As discussed above, the quadratic form $Q$ on  $V=(\F_q)^{2m+1}$ has trivial radical, while the radical  $V^\perp$ of its associated symmetric bilinear form $(\cdot,\cdot)$  is the nonsingular 1-subspace $\langle d\rangle$. Moreover, the  form $(\cdot,\cdot)$ induces a nondegenerate alternating form on $V/V^\perp$ given by $(u+V^\perp,v+V^\perp)=(u,v)$ for all $u,v\in V$.
Since $\GammaO_{2m+1}(q)$ acts faithfully on $V/V^\perp$ it follows that $\GammaO_{2m+1}(q)\cong\GammaSp_{2m}(q)$.

The families $\cU$ of subspaces we treat in this case (as a careful reading of Table <ref> reveals) are $\cP_k$ for $k\leq m$, $\cN_k^\eps$ for even $k\leq m$ and  $\eps\in\{+,-\}$, $\cN_1$ (all nonsingular $1$-subspaces apart from  $\langle d\rangle$),  and $\cN_{2m}^\eps$ for $\eps\in\{+,-\}$.

Suppose that  $H\le\GammaO_{2m+1}(q)$, where $m\geq1$ and $q$ is even, such that $\Omega_{2m+1}(q)'\not\le H$. Suppose that $H$ acts transitively on a set $\cU$ of subspaces of $V=(\F_q)^{2m+1}$ given by Table $\ref{T-defn}$. Then either $m=1$ and $H$ is given by Theorem $\ref{T:Omega3}$, or $m\geq2$ and one of the following holds:

* $\cU=\cP_1$,  $H_0\lhdeq H$ where $(H_0,m,q)$ is given in
Table ; or

Theorem <ref>(a),(c) subgroups $H_0\lhdeq H\le\GammaO_{2m+1}(q)$ where $H$ is transitive
on $\cP_1$  and on $\cN_1$ and $q$ is even.

$H_0$       $\Sp_{2a}(q^b)$    $\textup{G}_2(q^b)'$

$ab>a$   $3b$

* $\cU=\cP_m$,  and $\Omega_{2m}^{-}(q)\lhdeq H$ where $m\ge2$; or

* $\cU=\cN_1$, and $H_0\lhdeq H$ where $(H_0,m,q)$ is given in
Table ; or

* $\cU=\cN_2^\eps$ and $H_0\lhdeq H$ where $(H_0,m,q)$ is given in Theorem $\ref{T3}$(c); or

* $\cU=\cN_{2m}^-$, and either $H$ is described in  <cit.>*Lemma 4.6, or $q=2$ or $4$ and $\Omega_{2m}^+(q)\lhdeq H$, or $(m,q)=(2,2)$ and  $H$ is a transitive subgroup of $\Sym_6$; or

* $\cU=\cN_{2m}^{+}$ and $H_0\lhdeq H$ where $H_0$
is given in Table  and $b_4\in\{1,2,4\}$.

Theorem <ref>(f) subgroups $H_0\lhdeq H\le\GammaO_{2m+1}(q)$ where $H$ is transitive
on $\cN_{2m}^{+}(\F_q^{\;2m+1})$ and $q$ is even.

$H_0$       $\Sp_{2a}(q^b)$    $\SU_{a}(q^{b_4})$   $\Omega_{2a}^{-}(q^{b_4})$     $\G_2(q^b)'$   $\Sz(q^b)$

$ab$    $ab_4$    $ab_4$    $3b$   $2b$

$2,4\tnote{*}$   $2,4 \tnote{*}$   $q$   $2^\textup{odd}$  $q^b\geq 8$

$H_0$         $\PSL_2(17)$   $3_{+}^{1+2}$

$4$   $3$

$2$   $2$

* If $q=4$ then $b_4=1$; if $q=2$ then $b_4\in\{1,2\}$.

Let $G=\GammaO_{2m+1}(q)$. The possibilities for $\cU$ were given before the statement, and in each case $G$ leaves $\cU$ invariant.  Let $H\leqslant G$ act  transitively on  $\cU$  such that $\Omega_{2m+1}(q)\not\leqslant H$. If $m=1$ then the possibilities for $H$ are given by Theorem <ref>. We assume from now on that $m\geq2$.  Let $\overline{V}=V/V^\perp$ and for each $U\in \cU$ let $\overline{U}=(U+V^\perp)/V^\perp$.    Note that $\Omega_{2m+1}(q)=\SO_{2m+1}(q)=\GO_{2m+1}(q)\cong\Sp_{2m}(q)$, unless $(m,q)=(2,2)$, in which case $|\SO_5(q):\Omega_5(2)|=2$.  Hence, if $\Omega_{2m+1}(q)'\not\leqslant H$  then $H$ does not induce $\Sp_{2m}(q)'$ on $\overline{V}$.

Case $\cU=\cP_k$ for some $k\leq m$.  Since $U\in\cU$ is totally singular,  $U\cap V^\perp=\{0\}$ and so $\overline{U}$ is a totally isotropic $k$-subspace of $\overline{V}$.
Moreover, since $\Sp_{2m}(q)$ acts transitively on the set of totally isotropic  $k$-subspaces of $\overline{V}$ it follows that all such subspaces of $\overline{V}$ arise in this manner. Moreover, by <cit.>*p. 143, $U$ is the unique totally singular $k$-subspace of $U+V^\perp$ and so there is a one-to-one correspondence between $\cU$ and the set of totally isotropic $k$-subspaces of $\overline{V}$.
Thus $H$ acts transitively on $\cU$ if and only if it acts transitively on the set of totally isotropic $k$-subspaces of $\overline{V}$. Thus by Theorem <ref> we have $k=1$ or $m$ and $H$ is listed there. This gives parts (a) and (b) and note that $\Sp_{2a}(q^b)\cong \Omega_{2a+1}(q^b)$.

Case $\cU=\cN_1$. Here $U\in \cU$ is a nonsingular 1-subspace other than $V^\perp=\langle d\rangle$. Then $U+V^\perp$ contains a unique singular 1-subspace $\langle w\rangle$ and the nonsingular 1-subspaces of $U+V^\perp$ are $\langle d+\lambda w\rangle$ for $\lambda\in\F_q$. Moreover, any singular 1-subspace $\langle w\rangle$ in $V$ gives rise to $q-1$ nonsingular 1-subspaces in $V^\perp+\langle w\rangle$
distinct from $V^\perp$. Hence $H\leqslant G$ is transitive on $\cU$ if and only if $H$ is transitive on $\cP_1$ and for each $\langle w\rangle \in\cP_1$, the subgroup $H_{\langle w\rangle}$ acts transitively on the set of $1$-subspaces $\langle d+\lambda w\rangle$ for $\lambda\in\F_q\setminus\{0\}$, equivalently,  $H_{\langle w\rangle}$ acts transitive on the nonzero vectors in $\langle w\rangle$.   Hence by part (a),  $H_0\lhdeq H$ where $H_0$ is given in Table <ref>. Moreover, for both possibilities, $(H_0)_{\langle w\rangle}$ induces a cyclic group of order $q-1$ on $\langle w\rangle$ (see <cit.>*Lemma 4.1.2 when $H_0=\Sp_{2a}(q^b)$ and <cit.>*Section 4.3.5 for $H_0=G_2(q^b)'$). Thus both possibilities for $H_0$ are also transitive on $\cN_1$, proving part (c).

Case $\cU=\cN_k^\eps$ for some even $k\leq m$ and $\eps\in\{+,-\}$. Here $U\in\cU$ is a nondegenerate $k$-subspace. Hence $V^\perp\cap U=\{0\}$ and $\overline{U}$ is a nondegenerate $k$-subspace of $\overline{V}$. Since $\Sp_{2m}(q)$ is transitive on the set of nondegenerate $k$-subspaces of $\overline{V}$, all such subspaces of $\overline{V}$ arise in this way. Thus by Theorem <ref>, $k=2$ and $H$ is one of the groups in  Table <ref>.  In particular, $\cU=\cN_2^+$ or $\cN_2^-$. Let $W=U+V^\perp$, a nondegenerate $3$-subspace of $V$.  Now $H$ is transitive on $\cN_2^\eps$ if and only if $H_{W}$ is transitive on the set of $2$-subspaces of $W$ contained in  $\cN_2^\eps$.

