Proof. There exists a nonsingular matrix P such that

$$ \begin{align*}P^{-1} H P = \left\{\begin{pmatrix} A_h & B_h \\ 0 & C_h \end{pmatrix} \mid h\in H\right\},\quad P^{\mathsf{T}} H^{\mathsf{T}} P^{-\mathsf{T}} = \left\{\begin{pmatrix} A_h^{\mathsf{T}} & 0 \\[3pt] B_h^{\mathsf{T}} & C_h^{\mathsf{T}} \end{pmatrix} \mid h\in H \right\}.\end{align*} $$

Set $A=A_g$ , $B=B_g$ , $C=C_g$ and $k=\dim (U)$ . Under our assumption, $A\in \mathrm {GL}_k(\mathbb {F})$ does not have the eigenvalue $\lambda $ . Hence, the same is true for $A^{\mathsf {T}}$ . So, imposing

$$ \begin{align*}\begin{pmatrix} A^{\mathsf{T}} & 0 \\ B^{\mathsf{T}} & C^{\mathsf{T}} \end{pmatrix} \begin{pmatrix} w \\ \overline{w}\end{pmatrix}= \begin{pmatrix} A^{\mathsf{T}} w \\ B^{\mathsf{T}} w + C^{\mathsf{T}} \overline{w}\end{pmatrix}= \begin{pmatrix} \lambda w \\ \lambda\overline{w} \end{pmatrix},\quad w \in \mathbb{F}^k, \; \overline{w} \in \mathbb{F}^{n-k},\end{align*} $$

we get $w = 0$ and

$$ \begin{align*}V_\lambda(P^{\mathsf{T}} g^{\mathsf{T}} P^{-\mathsf{T}}) =\left\{ \begin{pmatrix} 0 \\ \overline{w}\end{pmatrix}\mid C^{\mathsf{T}} \overline{w}= \lambda \overline{w}\right\}\leq \overline{E} =\langle e_i\mid k+1\leq i\leq n \rangle.\end{align*} $$

Set $\overline {U} = P^{-\mathsf {T}} \overline {E}$ . Since $\overline {E}$ is invariant under $P^{\mathsf {T}} H^{\mathsf {T}} P^{-\mathsf {T}}$ , we get that $\overline {U}$ is $H^{\mathsf {T}}$ -invariant. From $V_\lambda (g^{\mathsf {T}} ) = P^{-\mathsf {T}} V_\lambda (P^{\mathsf {T}} g^{\mathsf {T}} P^{-\mathsf {T}} )$ , it follows that $V_\lambda (g^{\mathsf {T}} ) \leq \overline {U}$ .

Proof. From $g^{\mathsf {T}} J g =J$ , we get $g(J^{-1} \overline {s}) =J^{-1} g^{-\mathsf {T}} \overline {s}=\lambda ^{-1} (J^{-1}\overline {s})$ for all $\overline {s}\in V_{\lambda }(g^{\mathsf {T}})$ . It follows that $J^{-1} V_{\lambda }(g^{\mathsf {T}})\leq V_{\lambda ^{-1}}(g)$ . However, take $v \in V_{\lambda ^{-1}}(g)$ . Then, $g^{\mathsf {T}} J v = J g^{-1} v =\lambda J v$ gives $Jv \in V_{\lambda }(g^{\mathsf {T}})$ , whence $V_{\lambda ^{-1}}(g)\leq J^{-1} V_{\lambda }(g^{\mathsf {T}})$ .

If $g_{|U}$ does not have the eigenvalue $\lambda $ , we apply Lemma 2.1: so, there exists an $H^{\mathsf {T}}$ -invariant subspace $\overline {U}$ , with $\dim (\overline {U})=n-\dim (U)$ , such that $V_\lambda (g^{\mathsf {T}}) \leq \overline {U}$ . Set $W=J^{-1}\overline {U}$ . For any $h \in H$ , we have $h W = h (J^{-1}\overline {U})=J^{-1}h^{-\mathsf {T}}\overline {U} =J^{-1} \overline {U}=W$ . Hence, W is H-invariant and $\dim (W)=\dim (\overline {U})=n-\dim (U)$ . Finally, $V_{\lambda ^{-1}}(g) = J^{-1}V_\lambda (g^{\mathsf {T}})\leq J^{-1}\overline {U}=W$ .

Proof. Clearly, it is enough to show that, for $k\leq n-3$ , there exists $h\in H$ such that $he_1=e_k$ . Noting that $ye_1=e_2$ , $ye_2=e_3$ , $xe_3=e_4$ , $ye_4=e_5$ for $n=9$ , $xe_1=e_3$ , ${ye_3=e_4}$ , $ye_4=e_5$ , $xe_4=e_2$ for $n\in \{11,17\}$ , and $xe_1=e_2$ , $ye_2=e_3$ , $ye_3=e_4\ xe_4=e_5$ for $n=13$ , our claim is true for $k\leq 5$ .

Now, let $5 \leq \ell \leq n - 3$ be the largest integer for which, for all $1\leq i\leq \ell $ , there exists $h_i \in H$ such that $h_i e_1= e_i$ . If $\ell < n - 3$ , there exists $h \in \{x, y\}$ such that $he_\ell = e_{\ell +1}$ , which is a contradiction.

