Strong Gelfand Pairs of the Symplectic Group Sp₄(q) Where q Is Even

Abstract

A strong Gelfand pair $(G,H)$ is a finite group G together with a subgroup H such that every irreducible character of H induces to a multiplicity-free character of G. We classify the strong Gelfand pairs of the symplectic groups ${\mathrm{Sp}}_4\left(q\right)$ for even q.

1. Introduction

For a finite group G we let $\widehat{G}$ denote the set of irreducible characters of G. Then a multiplicity-free character of G is a character $\chi$ of G such that for $\psi \in \widehat{G}$, we have $\langle \chi ,\psi \rangle \le 1$. Here only complex characters are considered.

A Gelfand pair $(G,H)$ is a finite group G together with a subgroup H such that the trivial character of H induces a multiplicity-free character of G. The importance of Gelfand pairs is indicated by six equivalent conditions; see  [1,2,3,4].

A strong Gelfand pair $(G,H)$ is a finite group G and $H\le G$ such that for every $\psi \in \widehat{H}$ the induced character $\psi \uparrow G$ is multiplicity free. We will call H a strong Gelfand subgroup of G in this situation. Equivalently, $(G,H)$ is a strong Gelfand pair if and only if the Schur ring determined by the H-classes $g^H=\{g^h:h\in H\},g\in G$, is commutative [3,5,6]. Here our convention is: $g^h=h^{-1}gh$. Note that $(G,G)$ is always a strong Gelfand pair.

In this paper we continue our investigation of strong Gelfand pairs of groups that are close to being simple; in [1,2,7] we found all such pairs for $G=\mathrm{SL}(2,p^n),n\ge 1,$ p a prime, and  the symmetric groups. We refer to [2,3,7] for necessary background and to [8] for some of the latest results on strong Gelfand pairs.

We note that Gelfand pairs and strong Gelfand pairs have applications in representation theory; see [8,9,10,11] among many other references. As explained above, an equivalent condition for $(G,H)$ to be a strong Gelfand pair is that the Schur ring determined by the H-classes is commutative. This shows that a strong Gelfand pair determines a commutative Schur ring and so a commutative association scheme, which then indicates connections with algebraic combinatorics. One other application of strong Gelfand pairs is to random walks on finite groups: if $(G,H)$ is a strong Gelfand pair, then one can define a random walk on G using probabilities that are constant on the above mentioned H-classes. The commutativity property of the H-classes means that the random walk is `diagonalizable’ and so can be very well understood.

This paper will consider strong Gelfand pairs for the symplectic groups Sp₄(2ⁿ) as their irreducible characters are known [12]. In contrast, the irreducible characters of ${\mathrm{Sp}}_{2k}\left(q\right),k>2,$ are not understood, where q is a prime power. The  groups ${\mathrm{Sp}}_4\left(2^n\right),n>1,$ are simple and the representation theory of these groups is considered in [13]. The main result of this paper is:

Theorem 1. 

The only strong Gelfand pair (${\mathrm{Sp}}_4\left(2^n\right),H),n\ge 2,$ is where $H={\mathrm{Sp}}_4\left(2^n\right)$.

Throughout we will use the standard Atlas notation [14].

2. Preliminary Results

All groups considered in this paper will be assumed finite. For a group G, the total character of G, denoted ${\tau }_G$, is the sum of all the irreducible characters of G; see [15,16,17]. The following gives the `total character argument’ for showing that certain subgroups are not strong Gelfand subgroups:

Lemma 1 

(Lemma $3.3$ [7]). Let $H\le G$ be groups. If there is $\chi \in \widehat{G}$ with

$$deg\left({\tau }_H\right)<deg\left(\chi \right),$$

then $(G,H)$ is not a strong Gelfand pair.

The following indicates that it is important to determine which maximal subgroups are strong Gelfand pairs.

Lemma 2 

(Lemma $3.1$ [2]). Suppose we have groups $H\le K\le G$. If $(G,K)$ is not a strong Gelfand pair, then neither is $(G,H)$.

For $q=2^e$, $e>1$, we find from Table $8.14$ of [18] that the maximal subgroups of ${\mathrm{Sp}}_4\left(q\right)$ are as listed in

Table 1. Maximal subgroups of ${\mathrm{Sp}}_4\left(q\right)$, for $q=2^e,e>1$.

Group	Order	Conditions
$\phantom{E^{3^2}}{E}_q^3:{\mathrm{GL}}_2\left(q\right)$	$q^3·(q^2+q){(q-1)}^2$
$\phantom{E^{3^2}}{E}_q^3:{\mathrm{GL}}_2\left(q\right)$	$q^3·(q^2+q){(q-1)}^2$
${\mathrm{Sp}}_2\left(q\right)\wr 2$	$2q^2{(q^2-1)}^2$
${\mathrm{Sp}}_2\left(q^2\right):2$	$2q^2(q^4-1)$
${\mathrm{Sp}}_4\left(q_0\right)$	$q_0^4(q_0^2-1)(q_0^4-1)$	$q=q_0^r,r$ is prime
${\mathrm{SO}}_4^{+}\left(q\right)$	$2q^2{(q^2-1)}^2$
${\mathrm{SO}}_4^{-}\left(q\right)$	$2q^2(q^4-1)$
$\mathrm{Sz}\left(q\right)$	$q^2(q^2+1)(q-1)$	e odd

.