For $H$ as in the first and last columns of Table <ref>, it follows from <cit.>*3.2.1(a), p. 47 that  $\Hbar_{\overline{U}}^{\overline{U}}\rhd \Sp_2(q)$ and so $H_W^W\lhdeq\GO_3(q)$. By Witt's Lemma,  $\GO_3(q)$ is transitive on the set of nondegenerate 2-subspaces of $W$ of each isometry type, as required.
We note that in the last column, the $\Sp_2(q^2)$ is occurring as an extension field group and not as $\Omega_4^-(q)$ as it is not possible for the stabiliser of a nondegenerate hyperplane to be transitive on $\cN_2^\eps$.

For the remaining columns of Table <ref>, $\soc(H)=\G_2(q^{m/3})'$. By <cit.>*p. 27 and Lemma 5.4, $\G_2(q^{m/3})$ acts transitively on the set of nondegenerate 2-subspaces of a 6-dimensional space over $\F_{q^{m/3}}$ equipped with a nondegenerate alternating form such that the stabiliser $K$ of such a subspace $X$ is the maximal subgroup $\SL_2(q^{m/3})\times\SL_2(q^{m/3})$. Moreover, by
<cit.>*Section 4.3.6 we see that $K$ induces $\SL_2(q^{m/3})\cong\GO_3(q^{m/3})$ on the nondegenerate 3-subspace corresponding to $X$ of the 7-dimensional vector space over $\F_{q^{m/3}}$ equipped with a nondegenerate quadratic form.   Thus, if  $X=\langle e_1,f_1\rangle$, then the stabiliser in $K$ of the $\F_{q}$-span $Y$ of $\{e_1,f_1\}$ induces $\SL_2(q)\cong\Sp_2(q)$ on $Y$. Hence  $\GO_3(q)\lhdeq H_W^W$ acts transitively on the set of subspaces in $\cN_2^\eps$ contained in $W$ and so each group in Table <ref> that is transitive on the $\cN_2$ subspaces of $\Vbar$ is transitive on $\cN_2^\eps$, proving part (d).

Case $\cU=\cN_{2m}^\eps$ for some $\eps\in\{+,-\}$. Here $U$ is a nondegenerate $2m$-subspace and $(U+V^\perp)/V^\perp=\overline
{V}$.  By <cit.>*Theorem 1.5, $\Sp_{2m}(q)_U=\GO_{2m}^\eps(q)$ is maximal in $\Sp_{2m}(q)$  and so $G_U$ is maximal in $G$. We assume the same setup as in the proof of Theorem <ref>,
and in particular that $H\leqslant B$ where $\overline{B}$ is a core-free subgroup of $\PGammaSp_{2m}(q)$, and that $G^*=\overline{B}(G^*)_U$ is a core-free maximal factorisation, where $G^*$ is an almost simple group with socle $L=\PSp_{2m}(q)$. Moreover, $B=N_G(B\cap \Omega_{2m+1}(q))$.

Suppose first that $\cU=\cN_{2m}^{-}$. Here, when $(m,q)\neq (2,2)$,  the transitive
subgroups of $G$ not containing $\Sp_{2m}(q)$ were computed
in <cit.>*Lemmas 4.6 with reference to <cit.>*Lemma 4.5.  We note that however, the reference to <cit.>*Lemma 4.5 forgot to include the possibility that $\Omega_{2m}^+(q)\lhdeq H$ when $q=2$ or 4.
When $(m,q)=(2,2)$ we have $|\cN_4^-|=6$ and $\GO_5(2)\cong\Sym_6$. This gives part (e).

Finally, suppose that $\cU=\cN_{2m}^{+}$.
Suppose first that $(m,q)=(2,2)$. Then a Magma computation shows that either $\Alt_6=\Omega_5(2)$ or $\Alt_5$ are normal subgroups of $H$, or $H=F_{20}$. Moreover, both classes of $\Alt_5$-subgroups in $G$ are transitive. The two $\Alt_5$ subgroups correspond to $\Omega_4^-(2)$ and $\Omega_3(4)$, each of which appears in Table <ref>, while there is a unique conjugacy classes  of $F_{20}$ subgroups and so these must be normalising $\Omega_2^-(2^2)\cong \C_5$ as in the third column of Table <ref>.
Thus we may assume that $(m,q)\ne(2,2)$, and then
it follows from <cit.>*p. 10, p. 13 that we have the five cases
in Table <ref>.  For reasons which will become clear later we deal with (f1) last.

Subcases when $L=\PSp_{2m}(q)$ and $\cU=\cN_{2m}^{+}$.

(f1)    (f2)    (f3)    (f4)    (f5)

$B\cap \Sp_{2m}(q)$        $\Sp_{2a}(q^b)\ldotp b$    $\GO^{-}_{2m}(q)$    $\Sz(q)$    $\G_2(q)$    $\PSL_2(17)$

$m$      $ab>a$   $\ge2$   $2$   $3$   $4$

$q$      ${\rm even}$   $2,4$   $2^\textup{even}\ge8$   ${\rm even}$   $2$

(f2) Here $q=2$ or $4$ and $B=N_G(\Omega_{2m}^-(q))$.  We record $\Omega_{2m}^-(q)\lhdeq H$ in Table <ref>.  Suppose now that $\Omega_{2m}^-(q)\not\leqslant H$. Then we have a factorisation $B=HB_U$. By <cit.>*3.2.4(e), pp. 51–52 $(\GO_{2m}^-(q))_U$ is contained
in the stabiliser in $\GO^-_{2m}(q)$ of a nonsingular 1-subspace of the natural $(2m)$-dimensional module for $\GO^-_{2m}(q)$. Thus $H$ is given by Theorem <ref>. If $m\geq 4$ then either $\SU_m(q)\lhdeq H$ or $H$ is given in Table <ref>,  and hence appears in Table <ref>.  For $m=3$, Theorem <ref> implies that either $\SU_3(q)\lhdeq H$ or $q=2$ and  $3_+^{1+2}\lhd H$. Finally, if $m=2$ then note that we have already considered the case $q=2$, while for $q=4$ a Magma calculation shows that no examples exist.

(f3) In this case $m=2$ with $q\geq 8$ and $B=N_G(\Sz(q))$. By <cit.>*Table 2, $\Sz(q)$ is transitive on $\cU$ and so $\Sz(q)\lhdeq H$ is listed in Table <ref>. Suppose now that $\Sz(q)\not\leqslant H$. Then we have a factorisation $B=HB_U$. However, <cit.>*Table 5 implies that no such factorisation exists.

(f4) In this case $m=3$ and $B=N_G(\G_2(q))$. By <cit.>*Table 2, $\G_2(q)$ is transitive on $\cU$ and so $\G_2(q)'\lhdeq H$ is listed in Table <ref>. Moreover, a Magma computation shows that when $q=2$, the group $\G_2(2)'\cong\PSU_3(3)$ is also transitive. Suppose now that $\G_2(q)'\not\leqslant H$. Then we have a factorisation $B=HB_U$. By <cit.>*(5.2.3b) p. 111 we have that $(\G_2(q))_U=\SL_3(q).2$. However, by <cit.>*Table 5 there are no such factorisations of an almost simple group with socle $\G_2(q)$ for $q$ even.

(f5) In this case $G=\GO_9(2)$ and $B=\PSL_2(17)$. We record $\PSL_2(17)$ in Table <ref> and now suppose that $H<B$.
Now $|\cN_8^{+}|=136$ and as there is a single conjugacy class of subgroups of $B=\PSL_2(17)$ of
index $136$ namely dihedral groups $D_{18}$, we have
$B_U=D_{18}$. Hence we
have a factorisation $B=\PSL_2(17)=H B_U$. Note that $D_{18}$ is the subgroup of $\PSL_2(17)$ arising from $\GammaL_1(17^2)$. However, by <cit.> no such   factorisation exists (though one does exist of $\PGL_2(17)$).