Proof. Assume, for a contradiction, that U is a proper H-invariant subspace. Define

$$ \begin{align*}g_9=[x,y],\quad g_{11}=(xy^2)^3xy,\quad g_{13}= (xy^2)^2xy, \quad g_{17}= (xy^2)^6xy.\end{align*} $$

Under our hypotheses on a, for $n=9$ , we have $V_1(g_9)= \langle e_1\rangle $ . By Corollary 2.2, we may assume $e_1\in U$ and hence $e_1,\ldots ,e_{n-3}\in U$ by Lemma 3.1. Similarly, for $n=11$ , we have $V_1(g_{11})= \langle e_3\rangle $ , for $n=13$ , we have $V_1(g_{13})= \langle e_2\rangle $ and for $n=17$ , we have $V_1(g_{17})= \langle e_6\rangle $ . In all these cases, as above, we may assume $e_1,\ldots ,e_{n-3}\in U$ . Noting that $y e_{n-3}+e_{n-3} = -2 e_{n-2}$ , $y^2 e_{n-5} = e_{n-1}$ and $y^2 e_{n-2} = -\tfrac 12 e_n$ , we get the contradiction $U=V$ .

Proof. Suppose the contrary. By [Reference Bray, Holt and Roney-Dougal1, Tables 8.58 and 8.74] and [Reference Kleidman and Liebeck5, Proposition 4.5.8], we have either $M\cong \Omega _n(q_0)$ where $q=q_0^r$ and r is an odd prime, or $M\cong \mathrm {SO}_n(q_0)$ where $q=q_0^2$ . Thus, there exists $g\in \mathrm {GL}_{n}(\mathbb {F})$ such that $x^g=x_0$ , $y^g=y_0$ , with $x_0, y_0\in \mathrm {GL}_{n}(q_0)$ . From $\mathrm {tr}([x,y]^j)=\mathrm {tr}([x^g, y^g]^j )= \mathrm {tr}( [x_0,y_0]^j)$ , $j=1,2$ , it follows that $4a+\kappa _n = (\mathrm {tr}([x,y])-\varsigma _n)^2-\mathrm {tr}([x,y]^2)\in \mathbb {F}_{q_0}$ , whence $a \in \mathbb {F}_{q_0}$ . So, $\mathbb {F}_q=\mathbb {F}_p[a]\leq \mathbb {F}_{q_0}$ implies ${q_0=q}$ .

Proof. For the sake of contradiction, suppose that H is contained in a monomial subgroup $M\in \mathcal {C}_2$ of $\Omega _n(q)$ . In this case, we may assume $q=p$ and H acts monomially with respect to an orthonormal basis $\mathcal {B}=\{v_1,v_2,\ldots ,v_n\}$ , see [Reference Kleidman and Liebeck5, Proposition 4.2.15]. Moreover, by [Reference Bray, Holt and Roney-Dougal1, Tables 8.58 and 8.74] and [Reference Kleidman and Liebeck5, Proposition 4.5.8], the order of M divides $2^{n-1}|\mathrm {Sym}(n)|$ . In particular, any prime divisor $\varrho $ of $|H|$ should satisfy $\varrho \leq n$ . If we can show that $e_1\in \mathcal {B}$ , we easily get a contradiction. Indeed, from $e_1 \in \mathcal {B}$ , it follows that $e_i \in \mathcal {B}$ for all $1\leq i \leq n-3$ (see Lemma 3.1). Hence, we may assume ${v_i=e_i}$ for ${1\leq i\leq n-3}$ . In particular, $e_{n-3}\in \mathcal {B}$ . As $ye_{n-3}= -2 e_{n-2}-e_{n-3}$ is not an element of $\langle e_i \mid 1\leq i \leq n-3\rangle $ , $ye_{n-3}$ should be orthogonal to $v_{n-3}$ obtaining the contradiction $v_{n-3}^{\mathsf {T}} J y e_{n-3} =e_{n-3}^{\mathsf {T}} J y e_{n-3}= -1\neq 0$ .

So, we now show that $e_1\in \mathcal {B}$ . To this purpose, note that if $\mathrm {tr} (h)\neq 0$ , then h must fix at least one $\langle v_j\rangle $ . Moreover, given $h\in H$ of order k, $h\langle v_j\rangle =\langle v_j\rangle $ implies $h v_j=\lambda v_j$ , with $\lambda =\pm 1$ . So, consider the permutation $\zeta $ induced by h on the $\langle v_i\rangle $ . If $\zeta ^b$ acts as the identity on $\{\langle v_1\rangle ,\langle v_2\rangle , \ldots , \langle v_n\rangle \}$ for some $b\geq 1$ , then $h^b v_i = \pm v_i$ for every i. It follows that $\zeta $ has order k or ${k}/{2}$ . In particular, if h has odd order, it permutes $\mathcal {B}$ and its cycle structure is determined by its rational canonical form. Also, if $h\in H$ does not have the eigenvalue $-1$ , from $h\langle v_j\rangle =\langle v_j\rangle $ , we get $hv_j=v_j$ . Clearly, this applies to $h=y$ . Since y has order $3$ , setting $r=0$ if $n=9$ , $r=1$ if $n=13$ and $r=2$ if $n \in \{11,17\}$ , y fixes $v_j$ for $1\leq j\leq r$ and permutes the remaining vectors $v_j$ in $({n-r})/{3}$ orbits of length $3$ .