Table 1 has the maximal subgroup $E_q^3:{\mathrm{GL}}_2\left(q\right)$ listed twice because there are two non-conjugate maximal subgroups of ${\mathrm{Sp}}_4\left(q\right)$ which are isomorphic to $E_q^3:{\mathrm{GL}}_2\left(q\right)$.

From Lemma 2 Theorem 1 will follow if we can show that none of these maximal subgroups is a strong Gelfand subgroup. We consider each case separately.

The next two results will allow us to assume $e\ge 3$.

We first consider the symplectic group ${\mathrm{Sp}}_4\left(2\right)$; since ${\mathrm{Sp}}_4\left(2\right)\cong S_6$ the result here follows from our consideration of the symmetric groups in [1]:

Proposition 1. 

The only proper subgroups of ${\mathrm{Sp}}_4\left(2\right)$ which are strong Gelfand subgroups are the maximal subgroups.

Proposition 2. 

No proper subgroup of ${\mathrm{Sp}}_4\left(4\right)$ is a strong Gelfand subgroup.

Proof. 

We use the MAGMA [19] code given in the Appendix A to obtain this result.    □

In what follows we will often have the situation where $H\le G,|G:H|=2$. We introduce the following conventions. For $\psi \in \widehat{H}$ it is well-known [20] that either

(i)

$\psi \uparrow G$ is a sum of two distinct characters in $\widehat{G}$ (call this the splitting case); or

(ii)

$\psi \uparrow G$ is irreducible (call this the fusion case).

In the splitting case, if  $\psi \uparrow G={\chi }_1+{\chi }_2,{\chi }_1,{\chi }_2\in \widehat{G}$, then ${\chi }_i\downarrow H=\psi ,i=1,2.$

In the situation $|G:H|=2$ the relationship between ${\tau }_G$ and ${\tau }_H$ is given in:

Lemma 3. 

Let $H\le G,|G:H|=2$. Let $\mathcal{S}$ be the set of $\psi \in \widehat{H}$ that split and let $\mathcal{F}$ be the set of $\psi \in \widehat{H}$ that fuse. Then

$${\tau }_G\left(1\right)=2\sum _{\psi \in \mathcal{S}}\psi \left(1\right)+\sum _{\psi \in \mathcal{F}}\psi \left(1\right).$$

The character table for ${\mathrm{Sp}}_4\left(q\right)$ is given in [12] and we will use notation from [12].

Theorem 2 

([12]).  

(i)

The degree of the total character of ${\mathrm{Sp}}_4\left(q\right)$ is $q^6+q^4-q^2$ if q is even.

(ii)

The largest degree of an irreducible character of ${\mathrm{Sp}}_4\left(q\right)$ is $q^4+2q^3+2q^2+2q+1$ when $q\ge 4$ is a power of 2.

Proof. 

(i) We just sum the degrees of characters of ${\mathrm{Sp}}_4\left(q\right)$ as listed in [12]. (ii) follows directly from [12].    □

Lemma 4. 

If $q=2^e,e>1$, then ${\mathrm{Sp}}_2\left(q\right)\wr 2\cong {\mathrm{SO}}_4^{+}\left(q\right)$ and ${\mathrm{Sp}}_2\left(q^2\right):2\cong {\mathrm{SO}}_4^{-}\left(q\right)$.

Proof. 

See Proposition $7.2.1$ and Table $8.14$ of [18].    □

We now consider the maximal subgroups separately in the following sections.

3. The Maximal Subgroup ${\mathbf{Sp}}_{\mathbf{2}}\left(\mathit{q}\right)\wr \mathbf{2}$

Theorem 3. 

For $q=2^e,e>1,$ the maximal subgroup ${\mathrm{Sp}}_2\left(q\right)\wr 2\le {\mathrm{Sp}}_4\left(q\right)$ is not a strong Gelfand subgroup.

Proof. 

This proof will be a `total character argument’ and so we will need to find the total character of ${\mathrm{Sp}}_2\left(q\right)\wr 2$. We have ${\mathrm{Sp}}_2\left(q\right)\wr 2={\mathrm{Sp}}_2{\left(q\right)}^2$:2 and one way to represent the elements of ${\mathrm{Sp}}_2{\left(q\right)}^2$:2 is by $2\times 2$ blocks of $2\times 2$ matrices, where the cyclic subgroup 2 is generated by $\left[\begin{array}{cc}0& I_2\\ I_2& 0\end{array}\right]$ and $(a,b)\in {\mathrm{Sp}}_2{\left(q\right)}^2$ is represented as the block matrix $\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right].$

Now ${\mathrm{Sp}}_2\left(q\right)\cong {\mathrm{SL}}_2\left(q\right)$ has character table given in [21] (see also [7]); we reproduce it here in

Table 2. Character Table for ${\mathrm{SL}}_2\left(q\right)$ with q even.

