Group Actions and Permutation Representations

Abstract

In this chapter we link representation theory with the theory of group actions and permutation groups. Once again, we are only able to provide a brief glimpse of these connections; see [3] for more. In this chapter all groups are assumed to be finite and all actions of groups are taken to be on finite sets.

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In this chapter we link representation theory with the theory of group actions and permutation groups. Once again, we are only able to provide a brief glimpse of these connections; see  [3] for more. In this chapter all groups are assumed to be finite and all actions of groups are taken to be on finite sets.

7.1 Group Actions

Let us begin by recalling the definition of a group action. If X is a set, then S _X will denote the symmetric group on X. We shall tacitly assume | X | ≥ 2, as the case | X |  = 1 is uninteresting.

Definition 7.1.1 (Group action). 

An action of a group G on a set X is a homomorphism σ: G→S _X . We often write σ _g for σ(g). The cardinality of X is called the degree of the action.

Example 7.1.2 (Regular action). 

Let G be a group and define λ: G→S _G by λ _g (x) = gx. Then λ is called the regular action of G on G.

A subset Y ⊆ X is called G-invariant if σ _g (y) ∈ Y for all y ∈ Y , g ∈ G. One can always partition X into a disjoint union of minimal G-invariant subsets called orbits.

Definition 7.1.3 (Orbit). 

Let σ: G→S _X be a group action. The orbit of x ∈ X under G is the set G ⋅x = { σ _g (x)∣g ∈ G}.

Clearly, the orbits are G-invariant. A standard course in group theory proves that distinct orbits are disjoint and the union of all the orbits is X, that is, the orbits form a partition of X (cf.  [11]). Of particular importance is the case when there is just one orbit.

Definition 7.1.4 (Transitive). 

A group action σ: G→S _X is transitive if, for all x, y ∈ X, there exists g ∈ G such that σ _g (x) = y. Equivalently, the action is transitive if there is one orbit of G on X.

Example 7.1.5 (Coset action). 

If G is a group and H a subgroup, then there is an action σ: G→S _G ∕ H given by σ _g (xH) = gxH. This action is transitive.

An even stronger property than transitivity is that of 2-transitivity.

Definition 7.1.6 (2-transitive). 

An action σ: G→S _X of G on X is 2-transitive if given any two pairs of distinct elements x, y ∈ X and x′, y′ ∈ X, there exists g ∈ G such that σ _g (x) = x′ and σ _g (y) = y′.

Example 7.1.7 (Symmetric groups). 

For n ≥ 2, the action of S _n on {1, …, n} is 2-transitive. Indeed, let i≠j and k≠ℓ be pairs of elements of X. Let X = { 1, …, n} ∖ { i, j} and Y = { 1, …, n} ∖ { k, ℓ}. Then $\vert X\vert = n - 2 = \vert Y \vert$, so we can choose a bijection α: X→Y. Define τ ∈ S _n by

$$\tau (m) = \left \{\begin{array}{@{}l@{\quad }l@{}} k \quad &m = i\\ \mathcal{l} \quad &m = j \\ \alpha (m)\quad &\text{ else.} \end{array} \right.$$

Then τ(i) = k and τ(j) = ℓ. This establishes that S _n is 2-transitive.

Let us put this notion into a more general context.

Definition 7.1.8 (Orbital). 

Let σ: G→S _X be a transitive group action. Define σ²: G→S _{X ×X} by

$${\sigma }_{g}^{2}({x}_{ 1},{x}_{2}) = ({\sigma }_{g}({x}_{1}),{\sigma }_{g}({x}_{2})).$$

An orbit of σ² is termed an orbital of σ. The number of orbitals is called the rank of σ.

Let Δ = { (x, x)∣x ∈ X}. As σ _g ²(x, x) = (σ _g (x), σ _g (x)), it follows from the transitivity of G on X that Δ is an orbital. It is called the diagonal or trivial orbital.

Remark 7.1.9.

Orbitals are closely related to graph theory. If G acts transitively on X, then any non-trivial orbital can be viewed as the edge set of a graph with vertex set X (by symmetrizing). The group G acts on the resulting graph as a vertex-transitive group of automorphisms.

Proposition 7.1.10.

Let σ: G→S _X be a group action (with X ≥ 2). Then σ is 2-transitive if and only if σ is transitive and rank (σ) = 2.

Proof.

First we observe that transitivity is necessary for 2-transitivity since if G is 2-transitive on X and x, y ∈ X, then we may choose x′≠x and y′≠y. By 2-transitivity there exists g ∈ G with σ _g (x) = y and σ _g (x′) = y′. This shows that σ is transitive. Next observe that

$$(X \times X) \setminus \Delta =\{ (x,y)\mid x\neq y\}$$

and so the complement of Δ is an orbital if and only if, for any two pairs x≠y and x′≠y′ of distinct elements, there exists g ∈ G with σ _g (x) = x′ and σ _g (y) = y′, that is, σ is 2-transitive.

Consequently the rank of S _n is 2. Let σ: G→S _X be a group action. Then, for g ∈ G, we define

$$\mathrm{Fix}(g) =\{ x \in X\mid {\sigma }_{g}(x) = x\}$$

to be the set of fixed points of g. Let Fix²(g) be the set of fixed points of g on X ×X. The notation is unambiguous because of the following proposition.