(f1) Here $B= N_G(\Sp_{2a}(q^b))$. It is shown in <cit.>*3.2.1(d) that $(\Sp_{2a}(q^b).b)\cap \GO^+_{2m}(q) = \GO_{2a}(q^b).b$
(and the argument holds for all $b$). It follows that the subgroup $\Sp_{2a}(q^b)$ is also transitive on $\cN_{2m}^+$, and hence  if
$\Sp_{2a}(q^b)\lhdeq H$ then $H$ is transitive on $\cU$ and we have listed this in Table <ref>. Suppose now that $\Sp_{2a}(q^b)\not\leqslant H$. Then we have a factorisation $B=HB_U$. As just noted, $(\Sp_{2a}(q^b))_U=\GO^+_{2a}(q^b)$ and so the result follows by induction on $m$ and noting that $q^b\neq 2$.

§ ORTHOGONAL GROUPS  IN EVEN DIMENSION OF PLUS TYPE

Let $V=(\F_q)^{2m}$, with $m\geq4$, equipped with a nondegenerate quadratic form $Q$ of plus type and associated bilinear form $(\dot,\dot)$. We let $\{e_1,\ldots,e_m,f_1,\ldots,f_m\}$ be a basis for $V$ such that $Q(e_i)=Q(f_i)=0$ and $B(e_i,f_j)=\delta_{ij}$ for all $i,j$.

In Subsection <ref>, we find all subgroups of $\GammaO_{2m}^+(q)$ that are transitive on a set $\cU$ given by Table <ref>, namely $\cP_k$ for $k< m$, $\cP_m^\eps$ for  $\eps\in\{+,-\}$, $\cN_k^\eps$ for even $k\leq m$ and  $\eps\in\{+,-\}$, and if   $q$ is odd also  $\cN_k^\eps$ for odd $k\leq m$ and  $\eps\in\{+,-\}$, and if $q$ is even also $\cN_1$. The union $\cP_m=\cP_m^+\cup\cP_m^-$ is a $\GammaO_{2m}^+(q)$-orbit, and also  if $k$ and $q$ are both odd then   $\cN_k=\cN_k^+\cup\cN_k^-$ is a $\GammaO_{2m}^+(q)$-orbit. Subgroups acting transitively on these latter sets are determined in Subsection <ref>, namely in Proposition <ref> for $\cP_m$ and Proposition <ref> for $\cN_k$.    The eight-dimensional case needs special attention and we do some preparatory work in Subsection <ref>.

The subgroups of $\GammaO_{2m}^+(q)$ that are transitive on $\cN_1$ when $q$ is even, or on $\cN_1^\eps$, where $\eps\in\{+,-\}$, when $q$ is odd are determined in <cit.>*Lemma 4.5. However,  some of the groups in <cit.>*Lemma 4.5 are given as subgroups of $\PGammaO_{2m}^+(q)$ instead of $\GammaO_{2m}^+(q)$, and in addition $\Alt_6$ is incorrectly listed when $(m,q)=(4,3)$ (see the proof of Lemma <ref> (e)). We therefore give a revised version of this classification as part (e)  of Theorem <ref> (see, in particular, Table <ref>). See also Remark <ref>(e).

The formulas for $|\cN_1^\eps|, |\cP_1|,|\cP_m^\eps|$, and $|\cN_2^{\eps}|$ (see <cit.>*pp. 62, 63, 67, 70) are:

\begin{align}
|\cN_1|&=q^{m-1}(q^m-1) \quad \textrm{ for $q$ even}, &  |\cN_1^\eps| =\frac{q^{m-1}(q^m-1)}{2} \quad \textrm{ for $q$ odd,}\\
|\cP_m^\eps|&=\prod_{i=1}^{m-1}(q^i+1)=|\cP_{m-1}|, &|\cN_2^{\eps}|=\frac{q^{2m-2}(q^m-1)(q^{m-1}+\eps 1)}{2(q-\eps 1)}.\label{E:OPlus2}
\end{align}

The formula for $|\cP_1|$ factors to give $|\cP_1|=\frac{(q^m-1)(q^{m-1}+1)}{q-1}$. When $m=4$, we see that $|\cP_1|=(q+1)(q^2+1)(q^3+1)=|\cP_4^{\cmt{$\eps$}}|$.

§.§ Eight-dimensional orthogonal groups of  plus type

The groups $\POmega_8^+(q)$ have more outer automorphisms than their counterparts in higher dimensions and this affects both their subgroup structure and actions on subspace families. We first make some technical comments, in particular concerning our notation.

When $L=\POmega_8^{+}(q)$ we have  $|\Aut(L):\PGammaO_8^+(q)|=3$. Let $\tau$ be an outer automorphism of $L$ of order $3$ that is not in $\PGammaO_8^+(q)$. It is well-known that $\tau$ induces a  3-cycle $(\mathcal{P}_1, \mathcal{P}_4^+, \mathcal{P}_4^-)$.  We note that $\tau$ does not induce an automorphism of the matrix group $\Omega_8^+(q)$ when $q$ is odd. Indeed the stabiliser in $\Omega_8^+(q)$  of a totally singular 1-subspace is not isomorphic to the stabiliser of a totally singular 4-subspace.

There are two types of subgroups $\Omega_7(q)$ of $L=\POmega_8^{+}(q)$ which
we denote by $\Omega_7(q)^R$ and $\Omega_7(q)^I$, where $\Omega_7(q)^R$ acts reducibly on $V$ fixing a nonsingular 1-subspace,
and $\Omega_7(q)^I$ acts irreducibly on $V$. If $q$ is odd, there are four $L$-conjugacy classes of $\Omega_7(q)^I$ and two of $\Omega_7(q)^R$; we note that $\Omega_7(q)^R$ has preimage $2\times \Omega_7(q)$ in $\Omega_8^+(q)$ while $\Omega_7(q)^I$ has preimage $\nonsplit{2}{\Omega_7(q)}$ <cit.>*Table 8.50. When $q$ is even, there are two $L$-conjugacy classes of subgroups $\Omega_7(q)^I$ and one of $\Omega_7(q)^R$; and in this case each preimage is isomorphic to $\Omega_7(q)\cong\Sp_6(q)$.  Furthermore,  $\tau$ induces 3-cycles    $(\Omega_7(q)^I,\Omega_7(q)^I,\Omega_7(q)^R)$ on these classes, for all values of $q$. We use  ${}^\wedge\Omega_7(q)^I$ to denote the minimal preimage of $\Omega_7(q)^I$ in $\Omega_8^+(q)$. Hence ${}^\wedge\Omega_7(q)^I=\nonsplit{2}{\Omega_7(q)}$ when $q$ is odd,  and is equal to $\Omega_7(q)^I$ when $q$ is even.

Let $M_2$ denote an imprimitive subgroup of $L$ preserving a decomposition $V=E\oplus F$ into totally singular 4-subspaces
$E$ and $F$. There are two $L$-conjugacy classes of subgroups $M_2$ and one $L$-conjugacy class of subgroups $N_2^{+}$, and $\tau$ induces a 3-cycle $(M_2,M_2,N_2^{+})$ on these classes.
Finally, we note that $\Omega_8^+(q)$ has two conjugacy classes of subgroups isomorphic to $\SU_4(q)$. Letting $K_1$ and $K_2$ be the images in $L$ of two such non-conjugate subgroups, the outer automorphism $\tau$ induces a 3-cycle $(N_L(K_1),N_L(K_2),N_2^-)$.
These observations are important in the proof of Theorem <ref> when considering actions on $\cP_1$, $\cN_2^+$ and $\cN_2^-$.

Arguments for the group $\POmega_8^+(3)$ are so delicate that we have separated them out in the following lemma.

Let $H\leqslant \GammaO^+_8(3)$ with $\Omega_8^+(3)\not\leqslant H$.

* If $H$  acts transitively on a set $\cU$ of subspaces given in Table $\ref{T-defn}$ then $H$ is listed in Theorem $\ref{T10}$.

* If $H$ acts transitively on $\cP_4$ then $H$ is given in Proposition $\ref{P:Both}$.

* If $H$ acts transitively on $\cN_1$ then $H$ is given in Proposition $\ref{P2}$.