Class	1	c	${\mathit{a}}^{\mathit{t}}$	${\mathit{b}}^{\mathit{m}}$
Size	1	$q^2-1$	$q(q+1)$	$q(q-1)$
Tr	1	1	1	1
$\psi$	q	0	1	$-1$
${\chi }_s$	$q+1$	1	${\rho }^{st}+{\rho }^{-st}$	0
${\theta }_j$	$q-1$	$-1$	0	$-({\sigma }^{jm}+{\sigma }^{-jm})$

. Here the parameters $s,t,j,m$ satisfy $1\le s,t\le (q-2)/2$, $1\le j,m\le q/2$, $\rho$ is a primitive $(q-1)$-th root of unity and $\sigma$ a primitive $(q+1)$-th root of unity.

Here the conjugacy classes of ${\mathrm{SL}}_2\left(q\right)$ are represented by powers of the following elements:

$$1=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],c=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right],a=\left[\begin{array}{cc}\rho & 0\\ 0& {\rho }^{-1}\end{array}\right],$$

and an element b of order $q+1$. We also give the sizes of the classes in Table 2.

Since ${\mathrm{Sp}}_2\left(q\right)\wr 2\cong {\mathrm{Sp}}_2{\left(q\right)}^2$:2 the irreducible characters of ${\mathrm{Sp}}_2\left(q\right)\wr 2$ are easily found using Table 2. In

Table 3. Character degrees for ${\mathrm{Sp}}_2\left(q\right)\wr 2$ with q even.

Character	Degree	Multiplicity
$\phantom{{\left(\right)}^{2^2}}{(\mathrm{Tr}\times \mathrm{Tr})}_1$	1	1
${(\mathrm{Tr}\times \mathrm{Tr})}_2$	1	1
$\mathrm{Tr}\times \psi$	$2·q$	1
$\mathrm{Tr}\times {\chi }_i$	$2·(q+1)$	$(q-2)/2$
$\mathrm{Tr}\times {\theta }_j$	$2·(q-1)$	$q/2$
${(\psi \times \psi )}_1$	$q^2$	1
${(\psi \times \psi )}_2$	$q^2$	1
$\psi \times {\chi }_s$	$2·q(q+1)$	$(q-2)/2$
$\psi \times {\theta }_j$	$2·q(q-1)$	$q/2$
${({\chi }_s\times {\chi }_s)}_1$	${(q+1)}^2$	$(q-2)/2$
${({\chi }_s\times {\chi }_s)}_2$	${(q+1)}^2$	$(q-2)/2$
${\chi }_s\times {\chi }_{s^{\prime }}$	$2·{(q+1)}^2$	$(q-2)(q-4)/8$
${\chi }_s\times {\theta }_j$	$2·(q^2-1)$	$q(q-2)/4$
${({\theta }_j\times {\theta }_j)}_1$	${(q-1)}^2$	$q/2$
${({\theta }_j\times {\theta }_j)}_2$	${(q-1)}^2$	$q/2$
${\theta }_j\times {\theta }_{j^{\prime }}$	$2·{(q-1)}^2$	$q(q-2)/8$

we give the degrees of the irreducible characters of ${\mathrm{Sp}}_2\left(q\right)\wr 2$. These character degrees are obtained using ([20], [Proposition $20.9$, Theorem $19.18$]). Further, in Table 3 we are assuming that

$$1\le s,s^{\prime }\le (q-2)/2,1\le j,j^{\prime }\le q/2\mathrm{and}s\ne s^{\prime },j\ne j^{\prime }.$$

In Table 3 the suffices $1,2$ are written to indicate that these are the split cases. The lack of such a suffix indicates the fusion cases. In Table 3 each case has a certain `Multiplicity’ that is also indicated; this depends on the parameters involved. Then from Table 3 we obtain the degree of the total character of ${\mathrm{Sp}}_2\left(q\right)\wr 2$:

$$\begin{array}{cc}\hfill \left({\tau }_{{\mathrm{Sp}}_2\left(q\right)\wr 2}\right)\left(1\right)& =1+1+2q+(q+1)(q-2)+(q-1)·q+2q^2+q·(q+1)·(q-2)\hfill \\ & +q^2(q-1)+{(q+1)}^2·(q-2)+{(q+1)}^2·(q-2)·(q-4)/4\hfill \\ & +(q^2-1)·q·(q-2)/2+{(q-1)}^2·q+{(q-1)}^2·q·(q-2)/4\hfill \\ & =q^4+q^3-q.\hfill \end{array}$$

Now $q^4+q^3-q<q^4+2q^3+2q^2+2q+1$, and  by Theorem 2$q^4+2q^3+2q^2+2q+1$ is the degree of an irreducible character of ${\mathrm{Sp}}_4\left(q\right)$. Then by Lemma 1$({\mathrm{Sp}}_4\left(q\right),{\mathrm{Sp}}_2\left(q\right)\wr 2)$ is not a strong Gelfand pair.    □

By Lemma 4 and the fact that the above argument is a `total character argument’ (not dependent on the particular embedding of ${\mathrm{Sp}}_2\left(q\right)\wr 2$ in ${\mathrm{Sp}}_4\left(q\right)$) we see that we have now also dealt with the maximal subgroup ${\mathrm{SO}}_4^{+}\left(q\right)$ case from Table 1:

Corollary 1. 