Proposition 7.1.11.

Let σ: G→S _X be a group action. Then the equality

$$\mathrm{{Fix}}^{2}(g) =\mathrm{ Fix}(g) \times \mathrm{ Fix}(g)$$

holds. Hence |Fix ² (g)| = |Fix (g)| ².

Proof.

Let (x, y) ∈ X ×X. Then σ _g ²(x, y) = (σ _g (x), σ _g (y)) and so (x, y) = σ _g ²(x, y) if and only if σ _g (x) = x and σ _g (y) = y. We conclude Fix²(g) = Fix(g) ×Fix(g).

7.2 Permutation Representations

Given a group action σ: G→S _n , we may compose it with the standard representation $\alpha : {S}_{n}\rightarrow {GL}_{n}(\mathbb{C})$ to obtain a representation of G. Let us formalize this.

Definition 7.2.1 (Permutation representation). 

Let σ: G→S _X be a group action. Define a representation $\widetilde{\sigma }: G\rightarrow GL(\mathbb{C}X)$ by setting

$${\widetilde{\sigma }}_{g}\left ({\sum \nolimits }_{x\in X}{c}_{x}x\right ) ={ \sum \nolimits }_{x\in X}{c}_{x}{\sigma }_{g}(x) ={ \sum \nolimits }_{y\in X}{c}_{{\sigma }_{ g-1}(y)}y.$$

One calls $\widetilde{\sigma }$ the permutation representation associated to σ.

Remark 7.2.2.

Notice that ${\widetilde{\sigma }}_{g}$ is the linear extension of the map defined on the basis X of $\mathbb{C}X$ by sending x to σ _g (x). Also, observe that the degree of the representation $\widetilde{\sigma }$ is the same as the degree of the group action σ.

Example 7.2.3 (Regular representation). 

Let λ: G→S _G be the regular action. Then one has $\widetilde{\lambda } = L$, the regular representation.

The following proposition is proved exactly as in the case of the regular representation (cf. Proposition 4.4.2), so we omit the proof.

Proposition 7.2.4.

Let σ: G→S _X be a group action. Then the permutation representation $\widetilde{\sigma }: G\rightarrow GL(\mathbb{C}X)$ is a unitary representation of G.

Next we compute the character of $\widetilde{\sigma }$.

Proposition 7.2.5.

Let σ: G→S _X be a group action. Then

$${\chi }_{\widetilde{\sigma }}(g) = \vert \mathrm{Fix}(g)\vert.$$

Proof.

Let X = { x ₁, …, x _n } and let $[{\widetilde{\sigma }}_{g}]$ be the matrix of $\widetilde{\sigma }$ with respect to this basis. Then ${\widetilde{\sigma }}_{g}({x}_{j}) = {\sigma }_{g}({x}_{j})$, so

$${[{\widetilde{\sigma }}_{g}]}_{ij} = \left \{\begin{array}{@{}l@{\quad }l@{}} 1\quad &{x}_{i} = {\sigma }_{g}({x}_{j})\\ 0\quad &\text{ else.} \end{array} \right.$$

In particular,

$$\begin{array}{rcl}{ [{\widetilde{\sigma }}_{g}]}_{ii}& = \left \{\begin{array}{@{}l@{\quad }l@{}} 1\quad &{x}_{i} = {\sigma }_{g}({x}_{i})\\ 0\quad &\text{ else} \end{array} \right.& \\ & = \left \{\begin{array}{@{}l@{\quad }l@{}} 1\quad &{x}_{i} \in \mathrm{ Fix}(g)\\ 0\quad &\text{ else} \end{array} \right.& \\ \end{array}$$

and so ${\chi }_{\widetilde{\sigma }}(g) = \mathrm{Tr}([{\widetilde{\sigma }}_{g}]) = \vert \mathrm{Fix}(g)\vert$.

Like the regular representation, permutation representations are never irreducible (if | X |  > 1). To understand better how it decomposes, we first consider the trivial component.

Definition 7.2.6 (Fixed subspace). 

Let ϕ: G→GL(V ) be a representation. Then

$${V }^{G} =\{ v \in V \mid {\phi }_{ g}(v) = v\ \text{ for all}\ g \in G\}$$

is the fixed subspace of G.

One easily verifies that V ^G is a G-invariant subspace and the subrepresentation $\phi {\vert }_{{V }^{G}}$ is equivalent to dimV ^G copies of the trivial representation. Let us prove that V ^G is the direct sum of all the copies of the trivial representation in ϕ.

Proposition 7.2.7.

Let ϕ: G→GL(V ) be a representation and let χ ₁ be the trivial character of G. Then ⟨χ _ϕ ,χ ₁ ⟩ = dim V ^G.

Proof.

