Strong Gelfand subgroups of $F\wr S_n$

The multiplicity-free subgroups (strong Gelfand subgroups) of wreath products are investigated. Various useful reduction arguments are presented. In particular, we show that for every finite group $F$, the wreath product $F\wr S_\lambda$, where $S_\lambda$ is a Young subgroup, is multiplicity-free if and only if $\lambda$ is a partition with at most two parts, the second part being 0,1, or 2. Furthermore, we classify all multiplicity-free subgroups of hyperoctahedral groups. Along the way, we derive various decomposition formulas for the induced representations from some special subgroups of hyperoctahedral groups.


Introduction
Let K be a subgroup of a group G. The pair (G, K) is said to be a Gelfand pair if the induced trivial representation ind G K 1 is a multiplicity-free G representation. More stringently, if (G, K) has the property that ind G K V is multiplicity-free for every irreducible K representation V , then it is called a strong Gelfand pair. In this case, K is called a strong Gelfand subgroup (or a multiplicity-free subgroup). Clearly, a strong Gelfand pair is a Gelfand pair, however, the converse need not be true. The problem of finding all multiplicity-free subgroups of an algebraic group goes back to Gelfand and Tsetlin's works [9,10], where it was shown that GL n−1 (C) (resp. Spin n−1 (C)) is a multiplicity-free subgroup in SL n (C) (resp. in Spin n (C)).
(SL n (C), GL n−1 (C)) and (Spin n , Spin n−1 ) are strong Gelfand pairs. It was shown by Krämer in [15] that for simple, simply connected (complex or real) algebraic groups, there are no additional pairs of strong Gelfand pairs. If the ambient group G is allowed to be a reductive group and/or the underlying field of definitions is changed, then there are many more strong Gelfand pairs [1,22,14]. For arbitrary finite groups, there is much less is known about the strong Gelfand pairs. In this paper we will be exclusively concerned with representations in characteristic 0. Building on Saxl's prior work [18,19], the list of all Gelfand pairs of the form (S n , K), where S n is a symmetric group, is determined by Godsil and Meagher in [11]. Recently, it was shown by Anderson, Humphries, and Nicholson [2] that, for n 7, the only strong Gelfand pairs of the form (S n , K) are given by Also recently, in [23], Tout proved that for a finite group F , the pair (F S n , F S n−1 ) is a Gelfand pair if and only if F is an abelian group.
In this article, we consider the strong Gelfand pairs of the form (F S n , K), where F is a finite group. Although the strongest results of our paper are about the pairs with F = Z/2, we prove some general theorems when F is a finite (abelian) group. The main purpose of our article is two-fold. First, we give a formula for computing the multiplicities of the irreducible F S n representations in the induced representations ind F Sn Sn V , where V is any irreducible representation of S n . Secondly, we will determine all strong Gelfand pairs (F S n , H), where F = Z/2. Along the way, we will present various branching formulas for such pairs.
We are now ready to give a brief outline of our paper and summarize its main results. In Section 2, we collect some well-known results from the literature.
The first novel results of our paper appear in Section 3, where 1) we prove a key lemma that we use later for describing some branching rules in wreath products, 2) we describe the multiplicities of the irreducible representations in ind F Sn Sn S λ where F is an abelian group, and S λ is a Specht module of S n labeled by the partition λ of n. Roughly speaking, in Theorem 3.12, we show that the multiplicities are determined by the Littlewood-Richardson rule combined with the descriptions of the irreducible representations of wreath products. As a corollary, we show that the pair (F S n , S n ) is not a strong Gelfand pair for n 6. It is easy to find negative examples for the converse statement. For example, it is easy to check that if F = Z/2 and n ∈ {1, . . . , 5}, then (F S n , S n ) is a strong Gelfand pair.
It is not difficult to see that for any S n representation W there is an isomorphism of F S n representations, ind F Sn Sn W ∼ = C[F n ] ⊗ W ; see Remark 3.9 for some further comments and the reference. In particular, if W is the trivial representation, and F is an abelian group, then ind F Sn Sn W is a multiplicity-free representation of F S n . In other words, (F S n , S n ) is a Gelfand pair if F is abelian. From this we find the fact that (F S n , diag(F ) × S n ) is a Gelfand pair. Now we have two questions about (F S n , diag(F ) × S n ) here: 1) What happens if we take a nonabelian group F ? 2) Is (F S n , diag(F ) × S n ) a strong Gelfand pair? The answers of both of these questions are rather intriguing although both of them are negative. In [3], Benson and Ratcliff find a range where (F S n , diag(F ) × S n ) with F nonabelian fails to be a Gelfand pair. In this paper, we show that, for F = Z/2 and n 6, (F S n , diag(F ) × S n ) is not a strong Gelfand pair (see Lemma 6.15). Our proof can easily be adopted to the arbitrary finite abelian group case.
In Section 4, we prove that, for an arbitrary finite group F , a pair of the form (F S n , F (S n−k × S k )) is a strong Gelfand pair if and only if k 2. In fact, by assuming that F is an abelian group, we prove a stronger statement in one direction: (F S n , (F S n−k ) × S k ) is a strong Gelfand pair if k 2.
Let F and K be two finite groups, and let π G : F G → G denote the canonical projection homomorphism. If K is a subgroup of F G, then we will denote by γ K the image of K under π G . The purpose of our Section 5 is to prove the following important reduction result (Theorem 5.4): If (F G, K) is a strong Gelfand pair, then so is (G, γ K ). As a consequence of this result, we observe that (Corollary 5.5), for n 7, if (F S n , K) is a strong Gelfand pair, then γ K ∈ {S n , A n , S n−1 ×S 1 , S n−2 ×S 2 }. Moreover, we show a partial converse of Theorem 5.4: Let n 7, and let B be a subgroup of S n . Then (F S n , F B) is a strong Gelfand pair if and only if (S n , B) is a strong Gelfand pair. This is our Proposition 5. 6.
In Sections 6 and 7, we classify the strong Gelfand subgroups of the hyperoctahedral group, B n := F S n , where F = Z/2. The hyperoctahedral group is a type BC Weyl group, and it contains the type D Weyl group, denoted by D n , as a normal subgroup of index 2. Our list of strong Gelfand subgroups of B n is a culmination of a number of propositions. In Section 6 we handle the groups K B n with γ K ∈ {S n , A n }, and in Section 7, we handle the case of K B n with γ K ∈ {S n−1 × S 1 , S n−2 × S 2 }. An essential ingredient for our classification is the linear character group L n := Hom(B n , C * ), which is isomorphic to Z/2 × Z/2 = {id, ε, δ, εδ}.
Here, ε and δ are defined so that ker ε = Z/2 A n , where A n is the alternating subgroup of S n , and ker δ = D n . The kernel of εδ will be denoted by H n .
Let χ be a linear character of a group A, and let B be another group such that χ(A) B. We will denote by (A × B) χ the following diagonal subgroup of A × B: n is odd two non-direct product index 2 subgroups of B n−2 × D 2 if n is odd four non-direct product index 2 subgroups of B n−2 × D 2 if n is even two non-direct product index 2 subgroups of B n−2 × H 2 if n is odd four non-direct product index 2 subgroups of B n−2 × B 2 if n is odd Table 1: The strong Gelfand subgroups of hyperoctahedral groups.
We will denote the natural copy of S n in B n , that is, {(0, σ) ∈ F n × S n : σ ∈ S n } by S n .
There is a unique subgroup Z of B 2 that is conjugate to S 2 and Z = S 2 . We will denote this copy of S 2 in B 2 by S 2 . In Table 1, we list all strong Gelfand subgroups of the hyperoctahedral group, B n , collating the results of Propositions 6.21, 6.30, 7.43, and 7.73. The index n in this table is assumed to be at least 6 for the cases of γ K ∈ {S n , A n }, at least 7 for the case of γ K = S n−1 ×S 1 , and at least 8 for the case of γ K = S n−2 ×S 2 . In fact, there are some additional strong Gelfand subgroups for n 7; in Propositions 6.21 and 6.30 we list them explicitly for n 5 and γ K ∈ {S n , A n }. In proving Proposition 7.43 we found two further strong Gelfand subgroups, which are included therein. Though Table 1 does not provide an exhaustive list for n 7, we remark that all subgroups in the table are still strong Gelfand if n 7. The strong Gelfand subgroups of B 2 are given in Lemma 7.48; those of B 3 are given in Proposition 7.74, in which we also give the number of strong Gelfand subgroups of B n for each 4 n 7.
Acknowledgements. The first author is partially supported by a grant from the Louisiana Board of Regents. The authors thank Roman Avdeev for useful communications about the multiplicityfree representations of reductive algebraic groups, and thank Sinéad Lyle for advice on GAP.

Preliminaries
We begin with setting up our conventions.
Throughout our manuscript, we will assume without further mention that our groups are finite. By a representation of a group we always mean a finite-dimensional complex representation.. The group-algebra of a group G will be denoted by C [G]. If H is a subgroup of G, then we will write H G. The boldface 1 will always denote the one-dimensional vector space which is the trivial representation of every group. When we want to emphasize the group G that acts trivially on 1, we will write 1 G . We present some combinatorial notation that we will use in the sequel. For a positive integer n, a partition of n is a non-increasing sequence of positive integers λ := (λ 1 , . . . , λ r ) such that r i=1 λ i = n. In this case, we will write λ n. Also, we will use the notation |λ| for denoting the sum r i=1 λ i .

Semidirect products
Let G be a group, and let N and H be two subgroups in G such that 1. N is normal in G; 2. G = HN ; In this case, we say that G is the semidirect product of N and H, and we write G = N H. Let F be another group, and let X be a G-set. The set of all functions from X to F is denoted by F X . Since F is a group, this set has the structure of a group with respect to point-wise multiplication. As G acts on X, it also acts on F X , hence, we can consider the group structure on the direct product F X × G defined by In the sequel, when we think that it will not lead to a confusion, we will skip the multiplication sign * from our notation. The group F X × G whose multiplication is defined in (2.1) will be called the wreath product of F and G with respect to X; it will be denoted by F G. Next, we setup some conventions and terminology.
1. Let id G denote the identity element of G. The subgroup F X × {id G }, denoted by F X , is called the base subgroup of F G. In some places in the text, when confusion is unlikely, we will write F instead of F X .
2. The diagonal subgroup of the base group is isomorphic to F . By abusing the notation, we will denote this copy of F in F G either by diag(F ) or by F , depending on the context. Note that if F is an abelian group, then diag(F ) is a central subgroup in F G.
3. Let id F denote the identity element of F . Then the group G := {id F } × G is a subgroup of F G as well. The diagonal copy of F in F G intersects G trivially, therefore, F G is a subgroup of F G; it is isomorphic to F × G since both of the subgroups F and G are normal subgroups in F G. Some authors refer to G as the passive factor of F G.
4. If F is the trivial group, then F G ∼ = G. If G is the trivial group, then F G ∼ = F X .
5. If G is a subgroup of S n , then F G is defined with respect to the set X := {1, . . . , n}. In particular, we have F S 1 = F . We set as a convention that F S 0 = {id}.
We finish this subsection by reviewing some simple properties of the wreath products. The proofs of these facts can be found in [7,Proposition 2.1.3].
Let H be a subgroup of G. Then we have Let G 1 and G 2 be two finite groups, and for i ∈ {1, 2}, let X i be a G i -set. To form the wreath product F (G 1 × G 2 ), we use the set X := X 1 X 2 with the obvious action of G 1 × G 2 . In this case, it is easy to check that there is an isomorphism of groups, In the sequel, we will be concerned with the subgroups of F G of the form (F G 1 ) × G 2 , where the second factor G 2 is the passive factor of F G 2 .