The maximal subgroup ${\mathit{SO}}_4^{+}\left(q\right)<{\mathrm{Sp}}_4\left(q\right)$ is not a strong Gelfand subgroup.

4. The Maximal Subgroups ${\mathit{E}}_{\mathit{q}}^{\mathbf{3}}$:${\mathbf{GL}}_{\mathbf{2}}\mathbf{\left(}\mathbf{q}\mathbf{\right)}$

By Theorems 1 and 2 we may assume that $q>4$.

Theorem 4. 

For $q=2^e,e>2$, the maximal subgroup $E_q^3:{\mathrm{GL}}_2\left(q\right)\le {\mathrm{Sp}}_4\left(q\right)$ is not a strong Gelfand subgroup.

Proof. 

In [12] two isomorphic maximal subgroups are considered; they are denoted P and Q. The orders of P and Q are $q^3(q^2+q){(q-1)}^2$ and they are isomorphic to $E_q^3$:${\mathrm{GL}}_2\left(q\right)$. The character tables for these subgroups are given in [12].

We take the inner product of the character of P denoted by ${\chi }_5\left(k\right)$ in [12] with a character of ${\mathrm{Sp}}_4\left(q\right)$ restricted to P, namely ${\chi }_1(m,n)\downarrow P$. In what follows $A_i,A_{ij},C_j,D_k$ is the notation used in [12] for the classes of P; further, the sizes of these classes are also given in [12]. Using all of this information we obtain:

$$\begin{array}{cc}\hfill & \langle {\chi }_5\left(k\right),{\chi }_1(m,n)\downarrow P\rangle \hfill \\ & ={\displaystyle \frac{1}{\left|P\right|}}\left(\right|A_1|q(q^2-1){(q+1)}^2(q^2+1)+|A_2|q(q-1){(q+1)}^2+\left|A_{31}\right|(-q)(q+1){(q+1)}^2\hfill \\ \hfill & +|A_{32}|(-q)(2q+1)+|C_2\left(i\right)|(q-1){\alpha }_{ik}(q+1){\alpha }_{im}{\alpha }_{in}+\left|D_2\left(j\right)\right|(-{\alpha }_{jk}){\alpha }_{jm}{\alpha }_{jn})\hfill \\ & ={\displaystyle \frac{1}{q^4(q-1)(q^2-1)}}(q(q^2-1){(q+1)}^2(q^2+1)\hfill \\ \hfill & +(q^2-1)q(q-1){(q+1)}^2+(q-1)(-q)(q+1){(q+1)}^2\hfill \\ \hfill & +(q-1)(q^2-1)(-q)(2q+1)\hfill \\ \hfill & +\sum _{i=1}^{(q-2)/2}{q}^3(q+1)(q-1){\alpha }_{ik}(q+1){\alpha }_{im}{\alpha }_{in}+\sum _{j=1}^{(q-2)/2}{q}^3(q^2-1)(-{\alpha }_{jk}){\alpha }_{jm}{\alpha }_{jn})\hfill \\ \hfill & ={\displaystyle \frac{1}{q^7-q^6-q^5+q^4}}(q^7+2q^6+q^5-q^3-2q^2-q+q^6+q^5-2q^4-2q^3+q^2\hfill \\ \hfill & +q-q^5-2q^4+2q^2+q-2q^5+q^4+3q^3-q^2-q\hfill \\ \hfill & +\sum _{i=1}^{(q-2)/2}(q^6+q^5-q^4-q^3){\alpha }_{ik}{\alpha }_{im}{\alpha }_{in}+\sum _{j=1}^{(q-2)/2}(-q^5+q^3){\alpha }_{jk}{\alpha }_{jm}{\alpha }_{jn})\hfill \\ \hfill & ={\displaystyle \frac{1}{q^7-q^6-q^5+q^4}}\left(q^7+3q^6-q^5-3q^4+\left(q^6-q^4\right)\sum _{j=1}^{(q-2)/2}{\alpha }_{jk}{\alpha }_{jm}{\alpha }_{jn}\right)\hfill \\ & ={\displaystyle \frac{3+q}{q-1}}+{\displaystyle \frac{1}{q-1}}\left(\sum _{j=1}^{(q-2)/2}{\alpha }_{jk}{\alpha }_{jm}{\alpha }_{jn}\right).\hfill \end{array}$$

Here ${\alpha }_{ij}={\overline{\gamma }}^{ij}+{\overline{\gamma }}^{-ij}$ where $\langle \gamma \rangle ={\mathbb{F}}_q^{\times }$, and $\overline{\gamma }$ is the image of $\gamma$ under a fixed monomorphism from ${\mathbb{F}}_q^{\times }$ into ${\mathbb{C}}^{\times }$, making $\overline{\gamma }$ a $(q-1)$-th root of unity. For clarity of notation, in our calculations we omit the overline.