Write V = m ₁ V ₁ ⊕ ⋯ ⊕ m _s V _s where V ₁, …, V _s are irreducible G-invariant subspaces whose associated subrepresentations range over the distinct equivalence classes of irreducible representations of G (we allow m _i  = 0). Without loss of generality, we may assume that V ₁ is equivalent to the trivial representation. Let ϕ⁽ⁱ⁾ be the restriction of ϕ to V _i . Now if v ∈ V , then $v = {v}_{1} + \cdots + {v}_{s}$ with the v _i  ∈ m _i V _i and

$${\phi }_{g}v = {({m}_{1}{\phi }^{(1)})}_{ g}{v}_{1} + \cdots + {({m}_{s}{\phi }^{(s)})}_{ g}{v}_{s} = {v}_{1} + {({m}_{2}{\phi }^{(2)})}_{ g}{v}_{2} + \cdots + {({m}_{s}{\phi }^{(s)})}_{ g}{v}_{s}$$

and so g ∈ V ^G if and only if v _i  ∈ m _i V _i ^G for all 2 ≤ i ≤ s. In other words,

$${V }^{G} = {m}_{ 1}{V }_{1} \oplus {m}_{2}{V }_{2}^{G} \oplus \cdots \oplus {m}_{ s}{V }_{s}^{G}.$$

Let i ≥ 2. Since V _i is irreducible and not equivalent to the trivial representation and V _i ^G is G-invariant, it follows V _i ^G = 0. Thus V ^G = m ₁ V ₁ and so the multiplicity of the trivial representation in ϕ is dimV ^G, as required.

Now we compute $\mathbb{C}{X}^{G}$ when we have a permutation representation.

Proposition 7.2.8.

Let σ: G→S _X be a group action. Let ${\mathcal{O}}_{1},\ldots ,{\mathcal{O}}_{m}$ be the orbits of G on X and define ${v}_{i} ={ \sum \nolimits }_{x\in {\mathcal{O}}_{i}}x$ . Then v ₁ ,…,v _m is a basis for $\mathbb{C}{X}^{G}$ and hence $\dim \mathbb{C}{X}^{G}$ is the number of orbits of G on X.

Proof.

First observe that

$${\widetilde{\sigma }}_{g}{v}_{i} ={ \sum \nolimits }_{x\in {\mathcal{O}}_{i}}{\sigma }_{g}(x) ={ \sum \nolimits }_{y\in {\mathcal{O}}_{i}}y = {v}_{i}$$

as is seen by setting y = σ _g (x) and using that σ _g permutes the orbit ${\mathcal{O}}_{i}$. Thus ${v}_{1},\ldots ,{v}_{m} \in \mathbb{C}{X}^{G}$. Since the orbits are disjoint, we have

$$\langle {v}_{i},{v}_{j}\rangle = \left \{\begin{array}{@{}l@{\quad }l@{}} \vert {\mathcal{O}}_{i}\vert \quad &i = j \\ 0 \quad &i\neq j \end{array} \right.$$

and so {v ₁, …, v _s } is an orthogonal set of non-zero vectors and hence linearly independent. It remain to prove that this set spans $\mathbb{C}{X}^{G}$.

Suppose $v={\sum \nolimits }_{x\in X}{c}_{x}x \in \mathbb{C}{X}^{G}$. We show that if z ∈ G ⋅y, then c _y  = c _z . Indeed, let z = σ _g (y). Then we have

$${\sum \nolimits }_{x\in X}{c}_{x}x = v ={ \widetilde{\sigma }}_{g}v ={ \sum \nolimits }_{x\in X}{c}_{x}{\sigma }_{g}(x)$$

(7.1)

and so the coefficient of z in the left-hand side of (7.1) is c _z , whereas the coefficient of z in the right-hand side is c _y since z = σ _g (y). Thus c _z  = c _y . It follows that there are complex numbers c ₁, …, c _m such that c _x  = c _i for all $x \in {\mathcal{O}}_{i}$. Thus

$$v ={ \sum \nolimits }_{x\in X}{c}_{x}x ={ \sum \nolimits }_{i=1}^{m}{ \sum \nolimits }_{x\in {\mathcal{O}}_{i}}{c}_{x}x ={ \sum \nolimits }_{i=1}^{m}{c}_{ i}{ \sum \nolimits }_{x\in {\mathcal{O}}_{i}}x ={ \sum \nolimits }_{i=1}^{m}{c}_{ i}{v}_{i}$$

and hence v ₁, …, v _m span $\mathbb{C}{X}^{G}$, completing the proof.

Since G always has at least one orbit on X, the above result shows that the trivial representation appears as a constituent in $\widetilde{\sigma }$ and so if | X |  > 1, then $\widetilde{\sigma }$ is not irreducible. As a corollary to the above proposition we prove a useful result, commonly known as Burnside’s lemma, although it seems to have been known earlier to Cauchy and Frobenius. It has many applications in combinatorics to counting problems. The lemma says that the number of orbits of G on X is the average number of fixed points.

Corollary 7.2.9 (Burnside’s lemma). 

Let σ: G→S _X be a group action and let m be the number of orbits of G on X. Then

$$m = \frac{1} {\vert G\vert }{\sum \nolimits }_{g\in G}\vert \mathrm{Fix}(g)\vert.$$

Proof.

Let χ₁ be the trivial character of G. By Propositions 7.2.5, 7.2.7 and 7.2.8 we have

$$m =\langle {\chi }_{\widetilde{\sigma }},{\chi }_{1}\rangle = \frac{1} {\vert G\vert }\sum \nolimits g\in G{\chi }_{\widetilde{\sigma }}(g)\overline{{\chi }_{1}(g)} = \frac{1} {\vert G\vert }\sum \nolimits g\in G\vert \mathrm{Fix}(g)\vert $$

as required.