Case $n=11,13,17$ . Call s the number of vectors $u_j=e_j+ y e_j+y^2 e_j$ , with $ye_j\neq e_j$ , fixed by y. Then, any $v_1 \in V_1(y)$ can be written as

$$ \begin{align*}v_1=\sum\limits_{i=1}^r \alpha_i e_i+ \sum\limits_{j=1}^s \beta_j u_j.\end{align*} $$

Substituting $e_i$ by $\lambda _i e_i$ and $u_j$ by $\mu _j u_j$ if necessary, we may assume that all the coefficients $\alpha _i,\beta _j$ are in $\{0,1\}$ . Since y fixes $v_1$ , by the transitivity of H on the subspaces generated by the vectors of $\mathcal {B}$ , due to its irreducibility, we may also assume $v_3=x v_1$ , $v_4=yv_3$ , $v_5 = y v_4$ and $v_6= x v_5$ . Imposing $v_1^{\mathsf {T}} Jv_3=v_j^{\mathsf {T}} J v_6=0$ for all $j\in \{1,4,5\}$ , we get $v_1\in \{e_1,\ldots ,e_r\}$ , unless $n \in \{11,17\}$ , $q=3$ and $a=-1$ . In these exceptional cases, by direct computation, the order of $(xy)^2 x y^2$ is divisible by a prime $\varrho \geq 41$ , which is a contradiction as $\varrho $ does not divide $|\mathrm {Sym}(n)|$ , $n\leq 17$ (see the beginning of the proof).

Case $n=9$ . Take $h=[x,y]$ and suppose $a^2-a-1\neq 0$ . Then $V_1([x,y])=\langle e_1\rangle $ . We have

$$ \begin{align*}\mathrm{tr} (h)=\frac{(a^2+1)^2}{a^2} \quad \textrm{and} \quad \chi_h(-1)= \frac{-8(a^2+a+1)^2}{a^2}.\end{align*} $$

It follows $e_1\in \mathcal {B}$ unless, possibly, when $a^2+1=0$ or $a^2+a+1=0$ . As previously remarked, the order of any element of M, and hence a fortiori of H, if odd must belong to the set $\{ 1, 3, 5, 7, 9, 15 \}$ , and if prime must belong to $\{2,3,5,7\}$ . Assume $a^2+1=0$ . If $p\neq 5$ , we may take $h=[x,y]^2$ , as $V_1(h)=\langle e_1\rangle $ and $\mathrm {tr}(h)=-4a - 2\neq 0$ . If $q=5$ , then $a=2$ and $[x,y]$ has order $156=2^2\cdot 3 \cdot 13$ , which is a contradiction. So, assume ${a^2+a+1=0}$ . If $p\neq 3$ , the permutation induced by $xy$ on the $\langle v_i\rangle $ has order divisible by $21$ , which is a contradiction. If $q=3$ , then $a=1$ and $[x,y]^3y$ has order $41$ , which is a contradiction.

Proof. For the sake of contradiction, suppose that H is imprimitive. By Lemma 3.4, we may assume $H\leq M\cong \Omega _3(q)^3. 2^4. \mathrm {Sym}(3)$ , where M permutes a decomposition ${\mathbb {F}_q^9 = W_1 \oplus W_2 \oplus W_3}$ , with $\dim (W_i)=3$ . Set $h=(xy)^7$ and $N=\Omega _3(q)^3$ . From $\dim (V_1(h))=7$ , we get $V_1(h)\cap W_i\neq \{0\}$ , whence $hW_i=W_i$ for each $i=1,2,3$ . It follows that $(xy)^7 \in N$ . Since $7$ is coprime to the index of N in M, we get $xy \in N$ . Since y acts as a $3$ -cycle on $\{W_1,W_2,W_3\}$ , it follows that the elements $(xy)^iy$ , $1\leq i\leq 7$ , have trace equal to zero. Thus, $0=\mathrm {tr}(xy^2)=-(a+a^{-1})$ gives the condition $a^2+1=0$ . In this case, $\mathrm {tr}((xy)^3y)=1$ , which is a contradiction.

Now, suppose that H is contained in a maximal subgroup M in class $\mathcal {C}_7$ of $\Omega _n(q)$ . By [Reference Bray, Holt and Roney-Dougal1, Table 8.58], $M\cong \Omega _3(q)^2.[4]$ . Then, $h=(xy)^7$ belongs to $\Omega _3(q)^2$ . Suppose first that $xy$ is semisimple. Up to conjugation, $h=\mathrm {diag}(\beta _1,1,\beta _1^{-1})\otimes \mathrm {diag}(\beta _2, 1, \beta _2^{-1})$ for some $\beta _1,\beta _2 \in \mathbb {F}_q^*$ . In order that it has the eigenvalue $1$ with multiplicity (at least) $7$ , we need $\beta _1=\beta _2=1$ , which gives $h=\mathrm {I}_9$ , which is a contradiction. Finally, assume that $xy$ has order divisible by p. Up to conjugation and because of (2.1),

$$ \begin{align*}h=\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} \beta & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \beta^{-1} \end{pmatrix}, \quad \beta\in \mathbb{F}_q^*.\end{align*} $$

Hence, $\chi _{h}(t)=(t-1)^3(t-\beta )^3(t-\beta ^{-1})^3$ . Since h is a bireflection (that is, ${\dim (V_1(h))=7}$ ), we must have $\beta =1$ , in which case $\dim (V_1(h))=3$ , which is a contradiction.