(1) Suppose first that $\cU$ is a set of subspaces given in Table $\ref{T-defn}$. Let  $G$ be the stabiliser in $\GammaO^+_8(3)=\CO^+_8(3)$ of $\cU$ and let $H\leqslant G$ act transitively on $\cU$ such that $\Omega_8^+(3)\not\leqslant H$. Then $G=\CO^+_8(3)$ unless $\cU=\cP_4^+,\cP_4^-,\cN_1^\eps$ or $\cN_3^\eps$. In the first two of the exceptional cases $G$ is an index 2 subgroup of $\CO_8^+(3)$ containing $\SO_8^+(3)$ while in the last two cases we have $G=\GO_8^+(3)$.   For $X\leqslant \CO^+_8(3)$ denote the image of $X$ in $\PCO^+_8(3)$ by $\overline{X}$. We now deal with each possibility for $\cU$ from Table $\ref{T-defn}$, in most cases using Magma.

(a) First consider $\cU=\cN_4^\eps$, where $\eps\in\{+,-\}$, and $G=\CO^+_8(3)$. Let $L=\Omega_8^+(3) $ and note that $L$ is transitive on $\cU$. Now  by <cit.>*Table 4.1.2, $|\cU|=\frac12 3^8(3^2+1)^2(3^4+3^2+1)$ if $\eps=+$ and  $|\cU|=\frac12 3^8(3^2-1)(3^6-1)$ if $\eps=-$. In either case $|\cU|>3^{15}$, and since $H$ is transitive on $\cU$ we have
$|H|>3^{15}$.  Also (see <cit.>*Table 2.1C) $|G:L|=8$ and  $|\POmega_8^+(3)|=\frac14 3^{12}(3^6-1)(3^4-1)^2(3^2-1)
<\frac14 .3^{28}<3^{29}$, and it follows that $|L|<3^{30}$ and  $|H\cap L| \geq \frac{1}{8} |H|> 3^{13}>  |L|^{1/3}$. Thus $H\cap L$ is one of the groups from <cit.>*Table 7 or Proposition 4.23.  Moreover for both values of $\eps$ the order $|H\cap L|$ is divisible by $3^4+3^2+1 = 7\times 13$, and  the only one of these subgroups with this property is  $H\cap L\cong \Omega_7(3)$.  In particular $H\cap L$ is maximal in $L$ and since, for $U\in\cU$,  $L$ is transitive on $\cU$ and its overgroup $L_{\{U,U^\perp\}}$ is maximal in $L$, we have a maximal core-free factorisation $L= (H\cap L)( L_{\{U,U^\perp\} })$.  However there are no such factorisations by <cit.>*Table 4.

(b)  Next $\cU=\cN_3^\eps$ and $G=\GO^+_8(3)$. Note that $\cN_3^+$ and $\cN_3^-$ are fused in $\CO_8^+(3)$. By Magma and Lemma <ref>, we deduce that $\nonsplit{2}{\Omega_7(3)}\lhdeq H$, which appears in Theorem <ref> (g).

(c) Next consider $\cU=\cN_2^+$ and $G=\CO^+_8(3)$.  A Magma calculation and Lemma <ref> shows that $\nonsplit{2}{\Omega_7(3)}\lhdeq H$, which appears in Table <ref>.

(d) The next  case  is $\cU=\cN_2^-$ and $G=\CO^+_8(3)$.  By Magma and Lemma <ref>, we deduce that either $\nonsplit{2}{\Omega_7(3)}\lhdeq H$ (which appears in Table <ref>), or $3^6\rtimes \PSL_4(3)\lhdeq \oH$ and $H$ is contained in $P_4$. We note that since $\SL_4(3)$ is quasisimple, Lemma <ref> applied to the subgroup $\PSL_4(3)$ of $\oH$ implies that $3^6\rtimes \SL_4(3)\lhdeq H$, as in Theorem <ref> (f).

(e) Next consider $\cU=\cN_1^\eps$ and $G=\GO^+_8(3)$.  Note that $\cN_1^+$ and $\cN_1^-$ are fused in $\CO_8^+(3)$.  The possibilities for $H$ are given by  <cit.>*Lemma 4.5 but there only the projective versions are listed.  Combining  <cit.>*Lemma 4.5 and a Magma calculation we obtain the following list of examples:

* $H\rhdeq \SL_4(3)$, $\SU_4(3)$ or $\Sp_4(3)$,

* $H\rhdeq \nonsplit{2}{\Omega_7(3)}$, $\nonsplit{2}{\Omega_8^+(2)}$,  $\nonsplit{2}{\Sp_6(2)}$, $\nonsplit{2}{\SU_4(2)}$, $\nonsplit{2}{\Alt_9}$,

* $H\leqslant P_4$ and modulo the unipotent radical $H$ induces a subgroup of $\GL_4(3)$ which is transitive on 1-subspaces.

The subgroups in parts (i) and (ii) appear in Table <ref>, and those in part (iii) are as in Theorem <ref> (e). Note that in (ii), $\nonsplit{2}{\Omega_7(3)}$ is the group
${}^\wedge\Omega_7(3)^I$, and  $\PSU_4(2)\cong\PSp_4(3)$ but this occurs differently to the group $\Sp_4(3)$ in (i). Also no groups with normal subgroup $\Alt_6$, as listed in <cit.>*Lemma 4.5(vii)  for $m=4, q=3$, act transitively on $\cN_1^\eps$.

(f) Here $G=\CO_8^+(3)$ acts on $|\cP_1|=1120$ singular $1$-subspaces. A  Magma  calculation shows that $H$ contains one of the following as a normal subgroup $\nonsplit{2}\Omega_7(3)$, $  \nonsplit{2}{\Omega^+_8(2)}$, $\nonsplit{2}{\Alt_8}$, $\nonsplit{2}{\Alt_9}$, $\nonsplit{2}{\Omega_7(2)}$,  $\nonsplit{4}{\PSL_3(4)}$, $\SU_4(3)$, $2^{1+6}_+.\Alt_7$, or $2^{1+6}_+.\Alt_8$. These all appear in Table <ref>.

(g)  When $\cU=\cP_2$, a Magma calculation reveals that there are no possibilities for $H$.

(h) When $\cU=\cP_3$ we have  $G=\CO_8^+(3)$ and a Magma calculation shows that $H$ contains $\nonsplit{2}{\Omega_8^+(2)}$ as a normal subgroup, as in Theorem <ref> (c).

(i)  Finally suppose that $\cU=\cP_4^\eps$ and $G$ is the index two subgroup of $\CO^+_8(3)$ that fixes each $\Omega_8^+(3)$-orbit on totally singular 4-subspaces. Note that if $H$ is transitive on $\cP_1$ then $\oH^\tau$ is transitive on $\cP_4^+$ or $\cP_4^-$ where $\tau$ is a suitable triality automorphism. The $\CO_8^+(3)$-conjugacy class consisting of $\nonsplit{2}\Omega_7(3)$ subgroups that are transitive on $\cP_1$ splits into two conjugacy classes under $G$. Applying triality to their images in $\oG$ and pulling back to $G$ we find that both $\Omega_7(3)^R\lhdeq H$ and $\nonsplit{2}\Omega_7(3)\lhdeq H$ are transitive on $\cP_4^\eps$ for $\eps=\pm$  (and both lie in subgroup families in Table <ref>). Similarly, the  $\textup{CO}_8^+(3)$-conjugacy class of $\textup{SU}_4(3)$ subgroups that are transitive on $\cP_1$ splits into two conjugacy classes under $G$. Applying triality to their images in $\oG$ and pulling back to $G$ we find that both $\SU_4(3)\lhdeq H$ and $\Omega_6^-(3)^R\lhdeq H\leqslant N_2^-$,  are transitive on $\cP_4^\eps$ for $\eps=\pm$  (and both lie in subgroup families  in Table <ref>).
Also, using <cit.>*Table 8.50 and Magma we see that all of $2^7:\Alt_8$ and $2^{1+6}_+.\Alt_8$, and $2^6:\Alt_7$ and $2^{1+6}_+.\Alt_7$ are transitive on $\cP_4^\eps$, for $\eps=\pm$. Finally, a Magma <cit.> computation yields the following remaining examples for a normal subgroup of $H$: $\nonsplit{2}{\Omega^+_8(2)}$, $\Omega_7(2)$, $\nonsplit{2}{\Omega_7(2)}$ $\nonsplit{4}{\PSL_3(4)}$, $\nonsplit{2}{\PSL_3(4)}$, $\nonsplit{2}{\Alt_8}$, $\Alt_8$, $\nonsplit{2}{\Alt_9}$ and $\Alt_9$. All of these groups are in Table <ref>.