As a corollary, we obtain two formulas for the rank of σ.

Corollary 7.2.10.

Let σ: G→S _X be a transitive group action. Then the equalities

$$\mathrm{rank}(\sigma ) = \frac{1} {\vert G\vert }\sum \nolimits g\in G\vert \mathrm{Fix}(g){\vert }^{2} =\langle {\chi }_{\widetilde{ \sigma }},{\chi }_{\widetilde{\sigma }}\rangle$$

hold.

Proof.

Since rank(σ) is the number of orbits of σ² on X ×X and the number of fixed points of g on X ×X is | Fix(g) | ² (Proposition 7.1.11), the first equality is a consequence of Burnside’s lemma. For the second, we compute

$$\langle {\chi }_{\widetilde{\sigma }},{\chi }_{\widetilde{\sigma }}\rangle = \frac{1} {\vert G\vert }\sum \nolimits g\in G\vert \mathrm{Fix}(g)\vert \overline{\vert \mathrm{Fix}(g)\vert } = \frac{1} {\vert G\vert }\sum \nolimits g\in G\vert \mathrm{Fix}(g){\vert }^{2}$$

completing the proof.

Assume now that σ: G→S _X is a transitive action. Let v ₀ =  ∑ _x ∈ X x. Then $\mathbb{C}{X}^{G} = \mathbb{C}{v}_{0}$ by Proposition 7.2.8. Since $\widetilde{\sigma }$ is a unitary representation, ${V }_{0} = \mathbb{C}{v}_{0}^{\perp }$ is a G-invariant subspace (cf. the proof of Proposition 3.2.3). Usually, $\mathbb{C}{v}_{0}$ is called the trace of σ and V ₀ the augmentation of σ. Let $\widetilde{\sigma }'$ be the restriction of $\widetilde{\sigma }$ to V ₀; we call it the augmentation representation associated to σ. As $\mathbb{C}X = {V }_{0} \oplus \mathbb{C}{v}_{0}$, it follows that ${\chi }_{\widetilde{\sigma }} = {\chi }_{\widetilde{\sigma }'} + {\chi }_{1}$ where χ₁ is the trivial character. We now characterize when the augmentation representation $\widetilde{\sigma }'$ is irreducible.

Theorem 7.2.11.

Let σ: G→S _X be a transitive group action. Then the augmentation representation $\widetilde{\sigma }'$ is irreducible if and only if G is 2-transitive on X.

Proof.

This is a simple calculation using Corollary 7.2.10 and the fact that G is 2-transitive on X if and only if rank(σ) = 2 (Proposition 7.1.10). Indeed, if χ₁ is the trivial character of G, then

$$\begin{array}{rcl} \langle {\chi }_{\widetilde{\sigma }'},{\chi }_{\widetilde{\sigma }'}\rangle & =& \langle {\chi }_{\widetilde{\sigma }} - {\chi }_{1},{\chi }_{\widetilde{\sigma }} - {\chi }_{1}\rangle \\ & =& \langle {\chi }_{\widetilde{\sigma }},{\chi }_{\widetilde{\sigma }}\rangle -\langle {\chi }_{\widetilde{\sigma }},{\chi }_{1}\rangle -\langle {\chi }_{1},{\chi }_{\widetilde{\sigma }}\rangle +\langle {\chi }_{1},{\chi }_{1}\rangle.\end{array}$$

(7.2)

Now by Proposition 7.2.8 $\langle {\chi }_{\widetilde{\sigma }},{\chi }_{1}\rangle \,=\,1$ because G is transitive. Therefore, $\langle {\chi }_{1},{\chi }_{\widetilde{\sigma }}\rangle \,=\,1$. Also, ⟨χ₁, χ₁⟩ = 1. Thus (7.2) becomes, in light of Corollary 7.2.10,

$$\langle {\chi }_{\widetilde{\sigma }'},{\chi }_{\widetilde{\sigma }'}\rangle =\mathrm{rank}(\sigma ) - 1$$

and so ${\chi }_{\widetilde{\sigma }'}$ is an irreducible character if and only if rank(σ) = 2, that is, if and only if G is 2-transitive on X.

Remark 7.2.12.

The decomposition of the standard representation of S ₃ in Example 4.3.17 corresponds precisely to the decomposition into the direct sum of the augmentation and the trace.

With Theorem 7.2.11 in hand, we may now compute the character table of S ₄.

Example 7.2.13 (Character table of S ₄). 