Proof. Assume the contrary. From what is seen at the end of Section 2, up to conjugation, we may assume $H\leq \mathrm {Sym}(\ell )\leq \mathrm {GL}_\ell (p)$ , with $\ell = n +1, n + 2$ .

Case $\ell =n+1$ . Fix $h \in H$ such that $\dim (V_1(h))=1$ and call $\zeta $ its preimage in $\mathrm {Sym}(\ell )\leq \mathrm {GL}_\ell (p)$ . Then, $\zeta $ has at most two orbits. It follows that $\mathrm {tr}(\zeta )=0$ if $\zeta $ is an $\ell $ -cycle or the product of two cycles of length at least two. Otherwise, $\mathrm {tr}(\zeta )=1$ and $\zeta $ is a cycle of length $\ell -1$ . Note that $\zeta $ and h have the same order.

We may take $h=xy$ , as $\dim (V_1(xy))=1$ . Hence, $\mathrm {tr}(\zeta )-1=\mathrm {tr}(xy)=-(a+a^{-1})$ gives the following two cases: if $\mathrm {tr}(\zeta )=0$ , then $a^2-a+1=0$ ; if $\mathrm {tr}(\zeta )=1$ , then $a^2+1=0$ . In the second case, the characteristic polynomial $\chi _{xy}(t)$ is divisible by $t^2+1$ , and then $xy$ has order divisible by $4$ . However, $\zeta $ has odd order n, being an n-cycle, which is a contradiction.

So, assume $a^2-a+1=0$ . In this case, $t^2+t+1$ divides $\chi _{xy}(t)$ and hence the order of $\zeta $ is divisible by $3$ . Furthermore, $(xy)^{n-2}$ has order p when $n\in \{11,17\}$ . For $n=11$ , we get that the order of $\zeta $ is $6$ , $9$ or $12$ , in contrast with $(xy)^9$ of odd order p. For $n=17$ , the order of $\zeta $ is $9$ , $12$ , $15$ or $18$ . However, $(xy)^9\neq \mathrm {I}_{17}$ and the other values are in contrast with $(xy)^{15}$ of odd order p. For $n\in \{9,13\}$ , we apply the previous argument to other elements h such that $\dim (V_1(h))=1$ . For $n=9$ , we take $h=[x,y]$ whose trace is equal to $1$ , which is a contradiction. For $n=13$ , we take $h= (xy^2)^2xy$ , which has trace equal to $3$ . Since $\mathrm {tr}(h)=\mathrm {tr}(\zeta )-1 \in \{-1,0\}$ , we get an absurdity unless $p=3$ . However, in this case, $a=-1$ and $h^8$ has order $41$ , which is a contradiction as $h^8\in H\leq \mathrm {Sym}(14)$ .

Case $\ell =n+2$ . In this case, $q \mid \ell $ , and hence we need to consider only the following cases: (a) $(n,q)=(9,11)$ ; (b) $(n,q)=(11,13)$ ; (c) $(n,q)=(13,3)$ ; (d) $(n,q)=(13,5)$ ; (e) $(n,q)=(17,19)$ . Take $g=(xy)^3 (xy^2)^7$ in case (a); $g= xy (xy^2)^2$ in cases (b), (c) and (e); and $g= xy (xy^2)^3$ in case (d). By direct computation, in all these cases, the order of g is divisible by a prime $\varrho \geq n+4$ , which is a contradiction as $\varrho $ should divide $|\mathrm {Sym}(n+2)|$ .

Proof. By condition (ii), H is a subgroup of $\Omega _{n}(q)$ . By condition (iii), Lemmas 3.2, 3.4 and 3.5, the group H is absolutely irreducible and is neither contained in a maximal subgroup in class $\mathcal {C}_2$ of $\Omega _n(q)$ nor contained in any maximal subgroup in class $\mathcal {C}_7$ . Since it contains the bireflection $(xy)^{n-2}$ , we can apply [Reference Guralnick and Saxl3, Theorem 7.1] which, combined with condition (i) and Lemma 3.3, gives two possibilities: (a) H is an alternating or symmetric group of degree $\ell $ and $\mathbb {F}^n$ is the deleted permutation module of dimension $\ell -1$ or $\ell -2$ ; (b) $H=\Omega _{n}(q)$ . Case (a) is excluded by Lemma 3.6: we conclude that $H=\Omega _{n}(q)$ .

Finally, we have to prove that there exists an element a satisfying all the requirements. If $q=p$ , take $a=-1$ . Suppose now $q=p^f$ with $f\geq 2$ , and let $\mathcal {N}(q)$ be the number of elements $b\in \mathbb {F}_q^*$ such that $\mathbb {F}_p[b]\neq \mathbb {F}_q$ . By [Reference Pellegrini, Tamburini Bellani and Vsemirnov12], we have $\mathcal {N}(q)\leq p({p^{\lfloor f/2 \rfloor }-1})/({p-1})$ , and hence it suffices to check when $({p^f-1})/{2} - p({p^{\lfloor f/2 \rfloor }-1})/({p-1})> 4$ . This condition is fulfilled unless $q=3^2$ . So, assume $q=9$ and take $a\in \mathbb {F}_9^*$ whose minimal polynomial over $\mathbb {F}_3$ is $t^2+1$ . Then, $\mathbb {F}_3[a]=\mathbb {F}_9$ and $-a=(a+1)^2$ is a square.