Basic properties of induced representations.
There are many equivalent ways of defining induced representation. We provide a definition for completeness: If H In loose terms, the G representation afforded by ind G H V is called the induced representation.
The following fact, which is referred to as the tensor identity by some authors, is another useful fact that we will refer to later in the text. Lemma 2.4. Let V be a representation of G, and let W be a representation of the subgroup H. Then we have the following isomorphism of G representations: Fortunately, there is a favorable situation where we have a similar factorization.
Proof. Clearly, it suffices to prove our claim for = 2. But this case is proved in [

Mackey Theory
In this subsection, we will mention some useful results of Mackey describing the relationship between the induced representations of two subgroups of G. A good exposition of the main ideas of these results is given in [5].
Let H, K G be two subgroups with a system S of representatives for the double (H, K)cosets in G. Let (σ, V ) and (ρ, W ) be representations of H and K, respectively. For s in S, let G s denote H ∩ sKs −1 , and let W s denote the representation ρ s : G s → GL(W ) by setting ρ s (g)w = ρ(s −1 gs)w for all g ∈ G s , and w ∈ W . Mackey's formula states that (2.6) In this notation, a closely related fact, which is called Mackey's lemma, states that

Generalized Johnson schemes
Let h be an element of the set {0, . . . , n}, and let 1 denote the trivial representation of the Young subgroup S n−h × S h of the symmetric group S n . It is well-known that (S n , S h × S n−h ) is a Gelfand pair. Indeed, by using Pieri's rule [6, Corollary 3.5.14], it is easy to see that In particular, the pair (F S n , F (S h × S n−h )) is a Gelfand pair.

Characterizations of the Gelfand property.
We can take quotients by normal subgroups and preserve the Gelfand property.  Proof. Let X be a G-set such that the semidirect product of F X G is the wreath product F G, and F X H is the wreath product F H. Since F X is a normal subgroup of F G, the proof follows from Lemma 2.10.
In the spirit of Corollary 2.12, we can fix the second factor and choose a Gelfand pair in the first factor. If F is an abelian group, then (F, {id}) is a Gelfand pair. Hence, Lemma 2.13 implies that (F G, G) is a Gelfand pair for every finite abelian group F , and for every finite group G.

A brief review of representations of F S n
The purpose of this section is to review the construction of the irreducible representations of F S n , where F is a finite group. We loosely follow James and Kerber's book [12,Section 4.4].
Let W 1 , . . . , W r be the complete list of pairwise inequivalent and irreducible representations of F . If n denotes a positive integer, then every irreducible representation D * of F n is given by an outer tensor product of the form . . , r}, let n j denote the number of factors of D * that are isomorphic to D j . Of course, some of these numbers might be equal to zero, nevertheless, the terms of the sequence n := (n 1 , . . . , n r ) sum to n. We will call such a sequence of nonnegative integers a composition of n. The composition n will be called the type of D * . Since S n permutes the factors of F n , two irreducible representations of F n are S n -conjugate if and only if they have the same type. An important group theoretic invariant of the irreducible representation D * , called the inertia group of D * , is given by F S(n) := F (S n 1 × · · · × S nr ) = F S n 1 × · · · × F S nr . (2.14) For every irreducible representation D of F , we have a representation of type n = (n) of F S n , which is denoted by D (n) , and defined as follows: The underlying vector space of D (n) is V = D n . If v := v 1 ⊗ · · · ⊗ v n is a basis element for V , then the action of an element ((f 1 , . . . , f n ), π) of F n S n on v is given by At the same time, for every irreducible representation D of S n , we have a corresponding irreducible representation of F S n . It is defined as follows: If (f, π) is an element of F S n and v is a vector from D , then the action of (f, π) on v is given by (2.15) Remark 2.16. Another name for the representation that is defined in (2.15) is inflation. More generally, if N is a normal subgroup of a finite group M and ρ : M/N → GL(V ) is a representation, then we have an associated representationρ : M → GL(V ),ρ(g) := ρ(gN ), which is called the inflation of ρ. Since the canonical quotient map M → M/N is a surjective homomorphism, if ρ is irreducible, then so is its inflation.
We now consider the inner tensor product of D (n) and D , that is, on which F S n acts diagonally. In general, the inner tensor products of irreducible representations are reducible, however, (2.17) is an irreducible representation for F S n . Indeed, according to Specht (but see [12,Theorem 4.4.3]) the complete list of pairwise inequivalent and irreducible representations of F S n is comprised of representations of the form where n = (n 1 , . . . , n r ) is a composition of n, D i 's (1 i r) are pairwise inequivalent irreducible representations of F , and D i (1 i r) is an irreducible representation of S n i .

A Useful Lemma and Induction From the Passive Factor
The main goal of this section is to prove a technical but a useful lemma that we will use repeatedly in the sequel. Also, we will show that in general the pair (F S n , S n ) need not be a strong Gelfand pair.
We begin with a lemma which we will use several times in the sequel. To keep its statement simple, we introduce some of the notation of its hypothesis here: F will denote a group, and D will be an irreducible representation of F . For two nonnegative integers n and k such that 0 k n, we will denote by E (resp. E ) an irreducible representation of S n−k (resp. S k ). By U , we will denote the S n representation U := ind Sn S n−k ×S k E E . We assume that the decomposition of U into irreducible S n representations is given by Lemma 3.1. We maintain the notation from the previous paragraph. If A is the F S n representation defined by ind F Sn F S n−k ×F S k (D; E) (D; E ), then its decomposition into irreducible subrepresentations is given by In particular, if U is a multiplicity-free S n representation, then A is a multiplicity-free F S n representation.
Proof. By its definition, the induced representation ind F Sn F S n−k ×F S k (D; E) (D; E ) is given by The outer tensor product (D; E) (D; E ), as a representation of F S n−k × F S k , can be rewritten as an inner tensor product of representations of F (S n−k × S k ) as follows: where F (S n−k × S k ) acts on D (n) , as a subgroup of F S n , by F n ; it acts on E E by S n−k × S k . By substituting (3.4) in (3.3), we see that Since F (S n−k × S k ) acts on D (n) as a subgroup of F S n , by definition, D (n) is the restricted representation res F Sn F (S n−k ×S k ) D (n) . Thus, by Lemma 2.4, we have Since F n acts trivially on E E , we know that ind F Sn F (S n−k ×S k ) E E = m 1 E 1 ⊕ · · · ⊕ m r E r . Therefore, (3.6) is given by This finishes the proof of our first claim. Our second assertion follows from (3.2) by setting all of the m i 's (1 i r) to 1. This finishes the proof of our lemma.
Let us present a straightforward consequence of Lemma 3.1. We will use the following notation: Let D be an irreducible representation of F . Let n be a positive integer, and let b = (b 1 , . . . , b r ) be a composition of n. For i ∈ {1, . . . , r}, let E i be an irreducible representation of S b i , and let U denote the S n representation U := ind Sn S(b) ( r i=1 E i ) whose decomposition into irreducible constituents is given by where S λ is the Specht module indexed by the partition λ, and m λ ∈ Z 0 is the multiplicity of S λ in U .
Corollary 3.7. We maintain the notation of the previous paragraph. If A is the induced representation ind F Sn F S(b) (D; E 1 ) · · · (D; E r ), then its decomposition into irreducible constituents is given by In particular, if U is a multiplicity-free S n representation, then A is a multiplicity-free F S n representation.
Proof. We apply induction on r. The base cases r = 2 is already proven in Lemma 3.1. The general case follows from r − 1 by distributivity of the tensor products over direct sums.
Remark 3.9. Let b be the composition (1, . . . , 1) of n. In this case, we have Let us denote S 1 × · · · × S 1 by n S 1 . Clearly, n S 1 is the trivial subgroup of S n . Let U denote the representation ind Sn n S 1 1 · · · 1 = ind Sn {id} 1 ∼ = C[S n ]. Hence, every irreducible representation S λ of S n appears in U with multiplicity m λ = dim S λ . Corollary 3.7 shows that This observation is a special case of a more general, well-known isomorphism that is presented in Jantzen's textbook [13,Section 3.8]. In our special case, it implies that, for any F n representation N , there is an isomorphism of F S n representations: We also know from [13, Section 3.8] that, for any S n representation W , there is an isomorphism of F S n representations: Here, F n acts on C[F n ] via its left regular representation, S n acts on W the usual way, and it acts on C[F n ] by permuting the factors of F n .
Although the isomorphism in (3.11) provides us with the general structure of the induced representation ind F Sn Sn W , we still want to determine the multiplicities of the irreducible representations in it. We resolve this problem by our next theorem. Theorem 3.12. Let F be an abelian group, and let U be an irreducible representation of F S n of the form U := ind F Sn F S(a) (D 1 ; D 1 ) · · · (D s ; D s ), where D 1 , . . . , D s are some pairwise inequivalent irreducible representations of F , and D 1 , . . . , D s are some irreducible representations of S a 1 , . . . , S as , respectively.
Under these assumptions, if W is an irreducible representation of S n , then the multiplicity of U in ind F Sn Sn W is equal to the multiplicity of W in ind Sn S(a) D 1 · · · D s . Proof. Since we will work with a fixed number n, and a fixed abelian group F , to ease our notation, let us set G := F S n and K := S n .
The multiplicity of U in ind G K W is equal to the dimension of the vector space We will use Mackey's formula and Frobenius reciprocity to compute the dimension of M . For brevity, we will denote the inertia group of V := (D 1 ; D 1 ) · · · (D s ; D s ), that is F S(a), by H. Let S be a system of representatives for the (H, K)-double cosets in G. Since F is a normal subgroup of G, and since it is contained in H, we see that HK = G. In other words, S has only one element, S = {id}. Therefore, there is only one local group of the form G s = H ∩ sKs −1 , which is given by G id = F S(a) ∩ S n = S(a).
Now we apply Mackey's formula (2.6): where W id is the copy of W viewed as a representation of S(a), that is, W id = res K S(a) W . In our abelian case, D 1 , . . . , D s are one-dimensional representations of F , so, the dimension of each factor (D i , D i ) of V is equal to the dimension of D i . In particular, the factor (D i , D i ) can be identified, as a representation of S a i , with D i . Therefore, res H S(a) V is equivalent to D 1 · · · D s . Thus (3.13) is equivalent to This shows that dim M is given by the multiplicity of D 1 · · · D s in res K S(a) W . By applying Frobenius reciprocity, we see that This finishes the proof of our theorem. Proof. By transitivity of the induction, if (F S n , B) is a strong Gelfand pair, then so is (F S n , S n ). Therefore, it suffices to show that (F S n , S n ) is a strong Gelfand pair if and only if n 5. In particular, we can now use Theorem 3.12.
Let a be a composition of n, and let S(a) denote the corresponding subgroup S a 1 × · · · × S as of S n . Let D 1 · · · D s be an irreducible representation of S(a). Clearly, if n 5, then at most one of the factors D i (1 i s) is of the form S (1 a ) or S (a) . Then by Pieri's formula, ind Sn S(a) D 1 · · · D s is a multiplicity-free representation. For n 6, we can use the induced representation ind Sn S n−3 ×S 3 S (n−4,1) S (2,1) and note that S (n−3,2,1) appears as a summand with multiplicity 2 -this is an easy check on the number of Littlewood-Richardson tableaux of skew-shape (n − 3, 2, 1)/(n − 4, 1) and weight (2,1). This completes the proof.