First of all S ₄ has five conjugacy classes, represented by Id, (1 2), (1 2 3), (1 2 3 4), (1 2)(3 4). Let χ₁ be the trivial character and χ₂ the character of the sign homomorphism. As S ₄ acts 2-transitively on {1, …, 4}, Theorem 7.1.10 implies that the augmentation representation is irreducible. Let χ₄ be the character of this representation; it is the character of the standard representation minus the trivial character so ${\chi }_{4}(g) = \vert \mathrm{Fix}(g)\vert - 1$. Let χ₅ = χ₂ ⋅χ₄. That is, if τ is the representation associated to χ₄, then we can define a new representation ${\tau }^{{\chi }_{2}}: {S}_{4}\rightarrow G{L}_{3}(\mathbb{C})$ by ${\tau }_{g}^{{\chi }_{2}} = {\chi }_{2}(g){\tau }_{g}$. It is easily verified that ${\chi }_{{\tau }^{{\chi }_{2}}}(g) = {\chi }_{2}(g){\chi }_{4}(g)$ and ${\tau }^{{\chi }_{2}}$ is irreducible. This gives us four of the five irreducible representations. How do we get the fifth? Let d be the degree of the missing representation. Then

$$24 = \vert {S}_{4}\vert = {1}^{2} + {1}^{2} + {d}^{2} + {3}^{2} + {3}^{2} = 20 + {d}^{2}$$

and so d = 2. Let χ₃ be the character of the missing irreducible representation and let L be the regular representation of S ₄. Then

$${\chi }_{L} = {\chi }_{1} + {\chi }_{2} + 2{\chi }_{3} + 3{\chi }_{4} + 3{\chi }_{5}$$

so for Id≠g ∈ S ₄, we have

$${\chi }_{3}(g) = \frac{1} {2}\left (-{\chi }_{1}(g) - {\chi }_{2}(g) - 3{\chi }_{4}(g) - 3{\chi }_{5}(g)\right ).$$

In this way we are able to produce the character table of S ₄ in Table 7.1.

Table 7.1 Character table of S ₄

The reader should try to produce a representation with character χ₃. As a hint, observe that K = { Id, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} is a normal subgroup of S ₄ and that S ₄ ∕ K≅S ₃. Construct an irreducible representation by composing the surjective map S ₄→S ₃ with the degree 2 irreducible representation of S ₃ coming from the augmentation representation for S ₃.

7.3 The Centralizer Algebra and Gelfand Pairs

Let σ: G→S _X be a transitive group action. Our goal in this section is to study the ring ${\mathrm{Hom}}_{G}(\widetilde{\sigma },\widetilde{\sigma })$. We only scratch the surface of this topic in this section. Much more information, as well as applications to probability and statistics, can be found in  [3].

Let us assume that X = { x ₁, …, x _n }. Define a matrix representation $\phi : G\rightarrow {GL}_{n}(\mathbb{C})$ by ${\phi }_{g} = {[{\widetilde{\sigma }}_{g}]}_{X}$. Then $\phi \sim \widetilde{\sigma }$ and so ${\mathrm{Hom}}_{G}(\widetilde{\sigma },\widetilde{\sigma })\cong{\mathrm{Hom}}_{G}({\phi }_{g},{\phi }_{g})$. Next observe that

$$\begin{array}{rcl}{ \mathrm{Hom}}_{G}(\phi ,\phi )& =\{ A \in {M}_{n}(\mathbb{C})\mid A{\phi }_{g} = {\phi }_{g}A,\forall g \in G\} & \\ & =\{ A \in {M}_{n}(\mathbb{C})\mid {\phi }_{g}A{\phi }_{g}-1 = A,\forall g \in G\}.& \\ \end{array}$$

From now on we denote Hom _G (ϕ, ϕ) by C(σ) and call it the centralizer algebra of σ.

Proposition 7.3.1.

C(σ) is a unital subring of ${M}_{n}(\mathbb{C})$.

Proof.

Trivially, ϕ _g I _n ϕ _g  − 1 = I _n for all g ∈ G. If A, B ∈ C(σ), then

$${\phi }_{g}(A + B){\phi }_{g}-1 = {\phi }_{ g}A{\phi }_{g}-1 + {\phi }_{ g}B{\phi }_{g}-1 = A + B$$

for all g ∈ G, and similarly ${\phi }_{g}(AB){\phi }_{g}-1 = {\phi }_{g}A{\phi }_{g}-1{\phi }_{g}B{\phi }_{g}-1 = AB$. Thus C(σ) is indeed a unital subring of ${M}_{n}(\mathbb{C})$.

We aim to show that dimC(σ) = rank(σ) and exhibit an explicit basis. Let $V = {M}_{n}(\mathbb{C})$ and define a representation τ : G→GL(V ) by τ _g (A) = ϕ _g Aϕ _g  − 1. The reader should perform the routine verification that τ is indeed a representation. Notice that

$${V }^{G} =\{ A \in {M}_{ n}(\mathbb{C})\mid {\phi }_{g}A{\phi }_{g}-1 = A,\forall g \in G\} = C(\sigma ).$$

Let σ²: G→S _{X ×X} be as per Definition 7.1.8. We exhibit an explicit equivalence between τ and $\widetilde{{\sigma }^{2}}$. We can then use Proposition 7.2.8 to obtain a basis for C(σ).

Proposition 7.3.2.

Define a mapping $T : {M}_{n}(\mathbb{C})\rightarrow \mathbb{C}(X \times X)$ by

$$T({a}_{ij}) ={ \sum \nolimits }_{i,j=1}^{n}{a}_{ ij}({x}_{i},{x}_{j})$$

where we have retained the above notation. Then T is an equivalence between τ and $\widetilde{{\sigma }^{2}}$.

Proof.