Proof. For $n \in \{15,16, 19,20, 23\}$ , the element $[x,y]$ acts on $\mathcal {A}$ as the following permutation:

$$ \begin{align*}\begin{cases} (e_3,e_4) & \textrm{if } n =15,\\ (e_1,e_5)(e_2,e_4) & \textrm{if } n =16,\\ (e_1,e_5,e_4,e_2) (e_3,e_8,e_7) & \textrm{if } n =19,\\ (e_1, e_6, e_8, e_2)(e_3, e_4,e_9,e_5) & \textrm{if } n=20,\\ (e_1,e_6,e_5,e_3,e_4,e_9,e_8,e_2)(e_7,e_{12},e_{11}) & \textrm{if } n =23. \end{cases} \end{align*} $$

Otherwise, it acts on $\mathcal {A}$ as

$$ \begin{align*}\begin{cases} (e_1,e_4,e_3,e_2,e_7,e_6) & \textrm{if } n\equiv 0 ~\mathrm{(mod~ 6)},\\ (e_1,e_5,e_4,e_2) (e_3,e_8,e_7)(e_6,e_{11},e_{10}) & \textrm{if } n \equiv 1 ~\mathrm{(mod~ 6)},\\ (e_1, e_6, e_8, e_2)(e_3, e_4,e_9,e_5) (e_7,e_{12},e_{11},e_{10},e_{15},e_{14}) & \textrm{if } n\equiv 2~\mathrm{(mod~ 6)},\\ (e_2,e_7,e_6)(e_3,e_4) & \textrm{if } n\equiv 3~\mathrm{(mod~ 6)},\\ (e_1,e_5)(e_2,e_4)(e_3,e_8,e_7,e_6, e_{11},e_{10}) & \textrm{if } n \equiv 4 ~\mathrm{(mod~ 6)},\\ (e_1,e_6,e_5,e_3,e_4,e_9,e_8,e_2)(e_7,e_{12},e_{11})(e_{10}, e_{15}, e_{14}) & \textrm{if } n \equiv 5 ~\mathrm{(mod~ 6)}. \end{cases} \end{align*} $$

Finally, $[x,y]$ acts on each summand of $\mathcal B$ as the cycle $(e_{5+4r+3j}, e_{10+4r+3j}, e_{9+4r+3j} )$ .

Proof. We apply Corollary 2.2 to $g=[y, \tau ]$ and $\lambda =1$ . So, we may assume that the eigenvector $s=e_{n-8} - e_{n-7}$ is contained in U. Take the matrices $M_1,M_2$ , whose columns are the images of s under the following elements:

$$ \begin{align*}\begin{array}{rl} M_1: & \mathrm{I}_9,\; y, \; y^2, \; \tau y^2, \; \tau^2 y^2,\; y\tau y^2,\; y^2\tau y^2,\; y\tau^2 y^2,\; y^2\tau^2 y^2;\\ M_2: & \mathrm{I}_9,\; y, \; y^2, \; \tau y^2, \; \tau^2 y^2,\; y\tau y^2,\; y\tau^2 y^2,\; (\tau y^2)^2,\; \tau y^2\tau^2 y^2. \end{array}\end{align*} $$

Then, $\det (M_1)= -2^{35} a^{10} (4 a^2+3)$ and $\det (M_2)=-2^{35} a^{10} (28a^2 - 3)$ . Clearly, these two matrices cannot be both singular, whence $\dim (U)=9$ , which is a contradiction.

Proof. Recall that $\tau $ is an element of order p. Considering the order of the maximal subgroups M described in the statement and the conditions on q given in [Reference Bray, Holt and Roney-Dougal1, Tables 8.58 and 8.59], we may reduce to the following cases:

1. (i) $M=2^8: \mathrm {Alt}(9)$ and $q\in \{3,5\}$ ;

2. (ii) $M=2^8: \mathrm {Sym}(9)$ and $q=7$ ;

3. (iii) $M=\mathrm {Alt}(10)$ and $q \in \{3,7\}$ ;

4. (iv) $M=\mathrm {PSL}_2(17)$ and $q=9$ ;

5. (v) $M=\mathrm {Sym}(11)$ and $q=11$ ;

6. (vi) $M=\mathrm {PSL}_2(8)$ and $q=27$ .

Now, we look for an element of H whose order does not divide $|M|$ . In particular, it suffices to find an element of H whose order is divisible by a prime $\varrho>17$ in case (iv), $\varrho>11$ otherwise. Define $g_j=y\tau ^j$ . If $q\in \{3,9\}$ , then $g_1$ has order divisible by $41$ . If $q=5$ , then $g_3$ has order divisible by a prime $\varrho \geq 13$ . If $q=7$ , take $j=2$ when $a=\pm 2$ , and $j=3$ when $a\in \{\pm 1, \pm 3 \}$ . Then, the order of $g_j$ is divisible by a prime $\varrho \geq 43$ . If $q=11$ , take $j=2$ if $a= \pm 5$ and $j=1$ otherwise. Then the order of $g_j$ is divisible by a prime $\varrho \geq 19$ . Finally, if $q=27$ , then $g_2$ has order divisible by $37$ . In all these cases, we easily obtain a contradiction.