(2)  Next suppose that $\cU=\cP_4$. Then a Magma calculation shows that $H$ satisfies one of the following, and these are the groups occurring in Table <ref> for Proposition <ref>  when $(m,q)=(4,3)$.

* $H\rhdeq \Omega_7(3)^R$, $\Omega_6^-(3)^R$;

* $H=\nonsplit{2}{\GO_7(2)}$, $\nonsplit{2}{\GO^+(8,2)}$;

* $H\rhdeq \Alt_8$, $\Alt_9$, $\nonsplit{2}{\PSL_3(4)}$, $\nonsplit{4}{\PSL_3(4)}$

* $H\rhdeq 2^7:\Alt_8$,  $2^6:\Alt_7$,

(3)  Finally, suppose that $\cU=\cN_1$ and that $H$ is transitive on $\cU$. Then $H$ has an index two subgroup $H^+$ that is transitive on $\cN_1^+$ and $\cN_1^-$, and so $H^+$ has already been determined in part (1). A Magma calculation then shows that one of the following holds, and so $H$ is as given by Proposition <ref>.

* $H\rhdeq \SL_4(3)$, $\SU_4(3)$ or $\Sp_4(3)$;

* $H= \nonsplit{2}{\SO_7(3)}$;

* $H\leqslant P_4$, $H$ contains an element of $\CO_8^+(3)\backslash\GO_8^+(3)$, and  modulo the unipotent radical of $P_m$,  $H\cap\GO_8^+(8,3)$  induces a subgroup of $\GL_4(3)$ that is transitive on $1$-subspaces.

§.§ Actions on subspace families from Table

We now consider all the subspace families $\cU$ from Table <ref>, for dimensions $2m\geq 6$.

Suppose that  $H\le\GammaO_{2m}^{+}(q)$, where $m\geq 3$, such that $\Omega_{2m}^{+}(q)\not\le H$. Let
$V=(\F_q)^{2m}$ and suppose that $H$ acts transitively on a set $\cU$
of subspaces of $V$ given by Table $\textup{\ref{T-defn}}$.
Then one of the following holds:

* $\cU=\cP_1$ and $H_0\lhdeq H$ where $(H_0,m,q)$ is given in
Table $\ref{T:OPlusa}$; or

(a) Subgroups $H_0\lhdeq H\le\GammaO^+_{2m}(q)$ where $H$ is transitive
on $\cP_1$.

$H_0$       $\Alt_7$   $\Alt_9$   $\nonsplit{2}{\Alt_8}$   $\nonsplit{2}{\Alt_9}$   $2^{1+6}_+\ldotp\Alt_7$   $2^{1+6}_+\ldotp\Alt_8$

$3$   $4$   $4$   $4$   $4$   $4$

$2$   $2$   $3$   $3$   $3$   $3$

$H_0$       $\nonsplit{4}{\PSL_3(4)}$   $\nonsplit{2}{\Omega_7(2)}$   $\nonsplit{2}{\Omega_8^+(2)}$   ${}^\wedge\Omega_7(q)^I$   $\SU_m(q)$

$4$   $4$   $4$   $4$

$3$   $3$   $3$

* $\cU=\cP_2$, $m=3$, $q=2$ and $\Alt_7\lhdeq H$; or

* $\cU=\cP_3$, $m=4$, $q=3$ and $\nonsplit{2}{\Omega_8^+(2)}\lhdeq H$; or

* $\cU=\cP_m^\eps$, and either $m=3$ and $H\leqslant N_{A}(\Omega_2^-(q^2))\leqslant N_4^-$ where $A$ is the stabiliser in $\GammaO_6^+(q)$ of $\cP_3^+$, or $H_0\lhdeq H$ where $(H_0,m,q)$ is given in
Table $\ref{T:OPlusb}$; or

(d) Subgroups $H_0\lhdeq H\le\GammaO^+_{2m}(q)$
where $H$ is transitive on $\cP_m^+$ or $\cP_m^-$ (excluding the case $m=3$ and  $H\leqslant N_A(\Omega_2^-(q^2))\leqslant N_4^-$.)

$H_0$       $\Omega_{2m-1}(q)^R$    $\Omega^{-}_{2m-2}(q)^R$     ${}^\wedge\Omega_7(q)^I$   $\SU_4(q)$   $2^4\colon \C_5$   $2^4\ldotp\Alt_5$   $\Alt_5$   $\Alt_7$

$\ge3$   $\ge3$   $4$   $4$   $3$   $3$   $3$   $3$

$3$   $3$   $3$   $2$

$H_0$       $2^{1+6}_+\kern-2pt\ldotp\kern-2pt\Alt_7$    $2^6:\Alt_7$    $2^{1+6}_+\kern-2pt\ldotp\kern-2pt\Alt_8$   $2^7:\Alt_8$   $\nonsplit{2}{\Omega_8^+(2)}$   $\nonsplit{2}{\Omega_7(2)}$   $\Omega_7(2)$

$4$   $4$   $4$   $4$   $4$   $4$   $4$

$3$   $3$   $3$   $3$   $3$   $3$   $3$

$H_0$       $\nonsplit{4}{\PSL_3(4)}$   $\nonsplit{2}{\PSL_3(4)}$   $\nonsplit{2}{\Alt_8}$   $\Alt_8$   $\nonsplit{2}{\Alt_9}$   $\Alt_9$   $\Alt_9$

$4$   $4$   $4$   $4$   $4$   $4$   $4$

$3$   $3$   $3$   $3$   $3$   $3$   $2$

* $\cU=\cN_1$ (with $q$ even) or $\cU=\cN_1^\varepsilon$ (with $q$ odd),  and either $H_0\lhdeq H$ where $(H_0,m,q)$ is listed in Table $\ref{T:OPlusN1}$, or $H\leq P_m$ such that, modulo the unipotent radical of $P_m$, $H$ induces a subgroup of $\GammaL_m(q)$ that is transitive on $1$-subspaces; or

(e) Subgroups $H_0\lhdeq H\le\GammaO_{2m}^{+}(q)$, where $H$ is transitive on $\cU=\cN_1$ with $q$ even, or on $\cU=\cN_1^\varepsilon$ with $q$ odd.

$H_0$       $\SL_m(q)$    $\SL_{m/2}(q^2)$     $\SL_{m/4}(q^4)$     $\Sp_m(q)$     $\Sp_{m/2}(q^2)$    $\Sp_{m/4}(q^4)$

$2,4$    $2$        $2,4$    $2$

$H_0$       $\SU_m(q)$    $\SU_{m/2}(q^2)$     $\SU_{m/4}(q^4)$     $\Omega_m^+(q^2)$    $\Omega_{m/2}^+(q^4)$

$\geq 8$           $m/2\geq 2$

$2,4$    $2$    $2,4$    $2$

$H_0$        $\G_2(q)'$     $\G_2(q^2)$    $\G_2(q^4)$     $\SL_2(q^2)$    $\SL_2(q^4)$     $\SL_2(q^8)$

$6$    $12$     $24$    $4$    $8$    $16$

$2,4$    $2$         $2,4$    $2$

$H_0$        ${}^\wedge\Omega_7(q)^I$    $\Omega_7(q^2)$    $\Omega_7(q^4)$   $\nonsplit{2}{\Omega_8^-(q^{1/2})}$    $\Omega^-_8(q)$   $\Omega^-_8(q^2)$

$4$    $8$   $16$   $4$     $8$     $16$

$2,4$    $2$        $2,4$    $2$

$H_0$        $\Omega_9(q)$    $\Omega_9(q^2)$    $\Omega_9(q^4)$     $\Alt_6$    $\Alt_7$    $\Alt_9$