The map T is evidently bijective and linear with inverse

$${\sum \nolimits }_{i,j=1}^{n}{a}_{ ij}({x}_{i},{x}_{j})\longmapsto ({a}_{ij}).$$

Let us check that it is an equivalence. Let g ∈ G and let $A = ({a}_{ij}) \in {M}_{n}(\mathbb{C})$. Put B = τ _g A; say B = (b _ij ). Define an action γ: G→S _n by ${\sigma }_{g}({x}_{i}) = {x}_{{\gamma }_{g}(i)}$ for g ∈ G. Then

$${b}_{ij} ={ \sum \nolimits }_{k=1,\mathcal{l}=1}^{n}\phi {(g)}_{ ik}{a}_{k\mathcal{l}}\phi {(g-1)}_{ \mathcal{l}j} = {a}_{{\gamma }_{g}-1(i),{\gamma }_{g}-1(j)}$$

because

$$\phi {(g)}_{ik} = \left \{\begin{array}{@{}l@{\quad }l@{}} 1\quad &{x}_{i} = {\sigma }_{g}({x}_{k})\\ 0\quad &\text{ else} \end{array} \right.\quad \text{ and}\quad \phi {(g-1)}_{ \mathcal{l}j} = \left \{\begin{array}{@{}l@{\quad }l@{}} 1\quad &{x}_{\mathcal{l}} = {\sigma }_{g}-1({x}_{j}) \\ 0\quad &\text{ else.} \end{array} \right.$$

Therefore, we have

$$\begin{array}{rcl} T{\tau }_{g}A& ={ \sum \nolimits }_{i,j=1}^{n}{b}_{ij}({x}_{i},{x}_{j}) ={ \sum \nolimits }_{i,j=1}^{n}{a}_{{\gamma }_{g}-1(i),{\gamma }_{g}-1(j)}({x}_{i},{x}_{j}) & \\ & ={ \sum \nolimits }_{i,j=1}^{n}{a}_{ij}({\sigma }_{g}({x}_{i}),{\sigma }_{g}({x}_{j})) ={ \sum \nolimits }_{i,j=1}^{n}{a}_{ij}{\sigma }_{g}^{2}({x}_{i},{x}_{j}) =\widetilde{ {\sigma {}^{2}}}_{g}TA& \\ \end{array}$$

and so T is an equivalence, as required.

We can now provide a basis for C(σ). If Ω is an orbital of σ, define a matrix $A(\Omega ) \in {M}_{n}(\mathbb{C})$ by

$$A{(\Omega )}_{ij} = \left \{\begin{array}{@{}l@{\quad }l@{}} 1\quad &({x}_{i},{x}_{j}) \in \Omega \\ 0\quad &\text{ else.} \end{array} \right.$$

Corollary 7.3.3.

Let σ: G→S _X be a transitive group action. We retain the above notation. Let Ω ₁ ,…,Ω _r be the orbitals of σ where r = rank (σ). Then the set {A(Ω ₁ ),…,A(Ω _r )} is a basis for C(σ) and consequently dim C(σ) = rank (σ).

Proof.

Proposition 7.2.8 implies that a basis for $\mathbb{C}{(X \times X)}^{G}$ is given by the elements v ₁, …, v _r where

$${v}_{k} ={ \sum \nolimits }_{({x}_{i},{x}_{j})\in {\Omega }_{k}}({x}_{i},{x}_{j}).$$

Clearly, A(Ω _k ) = T − 1v _k . As T restricts to an equivalence of $C(\sigma ) = {M}_{n}{(\mathbb{C})}^{G}$ and $\mathbb{C}{(X \times X)}^{G}$ (cf. Exercise 7.7), it follows that {A(Ω ₁), …, A(Ω _r )} is a basis for C(σ), as required.

An important notion in applications is that of a Gelfand pair; the reader is referred to  [3, 4.7] and  [7, 3.F] where a Fourier transform is defined in this context and applied to probability theory.

Definition 7.3.4 (Gelfand pair). 

Let G be a group and H a subgroup. Let σ: G→S _G ∕ H be the coset action. Then (G, H) is said to be a Gelfand pair if the centralizer algebra C(σ) is commutative.

Example 7.3.5.

Let G be a group and let H = { 1}. The coset action of G on G ∕ H is none other than the regular action λ: G→S _G and so $\widetilde{\lambda }$ is none other than the regular representation L. We claim that C(λ)≅L(G). For this argument, we identify the centralizer algebra with the ring Hom _G (L, L).

Let T ∈ C(λ) and define ${f}_{T}: G\rightarrow \mathbb{C}$ by

$$T1 ={ \sum \nolimits }_{x\in G}{f}_{T}(x-1)x.$$

We claim that the mapping T↦f _T is an isomorphism ψ: C(λ)→L(G). First note that, for g ∈ G, one has

$$Tg = T{L}_{g}1 = {L}_{g}T1 = {L}_{g}{ \sum \nolimits }_{x\in G}{f}_{T}(x-1)x.$$