Now supposing that $q>5$, if we have ${\sum }_{i=1}^{(q-2)/2}{\alpha }_{ik}{\alpha }_{im}{\alpha }_{in}=q-5$, then the above gives $\langle {\chi }_5\left(k\right),{\chi }_1(m,n)\downarrow P\rangle =2$. We will now show that there is a choice of $k,m,n$ so that ${\sum }_{i=1}^{(q-2)/2}{\alpha }_{ik}{\alpha }_{im}{\alpha }_{in}$ is equal to $q-5$. We calculate:

$$\begin{array}{cc}\hfill & \sum _{j=1}^{(q-2)/2}{\alpha }_{jk}{\alpha }_{jm}{\alpha }_{jn}\hfill \\ \hfill & =\left(\sum _{j=1}^{(q-2)/2}{\gamma }^{jk}+{\gamma }^{-jk}\right)\left(\sum _{j=1}^{(q-2)/2}{\gamma }^{jm}+{\gamma }^{-jm}\right)\left(\sum _{j=1}^{(q-2)/2}{\gamma }^{jn}+{\gamma }^{-jn}\right)\hfill \\ \hfill & =\sum _{j=1}^{q-2}{\gamma }^{j(k+m+n)}+\sum _{j=1}^{q-2}{\gamma }^{j(k+m-n)}+\sum _{j=1}^{q-2}{\gamma }^{j(k-m+n)}+\sum _{j=1}^{q-2}{\gamma }^{j(k-m-n)}\hfill \end{array}$$

and notice that each of these four sums will be $q-2$ if $q-1$ divides j, and $-1$ otherwise. Suppose that $q>5$ and choose $k=q-4,m=1,n=2$. Then $m\ne n$ and $m+n\ne q-1$, as required. We also have that only one of $k\pm m\pm n$ is congruent to zero mod $q-1$. This gives

$$\begin{array}{cc}\hfill & \sum _{j=1}^{q-2}{\gamma }^{j(q-1)}+\sum _{j=1}^{q-2}{\gamma }^{j(q-3)}+\sum _{j=1}^{q-2}{\gamma }^{j(q-5)}+\sum _{j=1}^{q-2}{\gamma }^{j(q-7)}=q-5.\hfill \end{array}$$

Then for $k=q-4,m=1,n=2$ we have:

$$\begin{array}{cc}\hfill & \langle {\chi }_5(q-4),{\chi }_1(1,2)\downarrow P\rangle ={\displaystyle \frac{3+q+\left({\sum }_{j=1}^{(q-2)/2}{\alpha }_{j(q-4)}{\alpha }_{j1}{\alpha }_{j2}\right)}{q-1}}={\displaystyle \frac{3+q+q-5}{q-1}}=2,\hfill \end{array}$$

showing that $({\mathrm{Sp}}_4\left(q\right),P)$ is not a strong Gelfand pair if $q>5$.

A similar argument shows that $({\mathrm{Sp}}_4\left(q\right),Q),q>5,$ is also not a strong Gelfand pair.    □

5. The Maximal Subgroups ${\mathbf{Sp}}_{\mathbf{2}}\left({\mathit{q}}^{\mathbf{2}}\right)$:2 and ${\mathbf{Sp}}_{\mathbf{4}}\left({\mathit{q}}_{\mathbf{0}}\right)$

The elements of the field ${\mathbb{F}}_{q^2}$ can be represented as $2\times 2$ matrices over ${\mathbb{F}}_q$. This shows how ${\mathrm{Sp}}_2\left(q^2\right)\le {\mathrm{Sp}}_4\left(q\right)$. The action of the 2 in ${\mathrm{Sp}}_2\left(q^2\right)$:2 is the Galois action.

Theorem 5. 

For $q=2^e,e>1,$ the pair $\left({\mathrm{Sp}}_4\left(q\right),{\mathrm{Sp}}_2(q^2\right)$:$2)$ is not a strong Gelfand pair.

Proof. 

Let $G={\mathrm{Sp}}_2\left(q^2\right):2$ and $H={\mathrm{Sp}}_2\left(q^2\right)\le G$. Using Table 2 we get the character table for H; see

Table 4. Character degrees for ${\mathrm{Sp}}_2\left(q^2\right)$ with q even.