Some Strong Gelfand Subgroups of Wreath Products
In this section we will prove that, for an arbitrary group F , a pair of the form (F S n , F (S n−k × S k )) is a strong Gelfand pair if and only if k 2. Furthermore, we will prove that, for an abelian group F , (F S n , (F S n−k ) × S k is a strong Gelfand pair if k 2.

The nonabelian base group case.
Theorem 4.1. Let F be a group, and let n 2. If k is 1 or 2, then the pair (F S n , F (S n−k × S k )) is a strong Gelfand pair.
where E is an irreducible representation of F S n−k and D is an irreducible representation of F S k . Then there exists a composition c = (c 1 , . . . , c s ) of k such that . . , D s are some pairwise inequivalent irreducible representations of F , and D 1 , . . . , D s are some irreducible representations of S c 1 , . . . , S cs , respectively. Similarly for E, let b denote the composition of n − k such that where E 1 , . . . , E r are some pairwise inequivalent irreducible representations of F , and E 1 , . . . , E r are some irreducible representations of S b 1 , . . . , S br , respectively. Of course, if k = 2, then we have only two possibilities for c; they are given by c ∈ {(1, 1), (2)}. If k = 1, then c ∈ {(1)}. In any case, to ease our notation, let us denote (D 1 ; D 1 ) · · · (D s ; D s ) by D 0 , and let us denote (E 1 ; E 1 ) · · · (E r ; E r ) by E 0 .
As far as the irreducible representations E 1 , . . . , E r , D 1 , . . . , D s are concerned, we have two possibilities: We proceed with the first case. In this case, by Lemma 2.5, we have where bc is the composition obtained by concatanating b and c. It is easy to see that the representation ind F Sn F S(bc) E 0 D 0 is one of the irreducible representations of F S n as in (2.18). Next, we will handle the second case. Without loss of generality let us assume that D 1 = E r . By arguing as before, first, we write Notice that we have Now, by using by Lemma 2.5 and transitivity of the induction, we split our induction: We continue with the computation of the middle term, Notice here that we can apply Lemma 3.1. Indeed, since c 1 ∈ {1, 2}, D 1 is either a sign representation or the trivial representation of S c 1 , therefore, by Pieri's formula, the induced representation ind . . , L l denote its irreducible constituents. Then, by Lemma 3.1, (4.2) is equivalent to the F S br+c 1 representation (D 1 ; L 1 ) ⊕ · · · ⊕ (D 1 ; L l ), hence we have In fact, even if there is another such summand, since c has at most two parts, we can apply the same procedure to our decomposition (4.3) by the second (inequivalent) irreducible representation D 2 . Since the list of irreducible representations of F that appear in the final direct sum would all be distinct, in this case also, we get a multiplicity-free representation of F S n . This finishes the proof of our theorem.

Abelian base groups.
In this subsection, F denotes an abelian group. Also, by a slight abuse of notation, the subgroup (F S n−1 ) × S 1 F S n , where S 1 corresponds to the (trivial) subgroup of F S 1 , will be denoted by F S n−1 .
Proposition 4.4. If n 2, then (F S n , F S n−1 ) is a strong Gelfand pair.
Proof. Let K denote the subgroup F S n−1 F S n . Every irreducible representation of K is of the form E 1, where 1 is the trivial representation of S 1 , and E is an irreducible representation of the factor F S n−1 . Then there exist a composition b = (b 1 , . . . , b r ) of n−1 and an irreducible representation We now look closely at the tensor product E U j for j ∈ {1, . . . , s}. Since U j is an irreducible . . , r}, then there is exactly one such index i. Without loss of generality, let us assume that this index is r. Then by another application of Lemma 2.5 we get S br ×S 1 E r 1 is a multiplicity-free S br+1 representation, by Lemma 3.1, the representation ind is a multiplicity-free F S br+1 representation with irreducible summands of the form (E r ;Ẽ r ), whereẼ r is an irreducible S br+1 representation. It follows that E U j for j ∈ {1, . . . , s} is a multiplicity-free F S n−1 × F S 1 representation. But since E r is uniquely determined by U j (in fact, we assumed that E r = U j ), the representations E U j , where j ∈ {1, . . . , s}, do not have any irreducible constituent in common. Also, any irreducible representation that appears in E U j for j ∈ {1, . . . , s} induces up to an irreducible representation of F S n . Now applying ind F Sn F S n−1 ×F S 1 to (4.5) proves our claim at once; the representation ind F Sn K E 1 is multiplicity-free. This finishes the proof of our proposition.
We proceed with another example. Lemma 4.6. For every abelian group F , the pair (F S 2 , S 2 ) is a strong Gelfand pair.
Proof. The subgroup S 2 has two one-dimensional irreducible representations; they are given by 1 and the sign representation . On one hand, since F is abelian, (F S 2 , S 2 ) is a Gelfand pair, hence, ind F S 2 S 2 1 is multiplicity-free. On the other hand, we know from the construction of irreducible representations of wreath products F S 2 that the inflation of any irreducible representation of S 2 is an irreducible representation of F S 2 . In particular, the (irreducible) linear representation of S 2 extends to a linear representation of F S 2 . Then we know that the triplet (F S 2 , S 2 , ) is an example of a "twisted Gelfand pair" [   Proof. The of this proposition proof is very similar to the proof of Proposition 4.4. The only difference is that, instead of using the permutation representation ind F S 1 S 1 1, we use the repre- By Lemma 4.6, we know that they are multiplicity-free representations of F S 2 . Since the rest of the arguments are the same as in the proof of Proposition 4.4, we omit the details.

A Reduction Theorem
We begin with setting up some new notation that will stay in effect in the rest of our paper.
Let X be a finite G-set with n := |X|. Let F be a finite group. Although F is not necessarily an abelian group, for simplifying (the exponents in) our notation, the inverse of an element a of F will be denoted by −a. Accordingly, if f is an element of F X , or equivalently, if it is an element of the subgroup F X in F G, then we will write −f to denote its inverse in F X (respectively in F X ). In this notation, if (f, g) is an element in F G, then its inverse is given by If (f , g ) and (f, g) are two elements from F G, then their product is given by Let π G denote the canonical projection homomorphism onto G, that is, If K is a subgroup of F G, then we denote the image of K under π G by γ K . Equivalently, γ K is given by The following remark/notation will be useful in the sequel.
Indeed, we have the following short exact In other words, we have the inclusion Since the union of all left cosets of Γ id K covers K, the inclusions (5.2) are actually equalities of sets; every Γ g We now go back to the strong Gelfand pairs. The following characterization of the strong Gelfand pairs is easy to prove. We will apply this result to wreath products. The main result of this section is the following reduction result.
Theorem 5.4. Let F and G be two finite groups, and let K be a subgroup of F G. If (F G, K) is a strong Gelfand pair, then so is (G, γ K ).
Proof. We assume that (F G, K) is a strong Gelfand pair. By Lemma 5.3, we know that (F G × K, diag(K)) is a Gelfand pair.
Let X denote the finite G-set such that F G = F X G, and let H denote the following subset of F G × K: We claim that H is a subgroup. First, we will show that H is closed under products: Let Since the second and the fourth entries are the same, this product is contained in H, hence, H is closed under products. Next, we will show that the inverses of the elements of H exist: For Clearly, τ is an element of H. The product of σ and τ is given by hence, ((x, y), (z, y)) is the inverse of ((f, b), (a, b)). These computations show that H is a subgroup of F G × K.
Evidently, the diagonal subgroup diag(K) in F G × K is a subgroup of H. Since (F G × K, diag(K)) is a Gelfand pair, it follows that (F G × K, H) is a Gelfand pair as well. Now we will identify a normal subgroup of F G × K by the help of the following map: It easy to verify that φ is a homomorphism. It is also evident that, if an element ((x, y), (z, w)) from F G × K lies in the kernel of φ, then y = w = id. In particular, we see that N := ker(φ) H. This is the normal subgroup that we were seeking. By Remark 2.11, now we know that the pair (( ). Finally, by using Lemma 2.10 once again, we conclude that (G, γ K ) is a strong Gelfand pair. This finishes the proof of our theorem.
We mentioned earlier that, for n 7, there are only four (minimal) strong Gelfand subgroups in S n [2]. As a simple consequence of Theorem 5.4, we deduce a similar statement for the strong Gelfand subgroups in F S n .
Corollary 5.5. Let n 7, and let K be a subgroup of F S n . If (F S n , K) is a strong Gelfand pair, then γ K ∈ {S n , A n , S n−1 × S 1 , S n−2 × S 2 }.
We record a partial converse of Theorem 5.4. Proposition 5.6. Let n 7, and let B be a subgroup of S n . Then (F S n , F B) is a strong Gelfand pair if and only if (S n , B) is a strong Gelfand pair.
Proof. Let K be a subgroup of the form F B for some subgroup B S n . Then γ K = B. Now Corollary 5.5 gives the ⇒ direction. For the converse, by the main result of [2], we have four cases: B = S n , B = A n , B = S n−1 × S 1 , and B = S n−2 × S 2 . In the first case there is nothing to do. The second case follows from the fact that F A n is an index 2 subgroup of F S n . The last two cases follow from Theorem 4.1.