Thus T is determined by f _T and hence ψ is injective. It is also surjective because if $f : G\rightarrow \mathbb{C}$ is any function, then we can define $T \in \mathrm{End}(\mathbb{C}G)$ on the basis by

$$Tg = {L}_{g}{ \sum \nolimits }_{x\in G}f(x-1)x.$$

First note that T belongs to the centralizer algebra because if g, y ∈ G, then

$$T{L}_{y}g = Tyg = {L}_{yg}{ \sum \nolimits }_{x\in G}f(x-1)x = {L}_{ y}Tg.$$

Also, we have

$$T1 ={ \sum \nolimits }_{x\in G}f(x-1)x$$

and so f _T  = f. Thus ψ is surjective. Finally, we compute, for T ₁, T ₂ ∈ C(λ),

$$\begin{array}{rcl}{ T}_{1}{T}_{2}1& = {T}_{1}{ \sum \nolimits }_{x\in G}{f}_{{T}_{2}}(x-1)x ={ \sum \nolimits }_{x\in G}{f}_{{T}_{2}}(x-1){T}_{1}{L}_{x}1 & \\ & ={ \sum \nolimits }_{x\in G}{f}_{{T}_{2}}(x-1){L}_{x}{ \sum \nolimits }_{y\in G}{f}_{{T}_{1}}(y-1)y ={ \sum \nolimits }_{x,y\in G}{f}_{{T}_{1}}(y-1){f}_{{T}_{2}}(x-1)xy.& \\ \end{array}$$

Setting g = xy, u = x − 1 (and hence y − 1 = g − 1u − 1) yields

$${T}_{1}{T}_{2}1 ={ \sum \nolimits }_{g\in G}{ \sum \nolimits }_{u\in G}{f}_{{T}_{1}}(g-1u-1){f}_{{ T}_{2}}(u)g ={ \sum \nolimits }_{g\in G}{f}_{{T}_{1}} {_\ast} {f}_{{T}_{2}}(g-1)g.$$

Thus ${f}_{{T}_{1}{T}_{2}} = {f}_{{T}_{1}} {_\ast} {f}_{{T}_{2}}$ and so ψ is a ring homomorphism. We conclude that ψ is an isomorphism.

Consequently, (G, {1}) is a Gelfand pair if and only if G is abelian because L(G) is commutative if and only if L(G) = Z(L(G)). But dimZ(L(G)) =  | Cl(G) | and dimL(G) =  | G | , and so Z(L(G)) = L(G) if and only if G is abelian.

It is known that (G, H) is a Gelfand pair if and only if $\widetilde{\sigma }$ is multiplicity-free, meaning that each irreducible constituent of $\widetilde{\sigma }$ has multiplicity one  [3, Theorem 4.4.2]. We content ourselves here with the special case of so-called symmetric Gelfand pairs.

If σ: G→S _X is a transitive group action, then to each orbital Ω of σ, we can associate its transpose

$${\Omega }^{T} =\{ ({x}_{ 1},{x}_{2}) \in X \times X\mid ({x}_{2},{x}_{1}) \in \Omega \}.$$

It is easy to see that Ω ^T is indeed an orbital. Let us say that Ω is symmetric if Ω = Ω ^T. For instance, the diagonal orbital Δ is symmetric. Notice that A(Ω ^T) = A(Ω)^T and hence Ω is symmetric if and only if the matrix A(Ω) is symmetric (and hence self-adjoint, as it has real entries).

Definition 7.3.6 (Symmetric Gelfand pair). 

Let G be a group and H a subgroup with corresponding group action σ: G→S _G ∕ H . Then (G, H) is called a symmetric Gelfand pair if each orbital of σ is symmetric.

Of course, we must show that a symmetric Gelfand pair is indeed a Gelfand pair! First we provide some examples.

Example 7.3.7.

Let H ≤ G and suppose that the action of G on G ∕ H is 2-transitive. Then the orbitals are Δ and $(G/H \times G/H) \setminus \Delta$. Clearly, each of these is symmetric. Thus (G, H) is a symmetric Gelfand pair.

Example 7.3.8.

Let n ≥ 2 and let [n]² be the set of all two-element subsets of {1, …, n}. Then S _n acts on [n]² as follows. Define $\tau : {S}_{n}\rightarrow {S}_{{[n]}^{2}}$ by τ_σ({i, j}) = { σ(i), σ(j)}. This action is clearly transitive since S _n is 2-transitive on {1, …, n}. Let H be the stabilizer in S _n of {n − 1, n}. Notice that H is the internal direct product of S _n − 2 and S _{{n − 1, n}} and so H≅S _n − 2 ×S ₂. The action of S _n on [n]² can be identified with the action of S _n on S _n  ∕ H.

If Ω is a non-trivial orbital, then a typical element of Ω is of the form ({i, j}, {k, ℓ}) where these two subsets are different. There are essentially two cases. If i, j, k, and ℓ are all distinct, then (i k)(j ℓ) takes the above element to ({k, ℓ}, {i, j}) and so Ω is symmetric. Otherwise, the two subsets have an element in common, say i = k. Then (j ℓ) takes ({i, j}, {i, ℓ}) to ({i, ℓ}, {i, j}). Thus Ω is symmetric in this case, as well. We conclude (S _n , H) is a symmetric Gelfand pair.

The proof that a symmetric Gelfand pair is in fact a Gelfand pair relies on the following simple observation on rings of symmetric matrices.

Lemma 7.3.9.

Let R be a subring of ${M}_{n}(\mathbb{C})$ consisting of symmetric matrices. Then R is commutative.

Proof.