$8$     $16$    $32$    $4$   $4$   $4$

$2,4$    $2$   $2$   $2$   $2$

$H_0$       $\nonsplit{2}{\Omega_8^+(2)}$    $\nonsplit{2}{\Sp_6(2)}$   $\nonsplit{2}{\SU_4(2)}$    $\nonsplit{2}{\Alt_9}$   $\PSU_4(3)$     $\nonsplit{3}{M_{22}}$

$4$     $4$    $4$    $4$   $6$   $6$

$3$    $3$    $3$    $3$    $2$    $2$

$H_0$        $\nonsplit{3}{\textup{Suz}}$    $\textup{Co}_1$

$12$     $12$

$2$    $2$

* $\cU=\cN^\pm_2$ and $H_0\lhdeq H$ where $(H_0,m,q)$ is given in
Table , or  $\cU=\cN_2^-$ and $H\leq P_m$ such that, modulo the unipotent radical of $P_m$ either  $H$ induces at least $\SL_m(q)$ or $m=3$ and $H$ induces a subgroup that is transitive on $1$-subspaces; or

(f) Subgroups $H_0\lhdeq H\le\GammaO_{2m}^{+}(q)$ where $H$    is transitive on $\cU=\cN_2^\pm$. In the first case
$H$ equals $[2^{10}]\ldotp31\ldotp5$.

$\cU$      $\cN_2^-$   $\cN_2^-$   $\cN_2^+$   $\cN_2^\eps$    $\cN_2^\eps$    $\cN_2^-$   $\cN_2^-$

$H_0$       $[2^{10}]\ldotp31\ldotp5$   $\SL_m(q)$    $\SU_m(q)$   ${}^\wedge\Omega_7(q)^I$    $\Alt_9$     $\Alt_8$*     $\GO_8^-(q^{1/2})$**

$(5,2)$   $(\ge3,2\;{\rm or }\;4)$   $(\textup{even},2 \textup{ or }4)$   $(4,q)$    $(4,2)$    $(4,2)$     $(4,4 \textup{ or } 16)$

* Even though $\Alt_8\cong\SL_4(2)$, this group is a different subgroup than the one appearing in the second column when $(m,q)=(4,2)$. Indeed this group is self-normalising and contained in an $\Alt_9$ while the $\SL_4(2)$ in the second column is normalised by a group of twice its order.

** This is the image under triality of a subfield subgroup and existence is only confirmed for $q=4$. Moreover, when $q=16$ we have $|H:H_0|=2$. See Remark <ref>.

* $\cU=\cN_3^\eps$, $m=4$, $q$ is odd and ${}^\wedge\Omega_7(q)^I\lhdeq H$; or

* $\cU=\cN_4^-$, $m=4$, and either $q=2$ and $H=\Omega_7(2)^I$,  or $q=4$ and $\Omega_7(4)^I.2\lhdeq H$.

Let $G$ be the setwise stabiliser in $\GammaO^+_{2m}(q)$ of $\cU$ and let $H\le G$ act transitively on $\cU$ such that $\Omega_{2m}^{+}(q)\not\le H$. Note that $G=\GammaO^+_{2m}(q)$ unless $\cU=\cP_m^\eps$,  which consists of totally singular $m$-subspaces, or
$\cU=\cN_k^\eps$, the set  of nondegenerate $k$-subspaces of a given isometry type $\eps$ with $kq$ odd. In these exceptional cases $G$ is an index two subgroup of $\GammaO^+_{2m}(q)$, and $G\geqslant \Omega^+_{2m}(q)$ if $\cU=\P_m^\eps$, or $G\geqslant \GO_{2m}^+(q)$ if $\cU=\cN_k^\eps$. By Theorem <ref> we may assume that $m\geq4$, and by Lemma <ref> we may assume also that $(m,q)\neq (4,3)$.  If $\cU=\cN_1$ (with $q$ even) or $\cU=\cN_1^\varepsilon$ (with $q$ odd), then part (e) follows from  <cit.>*Lemma 4.5, see also Remark <ref>.

Next suppose that  $\cU=\cN_2^+$ and $q=2$. Then $U\in\cN_2^+$ contains a unique nonsingular vector $ v$ and so $G_U\leqslant G_v$. This implies that $H$ acts transitively on $\cN_1$ and so is given by part (e). By (<ref>), $|\cN_2^+|=2^{2m-3}(2^m-1)(2^{m-1}+1)$. Hence if $m>4$ then $|H|$ is divisible by a primitive prime divisor of $2^{2m-2}-1$. It follows that  either $m$ is even and $\SU_m(2)\lhdeq H$, or $m=6$ and $\nonsplit{3}{M_{22}}\lhdeq H$, or $m=4$. A Magma calculation shows that  $N_{G}(\nonsplit{3}{M_{22}})$ is not transitive on $\cN_2^+$ when $m=6$, while another shows that when $m=4$ either
$\Alt_9$, $\Omega_7(2)^I$ or $\SU_4(2)$ is normal in $H$. Moreover, Lemma <ref> shows that $\SU_m(2)$ is transitive on $\cN_2^+$ for all even $m$. Thus $H$ is as listed in Table <ref>.

We also note that a Magma calculation shows that, when $\cU=\cN_4^+$ and $(m,q)=(4,2)$, there are no possibilities for $H$.

The remaining families $\cU$ to be considered are therefore $\cP_k$ for $k< m$, $\cP_m^\eps$, $\cN_k^\eps$ for even
$k\leq m$ with $(\cU,q)\ne (\cN_2^+,2)$ and $(\cU,m,q)\ne (\cN_4^+,4,2)$, and
if $q$ is odd also $\cN_k^\eps$ with $k$ odd  and $3\leq k\leq m$.  Let $U\in \cU$.
If $\cU=\cN_m^\eps$ and $U^\perp$ is isometric to $U$ then $G_U$ is an index 2 subgroup of $G_{\{U,U^\perp\}}$, which by <cit.>  is maximal in $G$. In all other cases, $G_U=G_{\{U,U^\perp\}}$, and moreover $G$ acts primitively on $\cU$: if
$(\cU,q)= (\cN_2^+,3)$ this follows from KL,BHRD (recalling that here $G=\GammaO_{2m}^+(q)$), and in all other cases it follows from <cit.>, <cit.> and <cit.> (noting that $G$ contains either $\GO^+_{2m}(q)$ or $\Omega_{2m}^+(q)$).
Thus in particular, in all cases  $G_{\{U,U^\perp\}}$ is maximal in $G$.

Denote the image of any subgroup $X$ of $\GammaO^+_{2m}(q)$ in $\PGammaO_{2m}^+(q)$ by $\overline{X}$ and let $L=\POmega_{2m}^+(q)$. Then we have a factorisation $\oG=\oH\,\oG_{\{U,U^\perp\}}$. Since $\Omega_{2m}^+(q)$ is quasisimple, Lemma <ref> implies that $L\not\leqslant \oH$. Let $\oB$ be maximal among core-free subgroups of $G$ containing $\oH$ and let $G^*=\oB L$. Then by Lemma <ref> and the fact that $(m,q)\neq (4,3)$ we have that $\oB$ is maximal in $G^*$ and we have a core-free maximal factorisation $G^*=\oB(G^*)_{\{U,U\}}$.

Suppose first that $\cU=\cN_m^\varepsilon$   and $U$ is isometric to $U^\perp$.  Since we have a core-free maximal factorisation $G^*=\oB(G^*)_{\{U,U\}}$, <cit.> implies that $m=4$, $q=2$ or $4$, $\cU=\cN_4^-$, and $\oB=N_G(\Omega_7(q)^I)$.  A Magma calculation shows that when $q=2$ the only example we get is $H=\Omega_7(2)^I$. When $q=4$, <cit.>*Table 8.50 implies that $N_{\oG}(\Omega_7(4)^I)=\Omega_7(4)^I.2$, where the extra 2 is a field automorphism. Constructing an appropriate $\Omega_8^+(4).2$ as a subgroup of $\Omega_{16}^+(2)$ in Magma, enables us to calculate the intersection of $N_{\oG}(\Omega_7(4)^I)$ with the stabiliser of a subspace in $\cN_4^-$ and show that $N_{\oG}(\Omega_7(4)^I)$ is transitive on $\cN_4^-$ while $\Omega_7(4)^I$ is not. Thus $\Omega_7(4)^I.2\lhdeq H$. This gives part (h).