Hyperoctahedral Groups
From now on we will denote by F the cyclic group of order 2. To simplify our notation, we will use the additive notation, so, F = Z/2 = {0, 1}. If there is no danger for confusion, the identity element of F n (n ∈ N) will be denoted by 0 as well. The wreath product F S n will be denoted by B n . For i ∈ {1, . . . , n}, the element x ∈ F n which has 1 at its i-th entry and 0's elsewhere will be denoted by e i . The set {e 1 , . . . , e n } will be called the standard basis for F n . In this notation, for i ∈ {1, . . . , n}, if f ∈ F n , then f i will denote the coefficient of e i in f . When there is no danger for confusion, we will use 1 to denote the sum of the standard basis elements, 1 = e 1 + · · · + e n . (6.1) The wreath product B n is called the n-th hyperoctahedral group. It follows from the general description of the irreducible representations of wreath products that every irreducible linear representation of B n is equivalent to one of the induced representations, where 0 k n, and S λ (resp. S µ ) is the Specht module indexed by the partition λ of n − k (resp. by the partition µ of k). The character of S λ,µ will be denoted by χ λ,µ . Our goal in the rest of this section is to determine the strong Gelfand subgroups of B n . In light of Corollary 5.5, it will suffice to determine the Gelfand subgroups K with γ K ∈ {S n , A n , S n−1 × S 1 , S n−2 × S 2 } only. Before going through these cases, we will point out some well-known facts about the structures of certain subgroups of B n . Also, we will describe a branching rule for the subgroup B n−1 in B n .

Some special subgroups of B n .
First of all, we want to point out that B n has a distinguished index 2 subgroup, denoted D n . Actually, it is the Weyl group of type D n . To describe it, we will view B n as a subgroup of S 2n . Let (f, σ) be an element of B n . For i ∈ {1, . . . , n}, we will construct a permutationf i of {1, . . . , 2n} as follows: For j ∈ {1, . . . , 2n}, the value off i at j is defined bỹ Likewise, by using σ we define a permutationσ of {1, . . . , 2n} bỹ We set x = x(f, σ) :=f 1 · · ·f nσ ∈ S 2n . Then the map (f, σ) → x(f, σ) is an injective group homomorphism B n → S 2n . Abusing notation, we will denote the image of B n in S 2n by B n as well. Then our distinguished subgroup is given by the intersection D n = A 2n ∩ B n . In the sequel, we will identify D n in B n in a different way.
It is going to be important for our purposes that we know all index 2 subgroups of B n . Fortunately, they are fairly easy to find once we give the Coxeter generators of B n . Let s 1 , . . . , s n−1 denote the (simple) transpositions (0, (1 2)), (0, (2 3)), . . . , (0, (n − 1 n)); these elements are the Coxeter generators of the passive factor S n in B n . Let t denote the element ((1, 0, . . . , 0), id) ∈ B n . Then we have the following Coxeter relations: The group of linear characters of B n , that is L n := Hom(B n , C * ), is generated by the characters ε and δ defined by Thus, L n is isomorphic to F × F . Explicitly, the four linear characters 1, ε, δ, εδ correspond to B n representations as follows.
These facts allow us to conclude the following useful statements: 1. B n has exactly three subgroups of index 2, corresponding to the kernels of the homomorphisms ε, δ, and εδ.
2. B n has exactly one normal subgroup of index 4, denoted by J n , that is given by the intersection ker ε ∩ ker δ.
4. The kernel of ε is F A n .
Remark 6.4. Let G be a finite group. If there is a unique normal index 4 subgroup J of G, then we think that it would be appropriate to call it the Stembridge subgroup of G because of John Stembridge's seminal work on the projective representations [21] where such a subgroup is extensively used.