If A, B ∈ R, then $AB = {(AB)}^{T} = {B}^{T}{A}^{T} = BA$ since A, B, and AB are assumed symmetric.

And now we turn to the proof that symmetric Gelfand pairs are Gelfand.

Theorem 7.3.10.

Let (G,H) be a symmetric Gelfand pair. Then (G,H) is a Gelfand pair.

Proof.

As usual, let σ: G→S _G ∕ H be the action map. Denote by Ω ₁, …, Ω _r the orbitals of σ. Then because each Ω _i is symmetric, it follows that each matrix A(Ω _i ) is symmetric for i = 1, …r. Since the symmetric matrices form a vector subspace of ${M}_{n}(\mathbb{C})$ and {A(Ω ₁), …, A(Ω _r )} is a basis for C(σ) by Corollary 7.3.3, it follows that C(σ) consists of symmetric matrices. Thus C(σ) is commutative by Lemma 7.3.9 and so (G, H) is a Gelfand pair.

7.4 Exercises

Exercise 7.1.

Show that if σ: G→S _X is a group action, then the orbits of G on X form a partition X.

Exercise 7.2.

Let σ: G→S _X be a transitive group action with | X | ≥ 2. If x ∈ X, let

$${G}_{x} =\{ g \in G\mid {\sigma }_{g}(x) = x\}.$$

(7.3)

G _x is a subgroup of G called the stabilizer of x. Prove that the following are equivalent:

• 1. G _x is transitive on X ∖ { x} for somex ∈ X;

• 2. G _x is transitive on X ∖ { x} for allx ∈ X;

• 3. G acts 2-transitively on X.

Exercise 7.3.

Compute the character table of A ₄. (Hints:

• 1. Let K = { Id, (12)(34), (13)(24), (14)(23)}. Then K is a normal subgroup of A ₄ and ${A}_{4}/K\cong\mathbb{Z}/3\mathbb{Z}$. Use this to construct 3 degree one representations of A ₄.

• 2. Show that A ₄ acts 2-transitively on {1, 2, 3, 4}.

• 3. Conclude that A ₄ has four conjugacy classes and find them.

• 4. Produce the character table.)

Exercise 7.4.

Two group actions σ: G→S _X and τ : G→S _Y are isomorphic if there is a bijection ψ: X→Y such that ψσ _g  = τ _g ψ for all g ∈ G.

• 1. Show that if τ : G→S _X is a transitive group action, x ∈ X and G _x is the stabilizer of x (cf. (7.3)), then τ is isomorphic to the coset action $\sigma : G\rightarrow {S}_{G/{G}_{x}}$.

• 2. Show that if σ and τ are isomorphic group actions, then the corresponding permutation representations are equivalent.

Exercise 7.5.

Let p be a prime. Let G be the group of all permutations $\mathbb{Z}/p\mathbb{Z}\rightarrow \mathbb{Z}/p\mathbb{Z}$ of the form x↦ax + b with $a \in \mathbb{Z}/p{\mathbb{Z}}^{{_\ast}}$ and $b \in \mathbb{Z}/p\mathbb{Z}$. Prove that the action of G on $\mathbb{Z}/p\mathbb{Z}$ is 2-transitive.

Exercise 7.6.

Let G be a finite group.

• 1. Suppose that G acts transitively on a finite set X with | X | ≥ 2. Show that there is an element g ∈ G with no fixed points on X. (Hint: Assume that the statement is false. Use that the identity e has | X | fixed points to contradict Burnside’s lemma.)

• 2. Let H be a proper subgroup of G. Prove that

  $$G\neq {\bigcup \nolimits }_{x\in G}xH{x}^{-1}.$$

  (Hint: Use the previous part.)

Exercise 7.7.

Let ϕ: G→GL(V ) and ρ: G→GL(W) be representations and suppose that T : V →W is an equivalence. Show that T(V ^G) = W ^G and the restriction T : V ^G→W ^G is an equivalence.

Exercise 7.8.

Show that if Ω is an orbital of a transitive group action σ: G→S _X , then the transpose Ω ^T is an orbital of σ.

Exercise 7.9.

Suppose that G is a finite group of order n with s conjugacy classes. Suppose that one chooses a pair (g, h) ∈ G ×G uniformly at random. Prove that the probability g and h commute is s ∕ n. (Hint: Apply Burnside’s lemma to the action of G on itself by conjugation.)

Exercise 7.10.

Give a direct combinatorial proof of Burnside’s lemma, avoiding character theory.

Exercise 7.11.

Let G be a group and define Λ: G→GL(L(G)) by putting Λ _g (f)(h) = f(g − 1h).

• 1. Verify that Λ is a representation.

• 2. Prove that Λ is equivalent to the regular representation L.

• 3. Let K be a subgroup of G. Let L(G ∕ K) be the subspace of L(G) consisting of functions $f : G\rightarrow \mathbb{C}$ that are right K-invariant, that is, f(gk) = f(g) for all k ∈ K. Show that L(G ∕ K) is a G-invariant subspace of L(G) and that the restriction of Λ to L(G ∕ K) is equivalent to the permutation representation $\mathbb{C}(G/K)$. (Hint: show that L(G ∕ K) has a basis consisting of functions that are constant on left cosets of K and compute the character.)