Character	Degree	Multiplicity
Tr	1	1
$\psi$	$q^2$	1
${\chi }_s$	$q^2+1$	$(q^2-2)/2$
${\theta }_j$	$q^2-1$	$q^2/2$

where $1\le s\le (q^2-2)/2$ and $1\le j\le q^2/2$.    □

In order to find the degree of ${\tau }_G$, we will need to determine which characters of H split and which fuse; it will suffice to determine which characters of H induce to irreducible characters of G. Again from [20], since $|G:H|=2,$ we know that, by  inducing, every character in $\widehat{H}$ either splits into a sum of two irreducible characters or fuses pairwise into irreducible characters in $\widehat{G}$. We use Lemma 3 and Table 4 to give:

Proposition 3. 

Let $G={\mathrm{Sp}}_2\left(q^2\right):2\ge H={\mathrm{Sp}}_2\left(q^2\right)$. Then

(i)

${\mathit{Tr}}_H$ splits;

(ii)

ψ splits;

(iii)

all ${\theta }_j$ fuse;

The characters ${\chi }_s$ sometimes split, but not always:

(iv)

${\chi }_s\uparrow G$ is irreducible if $(q^2-1)\nmid s(q\pm 1)$; and

(v)

${\chi }_s\uparrow G$ is the sum of two irreducible characters if $(q^2-1)\mid s(q\pm 1)$.

Proof. 

(i)

It is clear that ${\mathrm{Tr}}_H$ splits.

(ii)

Since $\psi \left(1\right)=q^2$ and there is no other character of degree $q^2$ we see that $\psi$ cannot fuse.

(iii)

It will suffice to show that $\langle {\theta }_j\uparrow G,{\theta }_j\uparrow G\rangle =1$. Now a calculation shows that ${\theta }_j\uparrow {\mathrm{Sp}}_2\left(q^2\right):2$ is as described in the following table, where $\sigma$ is a primitive $(q^2+1)$-th root of unity.

	${\mathrm{Tr}}_H$	c	$a^t$	$b^m$	$G\setminus H$
${\theta }_j\uparrow {\mathrm{Sp}}_2\left(q^2\right):2$	$2q^2-2$	$-2$	0	$-({\sigma }^{jm}+{\sigma }^{-jm}+{\sigma }^{jmq}+{\sigma }^{-jmq})$	0

Now $({\theta }_j\uparrow G)(G\setminus H)=\left\{0\right\}$ and for $g\in H$ we have g and $g^{-1}$ are conjugate. Thus

$$\begin{array}{cc}\hfill \langle {\theta }_j\uparrow G,{\theta }_j\uparrow G\rangle & ={\displaystyle \frac{1}{\left|G\right|}}\sum _{g\in G}({\theta }_j\uparrow G)\left(g\right)·({\theta }_j\uparrow G)\left(g^{-1}\right)\hfill \\ & ={\displaystyle \frac{1}{\left|G\right|}}\sum _{g\in H}({\theta }_j\uparrow G)\left(g\right)·({\theta }_j\uparrow G)\left(g^{-1}\right)={\displaystyle \frac{1}{\left|G\right|}}\sum _{g\in H}{({\theta }_j\uparrow G)}^2\left(g\right).\hfill \end{array}$$

Using Table 2 again and taking $g_m\in {\left(b^m\right)}^G$ the above is equal to

$$\begin{array}{cc}& {\displaystyle \frac{1}{2(q^6-q^2)}}\left(\underset{\mathrm{Tr}}{\underbrace{{(2q^2-2)}^2}}+\underset{c}{\underbrace{{(-2)}^2(q^4-1)}}+\underset{a^t}{\underbrace{0}}+\underset{\mathrm{size}\mathrm{of}{\left(b^m\right)}^H}{\underbrace{(q^4-q^2)}}\sum _{m=1}^{q^2/2}{\left({\theta }_j\uparrow G\right)}^2\left(g_m\right)\right)\hfill \\ & ={\displaystyle \frac{1}{2(q^6-q^2)}}\left(8q^4-8q^2+(q^4-q^2)\sum _{m=1}^{q^2/2}{\left({\theta }_j\uparrow G\right)}^2\left(g_m\right)\right)\hfill \\ & ={\displaystyle \frac{1}{2(q^6-q^2)}}\left(8q^4-8q^2+(q^4-q^2)\sum _{m=1}^{q^2/2}{\left(-{\sigma }^{jm}-{\sigma }^{-jm}-{\sigma }^{jmq}-{\sigma }^{-jmq}\right)}^2\right)\hfill \\ & ={\displaystyle \frac{1}{2(q^6-q^2)}}(8q^4-8q^2+(q^4-q^2)\sum _{m=1}^{q^2/2}(4+({\sigma }^{2jm}+{\sigma }^{-2jm})+({\sigma }^{2jmq}+{\sigma }^{-2jmq})\hfill \\ & +(2{\sigma }^{jm(q+1)}+2{\sigma }^{-jm(q+1)})+(2{\sigma }^{jm(q-1)}+2{\sigma }^{-jm(q-1)})\left)\right)\hfill \end{array}$$

Now, since $1+{\sum }_{i=1}^{q^2/2}{\sigma }^i+{\sigma }^{-i}={\sum }_{i=0}^{q^2}{\sigma }^i$, the above is

$$\begin{array}{cc}& {\displaystyle \frac{1}{2(q^6-q^2)}}\left(8q^4-8q^2+(q^4-q^2)\sum _{m=1}^{q^2}\left(2+{\sigma }^{2jm}+{\sigma }^{2jmq}+2{\sigma }^{jm(q+1)}+2{\sigma }^{jm(q-1)}\right)\right)\hfill \\ & ={\displaystyle \frac{(8q^4-8q^2+2q^2(q^4-q^2)-6(q^4-q^2))}{2q^6-2q^2}}={\displaystyle \frac{2q^6-2q^2}{2q^6-2q^2}}=1\hfill \end{array}$$

as required for (iii).