The associators of index 2 subgroups of B n .
Our references for this subsection are the two papers [20,21] of Stembridge. The linear character group L n of B n acts on the isomorphism classes of irreducible representations of B n via V → τ ⊗ V , where τ is a one-dimensional representation corresponding to an element of L n . We continue with the assumption that V is an irreducible representation of B n . If V ∼ = τ ⊗ V , then we will say that V is self-associate with respect to τ ; otherwise, V and τ ⊗ V are said to be associate representations with respect to τ . In the sequel, when there is no danger for confusion, it will be convenient to denote τ ⊗ V by χ τ V , where χ τ is the linear character of τ .
Let H denote the kernel of χ τ : B n → C * , and let V be a self-associate representation with respect to τ . Then there exists an endomorphism S ∈ GL(V ) such that gSv = τ (g)Sgv for all g ∈ B n and v ∈ V . Furthermore, as a consequence of Schur's lemma, one knows that S 2 = 1, hence, S has two eigenvalues, ±1. Any of the two endomorphisms ±S is called the τ -associator of V . Let V + (resp. V − ) denote the S-eigenspace of eigenvalue +1 (resp. eigenvalue −1). Then V + and V − are irreducible pairwise inequivalent H representations. If V and τ ⊗ V are associate representations with respect to τ , then both of them are irreducible and isomorphic as H representations [20, Lemma 4.1]. Let us translate these statements to the 'induced/restricted representation' language. Let V be a self-associate representation with respect to τ . Then the restriction of V to the subgroup H splits into two inequivalent irreducible representations. If V is not a self-associate representation with respect to τ , then the restriction of V to H is an irreducible representation of H, and furthermore, ind Bn H res Bn H V = V ⊕ τ ⊗ V . Let S λ,µ be an irreducible representation of B n , and let χ λ,µ denote the corresponding character. Then we have 1. δχ λ,µ = χ µ,λ , hence, S λ,µ is self-associate with respect to δ if and only if λ = µ; 2. εχ λ,µ = χ λ ,µ , hence, S λ,µ is self-associate with respect to ε if and only if λ = λ and µ = µ ; 3. εδχ λ,µ = χ µ ,λ , hence, S λ,µ is self-associate with respect to εδ if and only if λ = µ .
If f is an element of F n , then we will denote by #f the number of 1's in f . For a subgroup K of B n , we define Note that m K may not exist, as we may sometimes have Γ id K = {(0, id)}. Clearly, if it exists, then m K is an element of the set {1, . . . , n}. We have five major cases for m K : 5. m K does not exist.
Although the above five cases are defined for K with γ K = S n , in the sequel the same cases will be considered for the subgroups K B n where γ K ∈ {A n , S 1 × S n−1 , S 2 × S n−2 }. We recall our notation from Remark 5.1: For g ∈ γ K , Γ g K is the preimage (π G | K ) −1 (g).
Remark 6.11. We see from Lemma 6.8 that among the many equivalent descriptions of D n we have D n = {(f, σ) ∈ B n : #f is even}.
Lemma 6.12. If K is a subgroup of B n with γ K = S n , then m K / ∈ {3, . . . , n − 1}. In other words, there is no subgroup K B n such that γ K = S n and 3 m K n − 1.
Proof. Assume towards a contradiction that there exists a subgroup K in B n such that m K is an element of {3, . . . , n − 1}. Let (f, id) be an element of Γ id K . Then there are some basis elements e i 1 , . . . , e ir (r ∈ {3, . . . , n − 1}) with 1 i 1 < · · · < i r n such that f = e i 1 + · · · + e ir . Without loss of generality we assume that i r = n. Let σ denote the transposition (i r n) so that σ(i r ) = n. Since γ K = S n , the element (0, σ) is contained in K. In particular, (0, σ) * (f, id) = (σ · f, σ) is an element of K. Then, (σ · f, σ) * (σ · f, σ) = (f + σ · f, id) is an element of K. But this last element is equal to (e ir + e n , id). Since #(e ir + e n , id) = 2, we obtained a contradiction to our initial assumption m K 3. This finishes the proof of our assertion. Remark 6.13. Let H 1 and H 2 be two subgroups of a group G. The assumption that H 1 is isomorphic to H 2 does not guarantee that the following implications hold: (G, H 1 ) is a strong Gelfand pair ⇐⇒ (G, H 2 ) is a strong Gelfand pair. (6.14) For example, let (G, H 1 , H 2 ) be the triplet (G, H 1 , H 2 ) := (B 2 , diag(F ), S 2 ). It is easy to verify that (G, H 1 ) is not a strong Gelfand pair but (G, H 2 ) is a strong Gelfand pair. Nevertheless, if H 1 and H 2 are isomorphic via an automorphism of G, then Remark 2.11 shows that (6.14) hold.
Lemma 6.15. If for a subgroup K B n with γ K = S n the integer m K is n, then K is conjugate-isomorphic to a subgroup of diag(F ) × S n . Moreover, these subgroups are strong Gelfand subgroups of B n if and only if n 5.
Proof. For m K = n, the fact that Γ id K = {(0, id), (1, id)} follows from definitions. Therefore, we have a copy of the central subgroup diag(F ) × {id Sn } in K. Since any element of K is of the form (f, σ), where f ∈ Γ id K and σ ∈ S n , we see that K is conjugate-isomorphic to a subgroup of diag(F ) × S n . Furthermore, since γ K = S n , we know that the index of K in diag(F ) × S n is at most 2. As the group of linear characters of diag(F ) × S n is isomorphic to Z/2 × Z/2, we see that K can be a conjugate of one of the following four subgroups: 1) diag(F ) × S n , 2) {id} × S n , 3) diag(F ) × A n , 4) a non-direct product subgroup of index 2. Notice that, in our case, the subgroups in 3) cannot occur since γ K = S n , and the subgroups in 2) cannot occur because it needs m K = n.
Next, we will show that, for n 6, K := diag(F ) × S n is not a strong Gelfand subgroup of B n ; applying Remark 6.13 will then handle the other possible subgroups in item 1). Those in item 4) follow by transitivity of induction.
Let As in the proof of Theorem 3.12, we will use Mackey's formula and Frobenius reciprocity to compute the dimension of M . Let V denote the representation (1; D 1 ) ( ; D 2 ). Then H is the inertia group of V . Let S be a system of representatives for the (H, K)-double cosets in G.
Since F is a normal subgroup of G, and since it is contained in H, we see that HK = G. In other words, S = {id}. Therefore, there is only one local group of the form G s = H ∩ sKs −1 , which is given by G id = diag(F ) × (S a × S b ). By Mackey's formula (2.6), where W id is the copy of W viewed as a representation of diag(F ) × (S a × S b ), that is, Recall from Subsection 2.6 that for a representation M of F , and for r ∈ N, the notation M (n) stands for the r-fold outer tensor product M · · · M . In this notation, we have is either 1 or , depending on the parity of b. Substituting (6.18) and (6.17) in (6.16), and using the Frobenius reciprocity, we see that In particular, we see that if W = (res When n 5, the result for subgroups conjugate to diag(F ) × S n follows from the fact that the passive factor S n is strong Gelfand, which we prove in Lemma 6.20. For the subgroups conjugate to an index 2 subgroup of this, the result can easily be checked by computer. This completes the proof. Remark 6.19. There is an easier, alternative proof of the second part of Lemma 6.15 for n 10. Indeed, since n 10, we always have an irreducible representation V of S n that induces up with multiplicity at least 3. For example, by using Theorem 3.12, we see that multiplicity of S ((n−7,2,1),(3,1)) = ind Bn F (S n−4 ×S 4 ) (1; S (n−7,2,1) ) ( ; S (3,1) ) in ind Bn Sn S (n−6,3,2,1) is equal to the multiplicity of S (n−6,3,2,1) in ind Sn S n−4 ×S 4 S (n−7,2,1) S (3,1) . As in the proof of Corollary 3.14, we can easily count that there are three Littlewood-Richardson tableaux of skew-shape (n − 6, 3, 2, 1)/(n − 7, 2, 1) and weight (3,1). Now, since S n is a subgroup of index 2 in Y n , ind Yn Sn V = V 1 ⊕ V 2 , where V 1 and V 2 are two irreducible representations of Y n . Let W be an irreducible representation of B n such that the multiplicity of W in ind Bn Sn V is at least 3. Then the multiplicity of W in one the induced representations ind Bn Yn V 1 or ind Bn Yn V 2 is at least 2. Hence, we conclude that (B n , diag(F ) × S n ) is not a strong Gelfand pair. Then Y n is isomorphic to the passive factor S n by an automorphism of B n . Furthermore, if for a subgroup K B n with γ K = S n the integer m K does not exist, then K is either conjugate to S n or it is conjugate to Y n . In this case, (B n , K) is a strong Gelfand pair if and only if n 5.
Proof. If m K does not exist, then Γ id K has only one element, Γ id K = {(0, id)}. Therefore, K is isomorphic to γ K . Clearly, S n is such a subgroup, and the conjugate subgroups (f, id) * S n * (f, id) −1 , where f ∈ F n \ {1}, are all different from each other. Likewise, it is easy to check that Y n is a subgroup of B n such that γ Yn = S n and m Yn does not exist. Furthermore, , are all different from each other. It is not difficult to show that these are all possible subgroups of B n with γ K = S n and m K does not exist. We omit the details of this part of the proof.
Next, we will show that the subgroups S n and Y n are strong Gelfand subgroups of B n if and only if n 5. We already proved the former case in Corollary 3.14. To prove that Y n is not a strong Gelfand subgroup, we introduce the following map: We claim that ψ is an automorphism of B n . Clearly, ψ is well-defined and one-to-one. Hence, it is a bijection. Let (f, σ) and (g, τ ) be two elements from B n . Then (f, σ) * (g, τ ) = (f +σ ·g, στ ). We have four cases: 1) σ, τ ∈ A n ; 2) σ ∈ A n , τ / ∈ A n ; 3) σ, τ / ∈ A n ; 4) σ / ∈ A n , τ ∈ A n . In each case, it is easily checked that ψ((f, σ)) * ψ((g, τ )) = ψ((f + σ · g, στ )). It follows that ψ is an automorphism. Moreover, we see from the description of ψ that it is an outer automorphism of B n . Evidently, we have ψ(S n ) = Y n . But then by Remark 2.11 we know that (B n , S n ) is a strong Gelfand pair ⇐⇒ (B n , Y n ) is a strong Gelfand pair. Therefore, by Corollary 3.14, Y n is a strong Gelfand subgroup of B n if and only if n 5. Applying Remark 6.13 finishes the proof of our lemma.
We now paraphrase the conclusions of the above lemmas in a single proposition. Proposition 6.21. Let n 6, and let K be a strong Gelfand subgroup of B n . Let ε : B n → C * and δ : B n → C * be the linear characters as defined in (6.3). If γ K = S n , then (up to a conjugation) K is one of the following subgroups: 2. D n = ker δ; 3. H n = ker(εδ).
For n 5, in addition to these three cases, we have the following possibilities: S n , Y n from Lemma 6.20 and diag(F ) × S n . 6.3 γ K = A n . Assumption 6.22. Unless otherwise noted, we assume that n 3.
In this case, K is equal to F A n , hence, we have a strong Gelfand subgroup.
Proof. We will argue as in Lemma 6.6: Since m K = 1, Γ id K contains an element of the form (e k , id) for some k in {1, . . . , n}. Let (f, (1 2 k)) ∈ K. Without loss of generality, we will assume that k > 2. Since Γ id K is a normal subgroup of K, we know that Then it is not difficult to see that (e l , id) ∈ Γ id K for every l ∈ {1, . . . , n}. It follows that, Γ id K is equal to F . This means that, for every (f, σ) ∈ K, we have (0, σ) = (−f, id) * (f, σ) ∈ K. In other words, the alternating subgroup of the passive factor S n is a subgroup of K. But this shows that F n A n is a subgroup of K. Since this an index 2 subgroup in B n , and since K is a proper subgroup, we have the equality F A n = K. In particular, K is a strong Gelfand subgroup of B n by Proposition 5.6. Lemma 6.25. If K is a subgroup of B n with γ K = A n and m K = 2, then Γ id K = {(f, id) ∈ B n : #f is even}, and hence K = J n = ker ε ∩ ker δ. Proof. First assume that n 4. Since m K = 2, we know that Γ id K contains an element of the form (e i + e j , id) for some i, j ∈ {1, . . . , n} with i < j. Let u be a subset u := {k, l} of {1, . . . , n} with k < l and u ∩ {i, j} = ∅. Let σ denote the (even) permutation (k i)(l j), and let x be an element from Γ (k i)(l j) K . Then x = (f, σ) for some f ∈ F n . Since conjugating by x gives x * (e i + e j , id) * x −1 = ((k i)(l j) · (e i + e j ), id) = (e k + e l , id), every element of the form (e r + e s , id), where 1 r < s n and {r, s} ∩ {i, j} = ∅ is contained in Γ id K . Note that, we already have (e i + e j , id) ∈ Γ id K . Now by the argument that we used in the proof of Lemma 6.8, we see that if (f, id) is an element of Γ id K , then #f is even. In particular, we see that |Γ id K | = 2 n−1 . Since K/Γ id K ∼ = A n , we see that |K| = 2 n−1 n!/2 = 2 n−2 n!. Thus, K is an index 4 subgroup of B n . Finally, it is easy to check that K is contained in both of the subgroups ker ε and ker δ. Therefore, K is equal to ker ε ∩ ker δ. This finishes the proof of (6.26). We now proceed to prove our second claim. As in Section 6.1.1, we let L n denote the linear character group of B n . Let V be a finite-dimensional irreducible representation of B n , and let L V denote the stabilizer of V in L n , that is, L V = {τ ∈ L n : V ∼ = τ ⊗ V }. In (the proof of) [21, Theorem 3.1], Stembridge describes in detail the decomposition of the restriction res Bn K V into K-representations. In particular, Stembridge's theorem shows that res Bn K V is not multiplicity-free if and only if 1. L V = L n = {id, ε, δ, εδ}, 2. the ε-associator S of V and the δ-associator of V anti-commute, ST = −T S.
Since these conditions require the existence of a δ-associator, which is possible only if n is even, we conclude that res Bn K V is a multiplicity-free representation if n is odd. We now proceed with the assumption that n is even. An irreducible representation V with the correspond character χ λ,µ is self-associate with respect to all of the three linear characters ε, δ and δε if and only if λ = µ = µ , where µ is the partition conjugate to µ. In other words, χ λ,µ = χ λ,λ and λ is a self-conjugate partition of n/2. In this case, it is known that, the associators S = S λ,λ and T = T λ anti-commute if and only if n/2 is an odd integer (see the paragraph after [21, Theorem 6.4]).
If n = 3, we can explicitly check that K = J 3 , which is a strong Gelfand subgroup of B 3 .
Lemma 6.27. Let K be a subgroup of B n with γ K = A n . Then m K / ∈ {3, . . . , n − 1}. In other words, there is no subgroup K B n with γ K = A n and 3 m K n − 1.
Notice that for j / ∈ {i 1 , i 2 , i r , n}, we have x j = y j = 0. If j ∈ {i 1 , i 2 }, then we have x j = y j ; if j ∈ {i r , n}, then we have x j = y j + 1. Now we consider the product (x, id) * (y, id) = (x + y, id) = (e ir + e n , id) ∈ K.
Since #(e ir + e n , id) = 2, we obtained a contradiction to our initial assumption that m K 3. This finishes the proof of our lemma. Lemma 6.28. If K is a subgroup of B n with γ K = A n and m K = n, then K is conjugate to a subgroup of diag(F ) × A n . In particular, K is not strong Gelfand, unless n 5.
Proof. Since m K = n, we have Γ id K = {(0, id), (1, id)}, which is a central subgroup of B n . Since γ K = A n , by the exact sequence in Remark 5.1, we see that K is conjugate to a subgroup of diag(F ) × A n . In particular, the index of K in (a conjugate of) diag(F ) × A n is at most 2. Without loss of generality, we assume that K is a subgroup of diag(F ) × A n . Then since the group of linear characters of diag(F ) × A n has order 2, and since K = A n , we see that K = diag(F ) × A n . Since K diag(F ) × S n , by Lemma 6.15, we find that K is not a strong Gelfand subgroup for n 6. For 3 n 5, we verified in GAP that diag(F ) × A n is a strong Gelfand subgroup in B n if and only if n = 3. Lemma 6.29. If for a subgroup K B n with γ K = A n the integer m K does not exist, then K is conjugate to the alternating subgroup of the passive factor S n . In this case, (B n , K) is not a strong Gelfand pair for n 4. If n = 3, then (B n , K) is a strong Gelfand pair.
Proof. The proof proceeds in a similar way to that of Lemma 6.28.
Since m K does not exist, we have Γ id K = {(0, id)}. Since π Bn ((0, (1 2))) = (1 2) is not contained in A n , the element (0, (1 2)) is not contained in K. Let H denote the subgroup of B n that is generated by (0, (1 2)) and K. Then it is easy to check that γ H = S n and that Γ id H = Γ id K , hence, m H does not exist. It follows that H is one of the subgroups that we considered in Lemma 6.20.
By conjugating H we may assume that H = S n or the group Y n defined therein, and that K S n or K Y n . But |K| = n!/2 and we know that γ K = A n , hence we infer that K = A n (in both S n and Y n ). When n 6, since H is not a strong Gelfand subgroup, nor is K. For n 5 we checked in GAP that K is not a strong Gelfand subgroup unless n = 3. This finishes the proof.
We paraphrase the conclusions of the above lemmas in a single proposition. Proposition 6.30. Let K be a strong Gelfand subgroup of B n . If γ K = A n , then K is one of the following subgroups: 1. F A n ; 2. the Stembridge subgroup of B n , that is, J n = ker ε ∩ ker δ, where ε and δ are two inequivalent nontrivial linear characters of B n , where n ≡ 2 mod 4 for n 4, 7 The Cases of γ K = S n−1 × S 1 and γ K = S n−2 × S 2 In this last part of our paper, we analyze the strong Gelfand subgroups K in B n , where γ K ∈ {S n−1 × S 1 , S n−2 × S 2 }. These two cases provide us with the most diversity. For our proofs we heavily use analogs of the "Pieri's formulas" for the hyperoctahedral groups. Throughout this section also, F will denote the cyclic group Z/2.