Proof. By Lemma 5.2, the group $K_{|S_9}$ is not monomial. So, suppose that $K_{|S_9}$ preserves a nonsingular decomposition $\mathbb {F}_q^9=W_1\oplus W_2\oplus W_3$ with $\dim (W_i)=3$ . Clearly, for each $k\in K_{|S_9}$ , its cube fixes each $W_i$ , preserving a nonsingular symmetric form. Thus, its eigenvalues are $\pm 1, \alpha _i, \alpha _i^{-1}$ . It follows that $k^3$ must have the eigenvalue $1$ with multiplicity at least $3$ , or the eigenvalue $-1$ with multiplicity at least $2$ . Assume first $p=3$ . We have $\chi _{(y\tau )^3}(t)=(t-1)f(t)$ , where $f(t)=t^8 + t^7 -(a^{12} +a^6 - 1) t^6 - (a^{12} - 1) t^5 - (a^6 - 1) t^4 -(a^{12} - 1) t^3 - (a^{12} + a^6 - 1) t^2 + t + 1$ . Then, $f(1)=-a^{12}\neq 0$ and $f(-1)=1$ , which is a contradiction. Next, assume $p\neq 3$ . From $\mathrm {tr}(\tau )=9\neq 0$ , we get that $\tau $ fixes each $W_i$ . By the irreducibility of $K_{|S_9}$ , the element y acts on $\{W_1,W_2,W_3\}$ as the $3$ -cycle $(W_1,W_2,W_3)$ . In this case, both $(y\tau )^2$ and $(y^2\tau )^2$ should have trace $0$ , in contrast with (5.1) which gives $0=\mathrm {tr}((y^2\tau )^2)-\mathrm {tr}((y\tau )^2)=2^{12}a^4$ .

Finally, suppose that $K_{|S_9}$ is contained in a maximal subgroup $M\cong \Omega _3(q)^2.[4]\in \mathcal {C}_7$ , and hence actually in $\Omega _3(q)^2$ . Up to conjugation, we may suppose $\tau =\bigg (\begin {smallmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end {smallmatrix}\bigg ) \otimes \bigg ( \begin {smallmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end {smallmatrix}\bigg )$ . The dimensions of the fixed point space of this tensor product and of $\tau $ are, respectively, $3$ and $5$ , which is a contradiction.

Proof. Suppose the contrary.

Case $M\cong \mathrm {PSL}_2(q).2$ . In this case, M arises from the representation $\Phi : \mathrm {GL}_2(q)\to \mathrm {GL}_9(q)$ obtained from the action of $\mathrm {GL}_2(q)$ on the space T of homogeneous polynomials of degree $8$ in two variables $t_1, t_2$ over $\mathbb {F}_q$ . Up to conjugation in $\mathrm {GL}_2(q)$ , we may assume

$$ \begin{align*}\tau=\Phi(\mathrm{I}_2+E_{1,2})=\begin{cases} t_1 \mapsto t_1,\\ t_2 \mapsto t_1+t_2. \end{cases}\end{align*} $$

Direct computation (with respect to the basis $t_1^8, t_1^7t_2, \ldots , t_2^8$ of T) gives that the fixed point space of this linear transformation is generated by $t_1^8$ . So, it has dimension $1$ , which is a contradiction as $\tau $ has a fixed point space of dimension $5$ .

Case $M\cong \mathrm {PSL}_2(q^2).2$ . To understand M, start from the representation $\psi :\mathrm {GL}_2(q^2)\to \mathrm {GL}_3(q^2)$ described in (2.1). Next, consider the subspace W of $\mathrm {Mat}_3(q^2)$ consisting of the matrices A such that $A^{\mathsf {T}}=(a_{i,j}^q)=A^\sigma $ . Clearly, W has dimension $9$ over $\mathbb {F}_q$ and we may consider the representation $\Phi : \mathrm {GL}_3(q^2)\to \mathrm {GL}_9(q)$ induced by $A\mapsto (\psi (g))^{\mathsf {T}} A (\psi (g))^\sigma $ for all $g\in \mathrm {GL}_3(q^2)$ . The group M arises from this representation. Again, up to conjugation in $\mathrm {GL}_2(q^2)$ , we may suppose $\tau =\Phi (\psi (\mathrm {I}_2+E_{1,2}))=\Phi \bigg (\bigg (\begin {smallmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end {smallmatrix}\bigg )\bigg )$ . Direct calculation gives that the fixed point space of $\Phi (\psi (\mathrm {I}_2+E_{1,2}))$ on $W\leq \mathrm {Mat}_3(q^2)$ is generated by $E_{2,2}, E_{3,3}, E_{2,3}+E_{3,2}$ . Thus, it has dimension $3$ , which is again a contradiction as $\tau $ has a fixed point space of dimension $5$ .