In all remaining cases  $G_U=G_{\{U,U^\perp\}}$, and $G_U$ is maximal in $G$. Then $\cU$ and the possibilities for $\oB$ are given by <cit.> and are collated in Table <ref> according to the case of the theorem in which the family $\cU$ is addressed.. Note that  the Case (g) appears in disguise in <cit.>*Table 4, p. 14
because when $m=4$ the subgroup $(\PSp_2(q)\otimes\PSp_4(q)).2$ is isomorphic to $ (\POmega_3(q)\times\POmega_5(q)).2$  and is mapped to an $N_3$-subgroup under triality. (In constructing Table <ref> recall that $(m,q)\ne (4,3)$ and also that our notation for the entries $P_m, P_{m-1}$ in the tables in <cit.> is $\cP_m^\eps$.)

Remaining possibilities for $\cU$ when $L=\POmega^+_{2m}(q)$.

${\rm (a)}$    ${\rm (d)}$    ${\rm (f)}$    ${\rm (g)}$

$\cU$      $\cP_1$   $\cP_m^\eps$   $\cN_2^\eps$   $\cN_3^\eps$

By Lemma <ref>, $\oB=N_{\oG}(\oB\cap L)$ and so letting $B$ be the full preimage of $\overline{B}$ in $\GammaO^+_n(q)$, the Correspondence Theorem implies that
$B=N_G(B\cap \Omega_{2m}^-(q))$.  Table <ref> lists the groups $B_0$ such that $B=N_G(B_0)$ for each possibility for $\cU$.  We provide the following extra justification for when $m=4$:
By <cit.>*Table 4, if $L=\Omega^+_8(4)$  then there is a
factorisation $G=AB$ where $A\cong N_2^-$ and $B\cong N_2^+$. However $N_2^-$
is not transitive on $\cN_2^+$ and similarly $N_2^+$ is not transitive on
$\cN_2^-$ as, for a given $U\in \cN_2^-$, there exist $W_1,W_2\in\cN_2^+$ such that $U\cap W_1=0$ and $\dim(U\cap W_2)=1$. Indeed, as observed in Remark <ref>, under
triality an $N_2^+$ subgroup becomes an $M_2$ subgroup, that is the stabiliser of a decomposition of $V$ into a pair of maximal totally singular $4$-subspaces, and an $N_2^-$ subgroup
becomes an $\SU_4(q)$ subgroup.
Thus the factorisation is as listed  in cases
$(\textrm{f}^-2)$ and $(\textrm{f}^+1)$ respectively.
Similarly, in cases (f$^+2$) and (g), the subgroup $B$ must be $N_G({}^\wedge\Omega_7(q)^I)$ as $\Omega_7(q)^R$ does not act transitively on $\cN_2^{+}$ or $\cN_3^\eps$.  Furthermore, <cit.>*Table 4 lists a factorisation $L=AB$ with $A\cong N_2^-$ and $B=P_1,P_4^+$, or $P_4^-$.  Moreover, there are two possible $P_i$ for each choice of $A$. Since the orthogonal complement of a subspace in $\cN_2^-$ contains totally singular $1$-subspaces, it is not possible for $N_2^-$ to be transitive on $\cP_1$. Thus $N_2^-$ is transitive on $\cP_4^-$ and $\cP_4^+$. This gives case (d2). Moreover, the image of $N_2^-$ under triality is a $\SU_4(q)$ subgroup and so such a subgroup is transitive on $\cP_1$ and either $\cP_4^-$ or $\cP_4^+$. This gives cases (a2) and (d4).

Options for $\cU$, $(m,q)$ and $B=N_G(B_0)$.

${\rm (a1)}$    ${\rm (a2)}$    ${\rm (a3)}$

$\cU$      $\cP_1$   $\cP_1$   $\cP_1$

$(m,q)$      $(4,2)$   $(\textup{even},{\rm all})$   $(4,{\rm all})$

$B_0$      $*$   $\SU_m(q)$   ${}^\wedge\Omega_7(q)^I$

${\rm (d1)}$    ${\rm (d2)}$    ${\rm (d3)}$    ${\rm (d4)}$    ${\rm (d5)}$

$\cU$      $\cP_m^\eps$   $\cP_m^\eps$   $\cP_m^\eps$   $\cP_m^\eps$   $\cP_m^\eps$

$(m,q)$      $(\ge4,{\rm all})$   $(\ge4,{\rm all})$   $(4,{\rm all})$   $(4,{\rm all})$   $(4,2)$

$B_0$      $N_1$   $N_2^{-}$   $\Omega_{7}(q)^I$    $\SU_4(q)$   $\Alt_9$

(f${}^{+}1$)    (f${}^{+}2$)    (f${}^{-}1$)   (f${}^{-}2$)   (f${}^{-}3$)     (f${}^{-}4$)     (f${}^{-}5$)

$\cU$      $\cN_2^{+}$   $\cN_2^{+}$   $\cN_2^{-}$   $\cN_2^{-}$   $\cN_2^{-}$   $\cN_2^{-}$   $\cN_2^{-}$

$(m,q)$      $(\textup{even},4)$   $(4,>2)$   $(\ge4,{\rm all})$   $(\ge4,2\;{\rm or }\;4)$*   $(4,{\rm all})$    $(4,2)$    $(4,4 \textup{ or} 16)$

$B_0$      $\SU_m(q)$   ${}^\wedge\Omega_7(q)^I$   $P_m$   $\GL_m(q)$   ${}^\wedge\Omega_7(q)^I$    $\Alt_9$     $\Omega_8^-(q^{1/2})$**

$\cU$      $\cN_3^{\eps}$

$(m,q)$       $(m,\textup{odd})$

$B_0$        ${}^\wedge\Omega_7(q)^I$

* $(m,q)\neq (4,2)$ as in this case $\GL_4(2).2\leqslant \Omega_7(2)^I$.

** This is the image under triality of a subfield subgroup and existence is only confirmed for $q=4$. See Remark <ref>.

Case $\cU=\cP_1$. We consider cases (a1) to (a3) in Table <ref>.

(a1) Here $\GammaO_8^+(2)=\textup{SO}_8^+(2)$ acts faithfully on $|\cP_1|=135$ singular vectors. We use Magma and find that there are eight conjugacy classes of transitive subgroups of $\GO_8^+(2)$. Moreover, if $\Omega_8^+(2)\not\leqslant H$ then $H$ normalises one of $\SU_4(2)\cong\PSp_4(3)$, $\Alt_9$ or $\Omega_7(2)^I$, as listed in  Table <ref>.

(a2) In this case $B=N_G(\SU_m(q))$ where $m$ is even and $H\le B$ is transitive on $\cP_1$. Consider <cit.>*3.6.3(a), p. 70, and the
argument on <cit.>*3.5.2(a), p. 60. Let $k=\F_q$ and $K=\F_{q^2}$.
Here $V=K^m$ admits a $K$-hermitian form $[\, ,\,]$ and the $k$-quadratic form $Q$ preserved by ${\rm O}_{2m}^+(q)$ satisfies $Q(v)=[v,v]$ for $v\in V$.
Let $U\in \cU$. Then $KU:=\{\lambda u\mid \lambda\in K, u\in U\}$
is a singular $1$-subspace of the hermitian space $V=K^m$.  Since $m\ge4$, the subgroup $T=\SU_m(q)$ is transitive on
the set of nonzero singular vectors of $K^m$ by <cit.>*Lemma 2.10.5, p. 49. Moreover, the stabilizer $T_{KU}$ is transitive
on the set of $q+1$ singular $1$-dimensional $k$-subspaces of $KU$. Thus $\SU_m(q)$ is transitive on $\cU$ as recorded in Table <ref>.
Suppose now that $\SU_m(q)\not\le H$. Then we have the factorisation $B=B_U\,H$. Since $B_U$ fixes $KU$, which is a 1-dimensional $K$-subspace that is totally singular with respect to the hermitian form, it follow that $H$ acts transitively on the set of totally singular 1-subspaces over $K$. Since $m$ is
even and $(m,q)\neq (4,3)$, it follows from  Theorem <ref> that there are no such factorisations.