(iv)

Now a calculation shows that ${\chi }_s\uparrow {\mathrm{Sp}}_2\left(q^2\right):2$ is as described in the following table, where $\rho$ is a primitive $(q^2-1)$-th root of unity.

	${\mathrm{Tr}}_H$	c	$a^t$	$b^m$	$G\setminus H$
${\theta }_j\uparrow {\mathrm{Sp}}_2\left(q^2\right):2$	$2q^2+2$	2	$-({\rho }^{jm}+{\rho }^{-jm}+{\rho }^{jmq}+{\rho }^{-jmq})$	0	0

We again examine $\langle {\chi }_s,{\chi }_s\rangle$ to see when we obtain 1. Taking $g_t\in {\left(a^t\right)}^G$ an argument similar to the ${\theta }_j$ case gives

$$\begin{array}{cc}\hfill & \langle {\chi }_s\uparrow G,{\chi }_s\uparrow G\rangle ={\displaystyle \frac{1}{\left|G\right|}}\sum _{g\in G}\left({\chi }_s\uparrow G\right)\left(g\right)·\left({\chi }_s\uparrow G\right)\left(g^{-1}\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{2q^6-2q^2}}\left(\sum _{g\in H}{\left({\chi }_s\uparrow G\right)}^2\left(g\right)\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{2q^6-2q^2}}\left({(2q^2+2)}^2+4(q^4-1)+(q^4+q^2)\sum _{t=1}^{(q^2-2)/2}{\left({\chi }_s\left(g_t\right)\right)}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{2q^6-2q^2}}\left(8q^4+8q^2+(q^4+q^2)\sum _{t=1}^{(q^2-2)/2}{\left({\rho }^{st}+{\rho }^{-st}+{\rho }^{stq}+{\rho }^{-stq}\right)}^2\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{2q^6-2q^2}}(8q^4+8q^2+(q^4+q^2)\sum _{t=1}^{(q^2-2)/2}(4+{\rho }^{2st}+{\rho }^{-2st}+{\rho }^{2stq}+{\rho }^{-2stq}\hfill \\ \hfill & +2{\rho }^{-st(q-1)}+2{\rho }^{-st(q+1)}\left)\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{2q^6-2q^2}}\left(8q^4+8q^2+\left(2(q^2-2)-2\right)(q^4+q^2)+(q^4+q^2)\sum _{t=1}^{q^2-2}(2{\rho }^{st(q+1)}+2{\rho }^{st(q-1)})\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{2q^6-2q^2}}\left(8q^4+8q^2+2q^6-4q^4-6q^2+(q^4+q^2)\sum _{t=1}^{q^2-2}(2{\rho }^{st(q+1)}+2{\rho }^{st(q-1)})\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{2q^6-2q^2}}\left(2q^6+4q^4+2q^2+(q^4+q^2)\sum _{t=1}^{q^2-2}(2{\rho }^{st(q+1)}+2{\rho }^{st(q-1)})\right).\hfill \end{array}$$

Here we used the facts that $(q^2-1)\nmid 2s$ and $(q^2-1)\nmid 2sq$, since $1\le s\le (q^2-2)/2$. Now, since $(q^2-1)\nmid s$, only one of $(q^2-1)\mid s(q+1)$ or $(q^2-1)\mid s(q-1)$ can be true, this shows that the above is equal to

$$\left\{\begin{array}{c}{\displaystyle \frac{1}{2q^6-2q^2}}\left(2q^6+4q^4+2q^2-4(q^4+q^2)\right)={\displaystyle \frac{2q^6-2q^2}{2q^6-2q^2}}=1\mathrm{if}(q^2-1)\nmid s(q\pm 1)\hfill \\ {\displaystyle \frac{1}{2q^6-2q^2}}\left(2q^6+4q^4+2q^2+2(q^4+q^2)(q^2-3)\right)={\displaystyle \frac{4q^6-4q^2}{2q^6-2q^2}}=2\mathrm{if}(q^2-1)\mid s(q\pm 1).\hfill \end{array}\right.$$