Proof. By Lemmas 5.1 and 5.3, $K_{|S_9}$ is absolutely irreducible, and is neither contained in a maximal subgroup in class $\mathcal {C}_2$ of $\Omega _9(q)$ nor contained in any maximal subgroup in class $\mathcal {C}_7$ . Furthermore, by Lemmas 5.2 and 5.4, either $K_{|S_9}=\Omega _9(q)$ or $K_{|S_9}$ is contained in a maximal subgroup $M\in \{\Omega _9(q_0), \mathrm {SO}_9(q_0)\}$ in class $\mathcal {C}_5$ , where $q=q_0^r$ for some prime $r\geq 2$ . Suppose there exists $g\in \mathrm {GL}_{9}(\mathbb {F})$ such that $\tau ^g=\tau _0$ , $y^g= y_0$ , with $\tau _0,y_0\in \mathrm {GL}_{9}(q_0)$ . From $ -2176 a^4 + 128 a^2 = \mathrm {tr}((y\tau )^2)= \mathrm {tr}((y^g\tau ^g)^2 )= \mathrm {tr}((y_0\tau _0)^2)$ , it follows that $17 a^4- a^2 \in \mathbb {F}_{q_0}$ . Similarly, from $\mathrm {tr}((y^2\tau )^2) = 1920 a^4 + 128 a^2$ , we obtain $15 a^4+a^2 \in \mathbb {F}_{q_0}$ . It follows that $32a^4\in \mathbb {F}_{q_0}$ and then $a^2 \in \mathbb {F}_{q_0}$ . Again, from $\mathrm {tr}( y^2\tau ^2 (y\tau )^2)= -49\,152 a^6 + 16\,384 a^5 + 3840 a^4 + 256 a^2 \in \mathbb {F}_{q_0}$ , we get $a\in \mathbb {F}_{q_0}$ . So, $\mathbb {F}_q=\mathbb {F}_p[a]\leq \mathbb {F}_{q_0}$ implies $q_0=q$ , which is a contradiction. We conclude that $K_{|S_9}=\Omega _9(q)$ .

Proof. By [Reference Bray, Holt and Roney-Dougal1, Proposition 1.5.42(ii)], when n is even, we have $H\leq \Omega _n^+(q)$ or ${H\leq \Omega _n^-(q)}$ according as ${n(q-1)}/{4}$ is even or odd, respectively. Let $\ell $ be maximal with respect to

$$ \begin{align*}K_\ell:=\mathrm{diag}(\mathrm{I}_{n-\ell}, \Omega_{\ell}^\epsilon (q))\leq H,\end{align*} $$

where $\epsilon \in \{\circ ,\pm \}$ . Noting that $K'=\mathrm {diag}(\mathrm {I}_{n-9}, \Omega _9(q))$ by the previous proposition, we have that $\ell $ is at least $9$ and we need to show that $\ell =n$ . For the sake of contradiction, assume $9\leq \ell < n$ .

Suppose first that $(r,\ell )\not \in \{(2,n-2), (2,n-1) \}$ and $(r,\ell )\not \in \{(1,n-4),(2,n-8)\}$ when n is even. Then:

1. (a) if $\ell \equiv 0 ~\mathrm {(mod~ 3)}$ , then x fixes the subspaces $S_{\ell -1}$ and $E_{n-\ell -1}$ , and acts as the transposition $(e_{n-\ell },e_{n-\ell +1})$ on $\langle e_{n-\ell }, e_{n-\ell +1} \rangle $ ;

2. (b) if $\ell \equiv j ~\mathrm {(mod~ 3)}$ , with $j=1,2$ , then y fixes the subspaces $S_{\ell -j}$ and $E_{n-\ell -3 +j}$ , and acts as $(e_{n-\ell -2+j},e_{n-\ell -1+j}, e_{n-\ell +j})$ on $\langle e_{n-\ell -2+j},e_{n-\ell -1+j}, e_{n-\ell +j}\rangle $ .

Setting $g=x$ in case (a), and $g=y$ in case (b), we claim that $K_{\ell +1}:=\langle K_\ell , K_{\ell }^g \rangle $ equals

(5-2) $$ \begin{align} \mathrm{diag}(\mathrm{I}_{n-\ell-1}, \Omega_{\ell+1}^{\overline{\epsilon}}(q)),\quad \overline{\epsilon}\in \{\circ, \pm \}. \end{align} $$

Noting that $g^{-1}S_\ell $ is obtained from $S_\ell $ by replacing $e_{n-\ell +1}$ by $e_{n-\ell }$ , one gets $\langle S_\ell , g^{-1} S_\ell \rangle = S_{\ell +1}$ . Thus, $K_{\ell +1}$ fixes $S_{\ell +1}$ , induces the identity on $E_{n-\ell -1}$ and fixes the restriction of J to $S_{\ell +1}$ , of determinant $1$ . If follows that $K_{\ell +1}$ is contained in the group (5-2). Call $\rho $ the matrix in $\mathrm {GL}_n(q)$ which acts according to $e_{n-\ell }\mapsto -e_{n-\ell }$ , $e_{n-4}\mapsto -2e_n$ , $e_n\mapsto -\tfrac 12 e_{n-4}$ and fixes the remaining vectors $e_i$ . Since $\rho $ has determinant $1$ and spinor norm $(\mathbb {F}_q^*)^2$ , it belongs to $K_\ell ^g$ , which induces $\Omega _{\ell }^\epsilon $ on $g^{-1}S_\ell $ . Now, $\langle \rho , K_\ell \rangle $ is the stabilizer in the group (5-2) of the nondegenerate subspace $\langle e_{n-\ell }\rangle $ . So, it is a maximal subgroup of the group (5-2). From $K_{\ell +1}\nleq \langle \rho , K_\ell \rangle $ , we get the final contradiction $K_{\ell +1}=\mathrm {diag}(\mathrm {I}_{n-\ell -1}, \Omega _{\ell +1}^{\overline {\epsilon }}(q))$ .