Some Pieri rules.
Our goal in this section is to explicitly compute the decomposition formulas for induced representations from various subgroups of B n . While some of these formulas are known [17], we could not locate all of the decomposition rules that we need for our purposes. Let k be an element of {0, 1, . . . , n−1}, let λ be a partition of n−1−k, and let µ be a partition of k. Let S λ,µ denote the corresponding B n−1 -Specht module. We begin our computations by studying the decomposition of ind Bn B n−1 ×S 1 S λ,µ 1 into its irreducible constituents. By using 1) the transitivity of the induction, 2) Lemma 2.5, 3) the additivity of the induction, we see that Recall that the B n−1 -Specht module S λ,µ is given by ind B n−1 B n−1−k ×B k (1; S λ ) ( ; S µ ). Note also that (1; 1) = ind B 1 B 1 (1; 1). We apply these observations to the first summand in (7.1): By the transitivity of the induced representations, we have Therefore, by Lemma 2.5, the following induced representation is equal to (7.2); Now by applying Lemma 3.1 to ind (1; 1), we re-express the formula (7.3), hence, the formula (7.2), more succinctly as follows: Next, we focus on ind Bn B n−1 ×B 1 S λ,µ ( ; 1). By adapting the above arguments to this case, we find that Notation 7.6. If λ is a partition of n − 1, then λ denotes the set of all partitions obtained from λ by adding a "box" to its Young diagram. Equivalently, we have Lemma 7.7. If S λ,µ is an irreducible representation of B n−1 , then we have the following decomposition rules: Proof. The first and the second identities follow from (7.4) and (7.5), respectively, where we decompose the S n−k (resp. S k+1 ) representation ind S k ×S 1 S µ 1) into irreducible constituents. We then use the additivity property for the induced representations. In light of the decomposition (7.1), the third identity follows from the first two identities.
Remark 7.8. Lemma 7.7 combined with the fact that (S n , S n−1 × S 1 ) is a strong Gelfand pair gives a second proof of the fact that (B n , B n−1 × S 1 ) is a strong Gelfand pair.
Our goal is to extend Lemma 7.7 to certain subgroups of B n−2 × B 2 , so, we setup some relevant notation: Notation 7.9. If λ be a partition of n − 2, thenλ andλ are the following sets of partitions of n:λ := {τ n : S τ is a summand of ind Sn S n−2 ×S 2 S λ 1 S 2 }, λ := {τ n : S τ is a summand of ind Sn where S 2 is the sign representation of S 2 . ( The irreducible representations of B 2 are easy to list. They are given by (2) . (7.10) In Table 7.1, we present the values of the characters of these representations; they are computed by using the formula (5.5) in [21]. We fix an integer n 4. For k ∈ {0, 1, . . . , n − 2}, let S λ,µ be an irreducible representation of B n−2 , where λ n − k − 2 and µ k. Our goal is to compute the irreducible constituents of ind Bn B n−2 ×B 2 S λ,µ W , where W is one of the representations in (7.10). The method of computation for all these five cases are similar, so, we will present details only in the first case.
The case of W = S (2),∅ . Recall that S λ,µ is equal to ind . By using Lemma 2.5 repeatedly, we transform our induced representation to another form:
Proof. As a subgroup of B 2 , D 2 is given by diag(F ) × S 2 , which is isomorphic to Z/2 × Z/2. Hence, D 2 has four irreducible representations. However, two of these irreducible representations induce up the same representation of B 2 . Indeed, by Clifford theory, and the tools of Section 6.1.1, the irreducible representations of D 2 are found as follows. An irreducible representation V of D 2 is given by either res B 2 D 2 S λ,µ , where λ and µ are two distinct partitions such that |λ| + |µ| = 2, or by one of the two irreducible constituents of res B 2 D 2 S (1), (1) . By Frobenius reciprocity, we see that ind B 2 D 2 V is one of the following three representations of B 2 : S (1),(1) , S (2),∅ ⊕ S ∅, (2) , or S (1 2 ),∅ ⊕ S ∅,(1 2 ) . The rest of the proof follows from Lemma 7.16.
Corollary 7.19. Let S λ,µ be an irreducible representation of B n−2 , and let W be an irreducible representation of H 2 . Then we have Proof. The group H 2 is isomorphic to Z/4, and hence it has four inequivalent irreducible representations. Arguing as in Corollary 7.18, these representations can be described in terms of the irreducible representations of B 2 . Any irreducible representation V of H 2 is either equal to res B 2 H 2 S λ,µ , where λ and µ are two distinct partitions such that λ = µ and |λ|+|µ| = 2, or it is one of the two irreducible constituents of the representation res B 2 H 2 S (1), (1) . By Frobenius reciprocity, ind Corollary 7.20. Let S λ,µ be an irreducible representation of B n−2 , and let W be an irreducible representation of S 2 . Then we have Consequently, we have ind Bn Proof. Our first claim follows from a direct computation by using Let H be an index 2 subgroup of a finite group G, and let η denote the sign representation of the quotient G/H ∼ = Z/2. If W is an irreducible representation of G with character χ, then we have two cases: We will apply this standard fact in the following special situation: G = A × B, where A and B are two finite groups. Let φ : G → Z/2 be a surjective homomorphism. We denote the restriction of φ to the subgroup A × {1} by φ 1 . Likewise, the restriction of φ onto {id} × B is denoted by φ 2 . Then for every (a, b) ∈ G, we have φ(a, b) = φ 1 (a, 1)φ 2 (id, b). In particular, we have three distinct possibilities for the kernel H of φ: (H1) φ 2 is the trivial homomorphism. In this case, φ 1 must be surjective. Otherwise we have both of the subgroups A × {1} and {1} × B as subgroups of H, hence, we have H = G, which is absurd. Now, since φ 1 is surjective, ker φ 1 is an index 2 subgroup of A × {1}. In particular, H = ker φ 1 B.
(H3) Both of the homomorphisms φ 1 and φ 2 are surjective. Then the kernel of φ is given by Clearly, the most complicated case is the case of (H3). We will refer to this case as the nonobvious index 2 subgroup case. Nevertheless, if B is Z/2 = {1, −1} (in multiplicative notation), then we can describe H quite explicitly,  {(a, b) Let V be an irreducible representation of H.

Case Let H be as in (H2.a)-(H2.c). Then
3. Case 3. Let H be as in (H3.a)-(H3.c). In particular, ker φ 1 × {1} is an index 2 subgroup of H; we have In fact, it is a normal subgroup of G, so, the irreducible representations of ker φ 1 × {1} are easy to describe. Consequently, we can effectively analyze ind G H V in relation with the induced representations ind G ker φ 1 ×{1} V , where V is an irreducible representation of ker φ 1 × {1}. We will do this in the sequel for the cases (H3.b) and (H3.c).  Assumption 7.24. In the rest of this subsection, K will denote a subgroup of B n such that γ K = S n−1 × S 1 , hence, K F (S n−1 × S 1 ). In addition, unless otherwise stated, the integer n will be assumed to be at least 11. Notation 7.25. We denote by φ the natural isomorphism φ : F (S n−1 × S 1 ) → B n−1 × B 1 . For λ ∈ F (S n−1 × S 1 ) and (a, b) ∈ B n−1 × B 1 such that φ(λ) = (a, b), we denote by λ α the element of F (S n−1 × S 1 ) such that φ(λ α ) = (a, id B 1 ). Similarly, we will denote by λ β the element of F (S n−1 × S 1 ) such that φ(λ β ) = (id B n−1 , b). Clearly, we have λ = λ α λ β .
In this notation, we now have the following two subgroups: For the sake of completeness, let us show that Λ α K is indeed a subgroup of B n . If λ = λ α λ β , then λ −1 = λ −1 α λ −1 β . Therefore, for λ ∈ K, we have λ −1 α ∈ Λ α K . Also, if λ, λ are two elements of K, then (λλ ) α = λ α λ α . Therefore, Λ α K is closed under products and taking inverses; hence, it is a subgroup. The fact that Λ β K is a subgroup of B n can be shown in a similar way.
Proof. We already know that B n−1 × B 1 is a strong Gelfand subgroup of B n , so, let us assume that K = B n−1 × B 1 . Recall that K is an index 1 or 2 subgroup of Λ α K Λ β K . If Λ β K = {id K }, then by Corollary 7.29 K is as in 2., 4., or 6. We proceed with the assumption that Λ β K = {id K }. In this case K can be a subgroup of the form K × Λ β K , where K is an index 2 subgroup of Λ α K . However, in this case, Λ α K can only be B n−1 ; otherwise, if Λ α K = D n−1 or H n−1 , we would have that K = J n−1 and thus that, γ K = A n−1 , which would contradict our assumption that γ K = S n−1 × S 1 . Thus, once again by Corollary 7.29, K can be as in 3. or 5. These options for K are the obvious options. For the non-obvious index 2 subgroups of Λ α K Λ β K , we apply our discussion from the beginning of this subsection.
The remaining possibilities for K are given by the non-obvious index 2 subgroups of G × B 1 , where G ∈ {B n−1 , D n−1 , H n−1 }. We already encountered them in Examples 7.21, 7.22, and 7.23. They account for the possibilities that we listed in the items 7., 8., 9. for G = B n−1 ; 10. for G = D n−1 ; and 11. for G = H n−1 . This finishes the proof of our lemma.
We now proceed to check the strong Gelfand property of the subgroups that we listed in Lemma 7.30. We will make use of several elementary results from Subsection 7.1.
By the discussion in Section 6.1.1, Every irreducible representation of D n−1 is either equal to res B n−1 D n−1 S λ,µ , where λ and µ are two distinct partitions with |λ| + |µ| = n − 1, or it is one of the two irreducible constituents of res B n−1 D n−1 S λ,λ , where 2|λ| = n − 1. By Frobenius reciprocity, for every irreducible representation V of D n−1 , we have exactly one of the following two cases: 1. ind B n−1 D n−1 V = S λ,λ is an irreducible representation of B n−1 if V is one of the two irreducible constituents of res B n−1 D n−1 S λ,λ for some partition λ such that 2|λ| = n − 1;
We will analyze the induced representations ind Bn D n−1 ×B 1 V 1 B 1 . By the transitivity of the induction, we have If V is as in item 1., then (7.31) gives ind Bn is a strong Gelfand subgroup of B n , the resulting induced representation is multiplicity-free. Likewise, if V is as in item 2., then (7.31) together with part 1. of Lemma 7.7 give In this case also, since λ = µ, we see that the irreducible representations of B n that appear in (7.32) are inequivalent. Thus, we proved that (B n , D n−1 × B 1 ) is a strong Gelfand pair. We now analyze the pair (B n , D n−1 × S 1 ). Once again, by the transitivity of the induction, we have If V is as in item 2. above, then (7.33) and part 3. of Lemma 7.7 give Here, λ and µ are two distinct partitions such that |λ| + |µ| = n − 1. If n − 1 = 2m + 1 for some m ∈ N, then we consider the partitions λ = (m) and µ = (m + 1). It is easily checked that the multiplicity of S (m+1),(m+1) in (7.34) is 2. On the other hand, if n − 1 = 2m for some m ∈ N, then it is easy to check that (7.34) is a multiplicity-free B n representation. Next, we consider the irreducible representations V of D n−1 as in item 1. In particular, n − 1 is even. Let λ be a partition such that 2|λ| = n − 1, let S λ,λ denote the corresponding irreducible representation of B n−1 . By part 3. of Lemma 7.7, we get Clearly, the irreducible constituents of (7.35) are inequivalent, hence, ind Bn D n−1 ×S 1 V 1 S 1 is a multiplicity-free representation of B n . In summary, we proved that (B n , D n−1 × S 1 ) is a strong Gelfand pair if and only if n is odd.