Since there are ${\displaystyle \frac{q}{2}}$ values of s for which $(q^2-1)\mid s(q+1)$ and ${\displaystyle \frac{q-2}{2}}$ values where $(q^2-1)\mid s(q-1)$, we see that ${\displaystyle \frac{2q-2}{2}}=q-1$ characters ${\chi }_s$ of H split in G. Then the remaining ${\displaystyle \frac{q^2-2q}{2}}$ characters fuse in G. Recall that $1\le j\le q^2/2$ and $1\le s\le (q^2-2)/2$. So

$$deg\left({\tau }_G\right)=2+2q^2+\left(2(q-1)+{\displaystyle \frac{q^2-2q}{2}}\right)(q^2+1)+{\displaystyle \frac{q^2}{2}}\left(q^2-1\right)=q^4+q^3+q.$$

By Theorem 2 $q^4+2q^3+2q^2+2q+1$ is the largest degree of an irreducible character of ${\mathrm{Sp}}_4\left(q\right),q\ge 4$. Since $G={\mathrm{Sp}}_2\left(q^2\right):2$ and

$$deg\left({\tau }_G\right)=q^4+q^3+q<q^4+2q^3+2q^2+2q+1$$

by Lemma 1 $({\mathrm{Sp}}_4\left(q\right),{\mathrm{Sp}}_2\left(q^2\right)$:$2)$ is not a strong Gelfand pair.    □

Similar to Corollary 1 we see that by Lemma 4 and the fact that the above argument is a `total character argument’ (not dependent on the particular embedding of ${\mathrm{Sp}}_2\left(q^2\right)$:2 in ${\mathrm{Sp}}_4\left(q\right)$) we have:

Corollary 2. 

The maximal subgroup ${\mathit{SO}}_4^{-}\left(q\right)<{\mathrm{Sp}}_4\left(q\right)$ is not a strong Gelfand subgroup.

Theorem 6. 

For $q=2^e,e>1$, and $q_0$ such that $q=q_0^r$ for a prime r, the maximal subgroup ${\mathrm{Sp}}_4\left(q_0\right)\le {\mathrm{Sp}}_4\left(q\right)$ is not a strong Gelfand subgroup.

Proof. 

By Theorem 2 $deg\left({\tau }_{{\mathrm{Sp}}_4\left(q\right)}\right)=q^6+q^4-q^2$ for all even q. Then $deg\left({\tau }_{{\mathrm{Sp}}_4\left(q_0\right)}\right)=q_0^6+q_0^4-q_0^2$ and since $q=q_0^r$ and $r\ge 2$ we see that

$$q^4+q^3+q^2+q=q_0^{4r}+q_0^{3r}+q_0^{2r}+q_0^r\ge q_0^8+q_0^6+q_0^4+q_0^2.$$

This shows that

$$deg\left({\tau }_{{\mathrm{Sp}}_4\left(q_0\right)}\right)=q_0^6+q_0^4-q_0^2<q^4+2q^3+2q^2+2q+1$$

and so by Lemma 1 $({\mathrm{Sp}}_4\left(q\right),{\mathrm{Sp}}_4\left(q_0\right))$ is not a strong Gelfand pair.    □

Theorem 7. 

For $q=2^{2n+1}$, with n a positive integer, the maximal subgroup $\mathrm{Sz}\left(q\right)$ in ${\mathrm{Sp}}_4\left(q\right)$ is not a strong Gelfand subgroup.

Proof. 

In [22] Suzuki gives the irreducible characters of $\mathrm{Sz}\left(q\right)$, where $q=2^{2n+1}$. They are:

(i)

the trivial character of degree 1;

(ii)

a doubly transitive character of degree $q^2$;

(iii)

$(q-2)/2$ characters of degree $q^2+1$;

(iv)

two complex characters of degree $2^n(q-1)$;

(v)

$(q+2^{n+1})/4$ characters of degree $(q-2^{n-1}+1)(q-1)$;

(vi)

$(q-2^{n+1})/4$ characters of degree $(q+2^{n-1}+1)(q-1)$.

This gives the following expression for $deg\left({\tau }_{\mathrm{Sz}\left(q\right)}\right)$:

$$\begin{array}{cc}\hfill & 1+q^2+\left({\displaystyle \frac{q-2}{2}}\right)\left(q^2+1\right)+2·2^n\left(q-1\right)+\left({\displaystyle \frac{q+2·2^n}{4}}\right)\left(q-2·2^n+1\right)\left(q-1\right)\hfill \\ \hfill & +\left({\displaystyle \frac{q-2·2^n}{4}}\right)\left(q+2·2^n+1\right)\left(q-1\right)\hfill \\ \hfill & =2^{n+1}(q-1)-q(q-1)+q^3.\hfill \end{array}$$

We now notice that the degree of the total character of $\mathrm{Sz}\left(q\right)$ is smaller than the maximal degree of an irreducible character in ${\mathrm{Sp}}_4\left(q\right)$ by Theorem 2. This shows by Lemma 1 that $({\mathrm{Sp}}_4\left(q\right),\mathrm{Sz}\left(q\right))$ is not a strong Gelfand pair when $q=2^{2n+1}$.    □

This completes consideration of all the maximal subgroups listed in Table 1 and so concludes the proof of Theorem 1.