\otimes V_{\rho }={\text{Sym}}^2{V}_{\rho }\oplus {\bigwedge }^2{V}_{\rho }$

with the symmetric and antisymmetric spaces corresponding to the $+1$ and $-1$ eigenspaces, respectively. Each of these spaces are stable under both $N$ and $\tau$. Moreover, each of these yield inequivalent extensions, both with dimension $d_{\rho }^2$ (since they act on $V_{\rho }\otimes V_{\rho }$). This means that for each $\rho \in {\hat{S}}_k$, we have 2 irreps of $W$ of dimension $d_{\rho }^2$.

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Multiset of Dimensions

Let’s now define $D=\{d_1,\dots ,d_n\}$ to be the degrees of the irreps of $S_k$ with $n=p(k)$, the partition number. This is because the partition numbers of $k$ are in one-to-one correspondence with the irreps of $S_k$, or equivalently, the Young tableaux of size $k$.

For each partition $\lambda ⊢k$, the dimension of the corresponding irrep is given by the hook-length formula:

$f^{\lambda }=\frac{k!}{{\prod }_{b\in \lambda }h(b)},$

where $h(b)$ is the hook-length of the box $b$ in the Young tableau of $\lambda$. This means that, in order to get the degrees, we need to

1. List all the partitions of $k$,
2. For each of these partitions, compute the hook-lengths, and
3. Apply the hook-length formula above to get the dimensions.

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Explicit Examples

Example 3: The $k=2$ case

We have two partitions of 2, which are $(2)$ and $(1,1)$. So we have 2 irreps of $S_2$.

Young tableaux and hook lengths:

• $(2)$: ┌─┬─┐ │2│1│ └─┴─┘ which has $\prod h(b)=2$.

• $(1,1)$: ┌─┐ │2│ ├─┤ │1│ └─┘ which has $\prod h(b)=2$.

Using the hook-length formula, we find

$f^{(2)}=1\text{and}{f}^{(1,1)}=1⟹d_{(2)}=1\text{ and }{d}_{(1,1)}=1.$

Diagonal terms: Each $d_i$ yields two irreps of dimensions $d_i^2$, so we have 2 from $d_{(2)}=1$ and 2 from $d_{(1,1)}=1$, and thus 4 irreps of dimension 1.

Off-diagonal terms: Each pair $i<j$ yields an irrep of dimension $2d_i{d}_j$. The only pair here is $(d_{(2)},d_{(1,1)})=(1,1)$ (with the ordering $(2)<(1,1)$), so we have 1 irrep of dimension 2.

This perfectly matches the GAP calculations.

Example 4: The $k=3$ case

We have three partitions of 3: $(3)$, $(2,1)$, and $(1,1,1)$.

Young tableaux and hook lengths:

• $(3)$: ┌─┬─┬─┐ │3│2│1│ └─┴─┴─┘ which has $\prod h(b)=6$.

• $(2,1)$: ┌─┬─┐ │3│1│ ├─┼─┘ │1│ └─┘ which has $\prod h(b)=3$.

• $(1,1,1)$: ┌─┐ │3│ ├─┤ │2│ ├─┤ │1│ └─┘ which has $\prod h(b)=6$.

Calculating $f^{\lambda }$, we find

$f^{(3)}=f^{(1,1,1)}=1\text{and}{f}^{(2,1)}=2.$

Therefore, $(d_{(3)},d_{(2,1)},d_{(1,1,1)})=(1,2,1)$.

Diagonal terms: 2 irreps of dimension 1, 2 irreps of dimension 1, and 2 irreps of dimension 4 implies that we have 4 irreps of dimension 1 and 2 irreps of dimension 4.

Off-diagonal terms: We have $\{(3),(2,1)\}$, $\{(3),(1,1,1)\}$, and $\{(2,1),(1,1,1)\}$. This gives us

$\{(3),(2,1)\}⟹\text{1 irrep of dimension }2\cdot 1\cdot 2=4,$ $\{(3),(1,1,1)\}⟹\text{1 irrep of dimension }2\cdot 1\cdot 1=2,$ $\{(2,1),(1,1,1)\}⟹\text{1 irrep of dimension }2\cdot 2\cdot 1=4,$

and thus we have 2 irreps of dimension 4 and 1 irrep of dimension 2.

This perfectly matches the GAP calculations.

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Summary

The dimensions produced by this procedure are given by

$\{2d_{\lambda }{d}_{\mu }\mid \lambda ,\mu ⊢k,\lambda <\mu \}\cup \{d_{\lambda }^2,d_{\lambda }^2\mid \lambda ⊢k\}.$

The number of irreps of $W$ is given, exactly, by

$\left(\genfrac{}{}{0ex}{}{p(k)}{2}\right)+2p(k)=\frac{p(k{)}^2+3p(k)}{2}.$

I have verified all of this up to $k=26$ on my MacBook, lol.