ind
H n−1 S λ,µ , where λ = µ and |λ| + |µ| = n − 1. Now let V be an irreducible representation of H n−1 as in 1. Since ind In this case also, it is easy to verify that (7.36) is a multiplicity-free representation of B n . Therefore, (B n , H n−1 × B 1 ) is a strong Gelfand pair. We now proceed to the case of (B n , H n−1 × S 1 ). In this case, if V is as in 2., then by part 3. of Lemma 7.7 we get Let n − 1 = 2m + 1 for some m ∈ N, and set λ = (m + 1) and µ = (1 m ). Then by Pieri's rule we see that the multiplicity of S (m+1),(1 m+1 ) in (7.37) is 2. Thus, if n is even, then (B n , H n−1 × S 1 ) is not a strong Gelfand pair. If n is odd, then it is easy to check that the induced representation (7.37) is a multiplicity-free B n representation.
Next, we consider the irreducible representations V of H n−1 as in 1. Then n is odd. By part 3. of Lemma 7.7, we get Clearly, the irreducible constituents of (7.38) are inequivalent, and hence ind Bn H n−1 ×S 1 V 1 S 1 is multiplicity-free. In summary, we proved that (B n , H n−1 × S 1 ) is a strong Gelfand pair if and only if n is odd.
We start with the case (B n , (B n−1 × B 1 ) δ ). To ease our notation, let us denote (B n−1 × B 1 ) δ by M . Let ν denote the linear character of B n−1 × B 1 such that ker ν = M . Then it is easy to check that ν| B n−1 ×{1} = δ B n−1 and ν| Let W = S λ,µ (D; 1) be an irreducible representation of B n−1 × B 1 . There are two possibilities: 1) W is a self-associate representation with respect to ν, or 2) W and νW are associate representations with respect to ν. However, since δ B n−1 S λ,µ = S µ,λ and δ 2 If n − 1 = 2m + 1 for some m ∈ N, then we fix a pair of partitions (λ, µ) such that λ is obtained from µ by removing a box from its Young diagram. Then it is easy to check that S µ,µ appears twice (7.39). If n is odd, then it is easy to check that (7.39) is multiplicity-free for any λ and µ. Therefore, we proved that (B n , (B n−1 × B 1 ) δ ) is a strong Gelfand pair if and only if n is odd. Next, we consider the pair (B n , (B n−1 × B 1 ) εδ ). To ease notation, we denote (B n−1 × B 1 ) εδ by N . We know that H n−1 × {1} is an index 2 subgroup of N , and N is an index 2 subgroup of B n−1 × B 1 . We will describe the irreducible representations of N . Let ν denote the linear character of B n−1 × B 1 such that ker ν = N . Then the restrictions of ν to the factors are given by ν| B n−1 ×{1} = εδ and ν| {1}×B 1 = δ B 1 . Let W = S λ,µ (D; 1) be an irreducible representation of B n−1 × B 1 . We have two possibilities here: 1) W is a self-associate representation with respect to ν, 2) W and νW are associate representations with respect to ν. However, since δS λ,µ = S µ ,λ and δ 2 where {D,D} = {1, }. Since {D,D} = {1, }, the representations S λ,µ (D; 1) and S µ ,λ (D; 1) are inequivalent. Thus, similarly to the previous case, there is no self-associate irreducible representation with respect to ν. Now let V be an irreducible representation of N . Then we have ind for some irreducible representation S λ,µ of B n−1 . By Lemma 7.7, ind Bn N V must be τ ∈λ If n − 1 = 2m + 1 for some m ∈ N, then we fix a pair of partitions (λ, µ) such that λ is obtained from µ by removing a box from its Young diagram. Then we see that the multiplicity of S µ ,µ in (7.40) is 2. If n is odd, then it is easy to check that (7.40) is multiplicity-free for any λ and µ. Therefore, we proved that (B n , (B n−1 × B 1 ) εδ ) is a strong Gelfand pair if and only if n is odd.
To ease our notation, let us denote (B n−1 ×B 1 ) ε by Z. We know that (F A n−1 )×{1} is an index 2 subgroup of Z, and Z is an index 2 subgroup of B n−1 × B 1 . We will describe the irreducible representations of Z. Let ν denote the linear character of B n−1 × B 1 such that ker ν = Z. Then the restrictions of ν to the factors are given by ν| B n−1 ×{id} = ε B n−1 and ν| {id}×B 1 = δ B 1 . Let W = S λ,µ (D; 1) be an irreducible representation of B n−1 × B 1 . We have two possibilities: 1) W is a self-associate representation with respect to ν, 2) W and νW are associate representations with respect to ν. But since there is no self-associate irreducible representation of B 1 with respect to δ B 1 , there is no self-associate irreducible representation of B n−1 × B 1 with respect to ν.
Let V be an irreducible representation of Z. Then for some irreducible representation S λ,µ of B n−1 , we have ind It is easy to see that this representation is multiplicity-free. Thus, (B n , Z) is a strong Gelfand pair.
To ease our notation, let us denote (D n−1 × B 1 ) εδ by K. Let ν denote the linear character of D n−1 × B 1 such that ker ν = K. Since K D n−1 × F , the restrictions of ν to the factors are given by ν| D n−1 ×{id} = (εδ) B n−1 and ν| {id}×B 1 = (εδ) B 1 = δ B 1 . Let U (D; 1) be an irreducible representation of D n−1 × B 1 , where D ∈ {1 F , F }. Since (1 F ; 1) and ( F ; 1) are δ B 1 -associate representations, there is no self-associate representations of D n−1 × B 1 with respect to ν. In particular, every irreducible representation of D n−1 × B 1 restricts to K as an irreducible representation. Thus, if V is an irreducible representation of K, then ind D n−1 ×B 1 K V is of the form U (1 F ; 1) ⊕ (εδU ) ( F ; 1). Let S λ,µ be the irreducible representation of B n−1 such that res B n−1 D n−1 S λ,µ = U , where λ and µ are two partitions such that |λ| + |µ| = n − 1. First we assume that λ = µ. Then we have Then by Lemma 7.7, we find that It is easy to check that this is a multiplicity-free representation of B n if n − 1 is even. If n − 1 is odd, then we choose λ = (m) and µ = 1 m+1 . Then S (m+1),(1 m+1 ) appears with multiplicity 2 in (7.42). Next, we assume that λ = µ. Of course, this choice is available only when n − 1 is even. Then we have Clearly, this representation is multiplicity-free if and only if λ = λ ; we can find self-conjugate partitions of n − 1 as long as n − 1 > 2. Therefore, (B n , K) is not a strong Gelfand pair for n > 3. If n = 3, then (B n , K) is a strong Gelfand pair. By a similar argument, one can also deduce that (B n , (H n−1 × B 1 ) δ ) is a strong Gelfand pair if and only if n = 3.