It remains to exclude the exceptional cases: in each of them, we get the same contradiction.

Case 1. $r=1$ , $\ell =n-4$ , n even. Let R be the stabilizer of $e_6$ in $K_{n-4}$ . Then, $\langle R^x, K_{n-4}\rangle = K_{n-3}$ , as it fixes the vectors $e_1,e_2,e_3$ and the subspace $E_3^\perp $ , inducing $\Omega _{n-3}(q)$ .

Case 2. $r=2$ , $\ell =n-8$ , n even. Let R be the stabilizer of $e_{10}$ in $K_{n-8}$ . Then, $\langle R^x, K_{n-8}\rangle = K_{n-7}$ , as it fixes the vectors $e_1,e_2,\ldots ,e_7$ and the subspace $E_7^\perp $ , inducing $\Omega _{n-7}(q)$ .

Case 3. $r=2$ , $\ell =n-2$ . Let R be the stabilizer of $e_{3}$ in $K_{n-2}$ . Then, $\langle R^x, K_{n-2}\rangle = K_{n-1}$ , as it fixes $e_1$ and $E_1^\perp $ , inducing $\Omega ^{\overline \epsilon }_{n-1}(q)$ .

Case 4. $r=2$ , $\ell =n-1$ . Similar to the above cases.

Proof. We have $s=e_{n-8} - e_{n-7}$ by the hypothesis $a^2\neq 3$ . Now, $\det (M_1)=-2^{35} a^6 (a^2 - 2) (4 a^4-13 a^2+16) $ and $\det (M_2)=-2^{35} a^6 (a^2 - 2) (28 a^4 - 83 a^2 - 16)$ . Since $a^2\neq 2$ , the matrices $M_1,M_2$ are both singular only if $p=13$ and $a^2=3$ , which is excluded by hypothesis.

Proof. If $q\in \{3, 5, 11\}$ proceed as in the proof of Lemma 5.2. If $q=7$ , take $j=1$ if $a=\pm 1$ and $j=3$ if $a=\pm 2$ ; take $\tilde g=\tau ^2 y \tau y$ if $a=\pm 3$ . Then, the order of $g_j$ and the order of $\tilde g$ are divisible by a prime $\varrho \geq 43$ . If $q=9$ , then $g_1$ has order divisible by $13$ , a prime that does divide $|\mathrm {PSL}_2(17)|$ ; if $q=27$ , then $g_2$ has order divisible by a prime $\varrho \in \{13, 73\}$ .

Proof. We proceed as in the proof of Lemma 5.3, describing only the necessary modifications to prove the primitivity of $K_{|S_9}$ . For $p=3$ , we have $\chi _{(y\tau )^3}(t)=(t-1)f(t)$ , where $f(t)=t^8 -t^7 - (a^{12} - a^6 +1) t^6 - a^{12} t^5 + (a^6 -1) t^4 - a^{12} t^3 - (a^{12} - a^6 + 1) t^2 -t + 1$ . Also in this case, $f(1)=-a^{12}$ and $f(-1)=1$ . If $p\neq 3$ , the product $y\tau $ should have trace $0$ , in contrast with $\mathrm {tr}(y\tau ) =-16$ .

Proof. By Lemmas 6.1 and 6.3, $K_{|S_9}$ is absolutely irreducible and is neither contained in a maximal subgroup in class $\mathcal {C}_2$ of $\Omega _9(q)$ nor contained in any maximal subgroup in class $\mathcal {C}_7$ . Furthermore, by Lemmas 6.2 and 6.4, either $K_{|S_9}=\Omega _9(q)$ or $K_{|S_9}$ is contained in a maximal subgroup $M\in \{\Omega _9(q_0), \mathrm {SO}_9(q_0)\}$ in class $\mathcal {C}_5$ , where $q=q_0^r$ for some prime $r\geq 2$ . Suppose there exists $g\in \mathrm {GL}_{9}(\mathbb {F})$ such that $\tau ^g=\tau _0$ , $y^g= y_0$ , with $\tau _0,y_0\in \mathrm {GL}_{9}(q_0)$ . From $\mathrm {tr}((y\tau )^2)=-2176a^4 + 6784a^2 - 224$ and $\mathrm {tr}((y^2\tau )^2)=1920a^4 - 5504a^2 - 288$ , we get that $-17a^4+53a^2$ and $15a^4-43a^2$ belong to $\mathbb {F}_{q_0}$ , whence $64a^2\in \mathbb {F}_{q_0}$ . We conclude that $K_{|S_9}=\Omega _9(q)$ .

Proof. Since $K_{|S_9}=\Omega _9(q)$ , we can repeat the argument of Corollary 5.6, proving that $H=\Omega _{n}^+(q)$ . For the second part of the statement, we have to prove that there exists an element a satisfying all the hypotheses. If $q=p$ , take $a=1$ . Suppose now $q=p^f$ with $f\geq 2$ and let $\mathcal {N}(q)$ be the number of elements $b\in \mathbb {F}_q^*$ such that $\mathbb {F}_p[b^2]\neq \mathbb {F}_q$ . By [Reference Pellegrini, Tamburini Bellani and Vsemirnov12, Lemma 2.7], it suffices to check that the condition $p^f - 2p ({p^{\lfloor f/2 \rfloor }-1})/({p-1})> 1$ is always fulfilled (the requirements $a^2\neq 2,3$ can be dropped).