Summary for
We now summarize the conclusions of the previous subsections in a single proposition. In particular, we maintain our notation from Lemma 7.30. Proposition 7.43. Let n 7. Let K be a subgroup of B n such that γ K = S n−1 × S 1 . In this case, (B n , K) is a strong Gelfand pair if and only if K is one of the following subgroups: Throughout this subsection, we will assume that n 8, K will be a subgroup of B n such that γ K = S n−2 × S 2 , and hence, K F (S n−2 × S 2 ). The idea of our analysis in this subsection is the same as the one that we used in the previous subsection.
Lemma 7.49. Let L denote the following subgroup of Λ α K : Then L is a normal subgroup of K, that is nontrivial if n 5. Furthermore, the following hold: Note that Cases 1. and 2. are not necessarily distinct. For example, an index 4 subgroup of B n−2 × B 2 might be an index 2 subgroup of D n−2 × B 2 .
We are now ready to determine all strong Gelfand subgroups of K B n such that γ K = S n−2 × S 2 . We will first examine subgroups as in Case 3.
Notation 7.52. For brevity, in the following subsections, we will denote the identity element (0, id B 2 ) of B 2 by 1, and we will denote the element (0, (1 2)) by −1. By Proposition 4.8 we know that (B n , B n−2 × S 2 ) is a strong Gelfand pair. Therefore, for any G such that S 2 G B 2 , the pair (B n , B n−2 × G) is a strong Gelfand pair. There is one more subgroup that we have to check, that is, G = H 2 . In this case, we see from Corollary 7.19 that (B n , B n−2 × G) is a strong Gelfand pair. (Of course, we could have used the same method for the subgroups G, where S 2 G B 2 .) First, we assume that n is an even number such that n − 2 = 2m for some m 3. We let λ denote the partition (m − 1, 1), and let µ denote the partition (m). Then V = res (1) is an irreducible representation of D n−2 × B 2 . By transitivity of induction, Lemma 2.5, and Lemma 7.16, part 3, we have Since the multiplicity of S (m,1),(m,1) in (7.53) is two, we see that (B n , D n−2 × B 2 ) is not a strong Gelfand pair. Next, we assume that n is odd. Let (λ, µ) be a pair of partitions such that |λ| + |µ| = n − 2. Then we have |λ| = |µ|. Hence, V = res B n−2 D n−2 S λ,µ is an irreducible representation of D m−2 , and furthermore, we have ind B n−2 D n−2 V = S λ,µ ⊕ S µ,λ . Let (a, b) be a pair of partitions such that |a| + |b| = 2. As before, by using the transitivity of the induction and Lemma 2.5, we get But since |λ| and |µ| are not equal, we see from Lemma 7.16 that (7.54) is multiplicity-free. In summary, we proved the following result Let n = 2m for some m 4. First, we assume that m is an even integer as well; m = 2k with k 2. Let λ = µ = (k + 1, 1 k−1 ), and note that λ = µ . Then V = res B n−2 H n−2 S λ,µ is an irreducible representation of H n−2 , and therefore, ind (1) is an irreducible representation of H n−2 × B 2 . By transitivity of induction, Lemma 2.5, and Lemma 7.16, part 3, we have Since the multiplicity of S (k+1,1 k ),(k+1,1 k ) in (7.53) is 2, we see that (B n , D n−2 × B 2 ) is not a strong Gelfand pair. Now suppose that m = 2k + 1 with k 2, and set λ := (k + 1, 1 k ) and µ := (k, 1 k+1 ). Clearly, λ is a self-conjugate partition and λ = µ. Then V = res It is easy to check that the multiplicity of S (k+2,1 k ),(k+1,1 k+1 ) in (7.57) is 2. Hence, if n is even, then (B n , D n−2 × B 2 ) is not a strong Gelfand pair. Next, we assume that n is odd. Then, by arguing as in the second part of the (B n , D n−2 ×B 2 ) case, it is easy to verify that, for every irreducible representation W of B 2 and for every pair of partitions (λ, µ) such that |λ| + |µ| = n, the induced representation ind Bn H n−2 ×B 2 S λ,µ W is multiplicity-free. In summary, similarly to the case of (B n , D n−2 × B 2 ), we proved the following result. Since D n−2 × D 2 is a subgroup of D n−2 × B 2 , if n is even, then by Lemma 7.55 (B n , D n−2 × D 2 ) is not a strong Gelfand subgroup. So, we proceed with the assumption that n = 2m + 1 for some m 4.
Let λ and µ be two partitions such that |λ| + |µ| = n − 2, and let S λ,µ denote the corresponding irreducible representation of B n−2 . Since |λ| = |µ|, the restricted representation V = res B n−2 D n−2 S λ,µ is an irreducible representation of D n−2 . Furthermore, we have ind ,∅ is an irreducible representation of D n−2 × D 2 . By transitivity of induction, Lemma 2.5, and Lemma 7.16, parts 1 and 5, we have Since ||λ| − |µ|| is odd, the representation (7.59) is multiplicity-free. By using similar arguments, we see that ind Bn D n−2 ×D 2 V S ∅, (2) , ind Bn D n−2 ×D 2 V S ∅,(1 2 ) , and ind Bn D n−2 ×D 2 V S (1 2 ),∅ are multiplicity-free representations of B n . Finally, we notice that ind Bn D n−2 ×D 2 V S (1),(1) = ind Bn D n−2 ×B 2 V S (1), (1) , hence, it is also multiplicity-free (by Lemma 7.55). Therefore, we proved the following result. The proof of this case is similar to that of Lemma 7.60. By Lemma 7.58, if n is even, then we know that (B n , H n−2 × D 2 ) is not a strong Gelfand pair. We proceed with the assumption that n is an odd number of the form n = 2m + 1 for some m 4. Let λ and µ be two partitions such that |λ| + |µ| = n − 2, and let S λ,µ denote the corresponding irreducible representation of B n−2 . Since λ = µ , V = res B n−2 H n−2 S λ,µ is an irreducible representation of H n−2 , and furthermore, we have ind From this point on, we argue as in the proof of Lemma 7.60. We omit the details but write the conclusion below. Since D n−2 × H 2 and H n−2 × H 2 are subgroups of D n−2 × B 2 and H n−2 × B 2 , respectively, if n is even, then by Lemmas 7.55 and 7.58 (B n , D n−2 × H 2 ) and (B n , H n−2 × H 2 ) are not strong Gelfand pairs. So, we proceed with the assumption that n = 2m + 1 for some m 4. In this case, the proofs of Lemmas 7.60 and 7.61 are easily modified, and we get the following result.   Proof. Suppose n − 2 = 2m + 1 for some m 4. Let λ = (m) and µ = (m + 1). Then S λ,µ ⊕ S µ,λ is a representation of B n−2 that is induced from an irreducible representation V of D n−2 . Let W denote the trivial representation of S 2 . By Corollary 7.20, we see that S (m+1,1),(m+1) has multiplicity 2 in ind Bn Next, suppose that n−2 = 2m for some m 4. Let λ = (m) and µ = (1 m ). Then S λ,µ ⊕S µ,λ is a representation of B n−2 that is induced from an irreducible representation V of D n−2 . Let W denote the trivial representation of S 2 . By Corollary 7.20, we see that S (m+1),(1 m+1 ) has multiplicity 2 in ind Bn D n−2 ×S 2 V W . This completes the proof.  Proof. Suppose n−2 = 2m+1 for some m 4. Let λ = (m) and µ = (1 m+1 ). Then S λ,µ ⊕S µ ,λ is a representation of B n−2 that is induced from an irreducible representation V of H n−2 . Let W denote the trivial representation of S 2 . By Corollary 7.20, we see that S (m+1,1),1 (m+1) has multiplicity 2 in ind Bn Next, we assume that n − 2 = 2m for some m 4. Let λ = (m − 1, 1) and µ = (1 m ). Then S λ,µ ⊕ S µ ,λ is a representation of B n−2 that is induced from an irreducible representation V of H n−2 . Let W denote the trivial representation of S 2 . By Corollary 7.20, we see that S (m,1),(2,1 m−1 ) has multiplicity 2 in ind Bn H n−2 ×S 2 V W . This completes the proof. There are two non-direct product index 2 subgroups K B n−2 × S 2 such that γ K = S n−2 × S 2 : , δ(a)) : a ∈ B n−2 }; We begin with the case K = (B n−2 × S 2 ) δ . Let ν denote the linear character of B n−2 × S 2 such that ker ν = K. Then the restrictions of ν to the factors are given by ν| B n−2 ×{1} = δ and ν| {id}×S 2 = ε. Let W = S λ,µ D be an irreducible representation of B n−2 × S 2 . Since δS λ,µ = S µ,λ , ε = 1, and ε1 = , we have where {D,D} = {1, }. In particular, the representations S λ,µ D and S µ,λ D are inequivalent. Hence, there is no self-associate irreducible representation with respect to ν.
We are now ready to describe the induction from K to B n−2 × S 2 by using Frobenius reciprocity. Let V be an irreducible representation of K. Then we have for some irreducible representation S λ,µ of B n−2 . Here, λ and µ may be any partitions with |λ| + |µ| = n − 2. It is now easy to see from Corollary 7.20 that if we induce the representation in (7.65), then we will get a non-multiplicity-free representation of B n . Indeed, for n = 2m, we can choose λ = µ, and for n = 2m + 1 we can choose λ = (m) and µ = (m + 1). Therefore, (B n , (B n−2 × S 2 ) δ ) is not a strong Gelfand subgroup. Next, we focus on the case K = (B n−2 × S 2 ) εδ . We know that H n−2 × {1} is an index 2 subgroup of K, and K is an index 2 subgroup of B n−2 × S 2 . We begin with describing the irreducible representations of K. Let ν denote the linear character of B n−2 × S 2 such that ker ν = K. Then the restrictions of ν to the factors are given by ν| B n−2 ×{1} = εδ and ν| {id}×S 2 = ε.
Let W = S λ,µ D be an irreducible representation of B n−2 × S 2 . Since εδS λ,µ = S µ ,λ , ε = 1, and ε1 = , we have where {D,D} = {1, }. Since D =D, the representations S λ,µ D and S µ ,λ D are inequivalent. Hence, we conclude that there is no self-associate irreducible representation with respect to ν. We are now ready to describe the induction from K to B n−2 × S 2 by using Frobenius reciprocity. Let V be an irreducible representation of K. Then we have for some irreducible representation S λ,µ of B n−2 . Here, λ and µ can be any partitions with |λ| + |µ| = n − 2. It is now easy to see from Corollary 7.20 that if we induce the representation in (7.66), then we will get a non-multiplicity-free representation of B n . Indeed, for n = 2m, can choose V with λ = µ , and for n = 2m + 1 we can choose λ = (1 m ) and µ = (m + 1). Therefore, (B n , (B n−2 × S 2 ) εδ ) is not a strong Gelfand subgroup.
Lemma 7.67. If n 8, then there is no strong Gelfand pair of the form (B n , K), where K is a non-direct product index 2 subgroup of B n−2 × S 2 such that γ K = S n−2 × S 2 .
Therefore, the restrictions of both of these representations to K give the same irreducible representation, C := res B n−2 ×D 2 K S λ,µ V = res B n−2 ×D 2 K S µ,λ Ṽ .
If n is odd, then the sum is easily seen to be multiplicity-free, by considering parities of the partitions involved. If C is as in b), then ind Bn K C = τ ∈λ,ρ∈μ S τ,ρ ⊕ α∈μ,β∈λ S α,β .
If n is even, then the sum is easily seen to be multiplicity-free, by considering parities of the partitions involved.
The case where ker ν| B n−2 ×{1} = H n−2 × {1} is almost identical, as is the result in that case. We summarize these below.
For the action of χ 2 on {V 0 , . . . , V 3 } we have χ 2 V 0 ∼ = V 2 and χ 2 V 1 = V 3 . We proceed with the assumption that V = V 0 andṼ = V 2 in E. Then, by Corollary 7.19 we have If n is even, then we may set λ = (m − 1) and µ = (m, 1), and see that S (m,1),(m,1) appears with multiplicity 2. If n is odd, then the sum is easily seen to be multiplicity-free, by considering parities of the partitions involved.
We proceed with the assumption that V = V 1 andṼ = V 3 in E. Then, by Corollary 7.19 we have If n is even, then we may set λ = µ. Then the sum is not multiplicity-free. If n is odd, then the sum is easily seen to be multiplicity-free, by considering parities of the partitions involved.
The case where ker ν| B n−2 ×{1} = H n−2 × {1} is almost identical, as is the result in that case. We summarize these below.
Lemma 7.69. Let n 8, and let K be a non-direct product index 2 subgroup of B n−2 × H 2 with γ K = S n−2 × S 2 . If n is even, then K is not a strong Gelfand subgroup. If n is odd, there are two such subgroups K that are strong Gelfand subgroups. 7.3.12 Non-direct product index 2 subgroups of B n−2 × B 2 .
2) K is a non-direct product index 2 subgroup of K := H n−2 × B 2 . This case develops essentially in the same way as the previous case does; K is not a strong Gelfand subgroup of B n . We omit the details for brevity.
3) K is an index 2 subgroup of a non-direct product index 2 subgroup K of B n−2 × B 2 .

Exceptional cases
Finally, we add some closing remarks regarding the missing cases for small n. Note that the strong Gelfand subgroups that we have found in Sections 6 and 7, and collected in Table 1, are all still strong Gelfand pairs in these small cases. Our lower bounds on n come in when reducing the possible cases to check, and are thus required only in order for the corresponding parts of Table 1 to be exhaustive. For γ K = S n or A n , we have filled in these extra cases along the way, and they appear in Propositions 6.21 and 6.30. With more work, we could have considered these extra subgroups when γ K = S n−1 × S 1 or S n−2 × S 2 , for example extending Corollary 7.29 and Lemma 7.30 to pick up extra possible subgroups for n 6.
Further to these, there are also missing cases for n = 4, 5, and 6 arising due to the extra possibilities for γ K -see [2,Theorem 4.13].
Instead of an extensive body of work to explicitly give a long list of all strong Gelfand subgroups in these few small cases, we have instead computed them in GAP, and the following proposition (along with Lemma 7.48) summarize the number of these in each case, with finer detail for B 3 .
Up to conjugation, the other small hyperoctahedral groups have the following numbers of strong Gelfand subgroups.