(a3) In this case $m=4$ and $G=P_1\,B$ where $B=N_G({}^\wedge\Omega_7(q)^I)$. Write $P_1=G_U$ where $U\in\cP_1$.
In light of Case (a1) and our assumption that $(m,q)\neq (4,3)$, we will assume $q\ge4$. By <cit.>*Table 4 we see that ${}^\wedge\Omega_7(q)^I$ is transitive on $\cP_1$ and so
we obtain the example ${}^\wedge\Omega_7(q)^I\lhdeq H$ in Table <ref>.
Suppose now that ${}^\wedge\Omega_7(q)^I\not\le H$. Since $H\le B$, and $H$ is transitive on $\cP_1$, we have $B=B_U\,H$ and $\overline{B}=\overline{B}_U\oH$.
Set $S:=\overline{B}^{(\infty)}=\Omega_7(q)^I$ and let $X$ be the natural 7-dimensional module for $S$ over $\F_q$. The formula (<ref>) for $|\cP_1|$ gives

\[
\]

Since $\cU=|S:S_U|$ is coprime to $q$, it follows from <cit.>*(1.6) that any maximal subgroup of $S$ containing $S_U$ is a parabolic subgroup. Comparing $|\cU|$ with the indices of parabolic subgroups given in (<ref>), we deduce that $S_U$ is the stabiliser in $S$ of a totally singular $3$-subspace of $X$. Thus $\oH$ acts transitively on the set of totally singular $3$-subspaces of $X$.

Since $q\geq 4$, Theorem <ref>(a) (see Table <ref>) when $q$ is odd and  Theorem <ref> when $q$ is even, imply that $\Omega_6^-(q)\lhdeq \oH$. In particular, $H$ fixes a unique nondegenerate subspace $W$ of $X$. Thus letting $\tau$ be a triality automorphism of $\POmega_8^+(q)$ such that $\oB^\tau$ normalises an $\Omega_7(q)^R$ subgroup we have that $\Omega_6^-(q)^\tau\lhdeq \oH^\tau\leqslant N_2^-$.  Since $\tau$ cyclically permutes $N_2^-$ with two $\mathcal{C}_3$-subgroups of type $\SU_4(q)$ (see Remark <ref>), $\Omega_6^-(q)\cong \SU_4(q)/\langle \pm I\rangle$, and ${}^\wedge\Omega_7(q)=\nonsplit{2}{\Omega_7(q)}$ when $q$ is odd,
it follows that $\SU_4(q)\lhdeq H$. Thus $H$ is already listed in Table <ref>.

Case $\cU=\cP_m^\eps$. There are two orbits of $L$ on the set of totally singular $m$-subspaces:
one is labelled $\cP_m^+$ and the other $\cP_m^-$.  The two orbits are fused in $\PGammaO^+_{2m}(q)$. Hence there is no loss of generality in assuming that
$\cU=\cP_m^+$ and $U\in\cP_m^+$,  a (maximal) totally singular $m$-subspace of $V$.
When $m=4$,  triality induces a 3-cycle $(\cP_m^+,\cP_m^-,\cP_1)$ by Remark <ref>. Thus a suitable image of $\oH$ under triality is transitive on $\cP_1$ and so is given in part (a). Recall that  triality induces the 3-cycles $(\Omega_7(q)^I,\Omega_7(q)^I,\Omega_7(q)^R)$ and $(N_L(K_1), N_L(K_2),N_2^-)$ on subgroups of $\POmega_8^+(q)$, where $K_1$ and $K_2$ are the images in $L=\POmega_8^+(q)$ of subgroups isomorphic to $\SU_4(q)$. Then also using <cit.>*Table 8.50 to obtain the correct structure for $H$ in the other cases, we list all possibilities obtained in this way in Table <ref>.
Thus from now on we may assume that $m\geq 5$, and in particular we only need to deal with the cases (d1) and (d2) in Table <ref>.

(d1) In this case  $B=G_{\langle v\rangle}$ for some nonsingular vector $v$. Set $W=\langle v\rangle^\perp$.
Observe that $(v,v)=Q(v+v)-Q(v)-Q(v)=2Q(v)=0$ if and only if $q$ is even.
Hence $N_1$ preserves the decomposition
$V=\langle v\rangle\oplus W$ when $q$ is odd, and preserves the chain $0<\langle v\rangle<W<V$ when
$q$ is even. In both cases the restriction $N_1^W$ of $N_1$ to $W$
contains $\Omega_{2m-1}(q)$ by Witt's lemma, and we note that $\Omega_{2m-1}(q)\cong\Sp_{2m-2}(q)$ when $q$ is even. Moreover, $\Omega_{2m-1}(q)=B^{(\infty)}=(B^W)^{(\infty)}$, and so if $\Omega_{2m-1}(q)\leqslant H^W$ then Lemma <ref> implies that $\Omega_{2m-1}(q)^R\lhdeq H$, as listed in Table <ref>.

Suppose now that $\Omega_{2m-1}(q)\not\le H^W$.
Now $Y:=\langle v\rangle^\perp\cap U$ is a totally singular
$(m-1)$-subspace that is stabilised by $B\cap P_m$.
Hence $B\cap P_m\le B_Y$ and $(B\cap P_m)^W\le (B_Y)^W=P_{m-1}(B^W)$.
Thus we have a factorisation $B^W=(B\cap P_m)^W\,H^W=P_{m-1}(B^W)\,H^W$ and therefore $H^W$ is transitive on the set of totally singular $(m-1)$-subspaces of $W$.
When $q$ is odd and recalling that $m\geq 5$, we deduce from Theorem <ref> that $\Omega_{2m-2}^-(q)\lhdeq H^W$. Similarly, when $q$ is even we note that $\Omega_{2m-1}(q)\cong\Sp_{2m-2}(q)$ and we deduce from Theorem <ref>(b) that again  $\Omega_{2m-2}^-(q)^R\lhdeq H^W$. In both cases, it follows from Lemma <ref> that $\Omega_{2m-2}^-(q)^R\lhdeq H$, as listed in Table <ref>.

(d2) In this case $B= N_2^{-}=G_{W^\perp}=G_W$ for some
nondegenerate $(2m-2)$-subspace $W$ such that $W^\perp\in \cN_2^{-}$. In particular, $N_2^{-}$ preserves the
decomposition $V=W\oplus W^\perp$ and by  <cit.>*Lemma 4.1.1, $N_2^{-}$ contains $\Omega_2^{-}(q)\times \Omega_{2m-2}^{-}(q)$ as a normal subgroup.
Since $\GO_2^{-}(q)$ is solvable it follows that $B^{(\infty)}=(B^W)^{(\infty)}=\Omega_{2m-2}^-(q)$. Thus by Lemma <ref>, if $\Omega_{2m-2}^-(q)\leqslant H^W$ it follows that $\Omega_{2m-2}^-(q)^R\lhdeq H$, as listed in Table <ref>.

Suppose next that $\Omega_{2m-2}^-(q)\not\leqslant H^W$, so we have a factorisation $B^W=(B\cap P_m)^WH^W$.
Observe that $U\cap W^\perp=0$ as $U\in\cP_m$ and $W^\perp\in\cN_2^{-}$. Therefore $\dim(U\oplus W^\perp)=m+2$, and hence also
$\dim(U\cap W)=m-2$ as $(U\oplus W^\perp)^\perp=U^\perp\cap W=U\cap W$. Since $\dim(W)=2m-2$
and $W$ has Witt index $\frac{2m-2}{2}-1=m-2$, it follows that $Y:=U\cap W$ is a (maximal)
totally singular $(m-2)$-subspace of $W$. Thus $B\cap P_m\leqslant B_Y$ and we have a factorisation $B^W=(B_Y)^WH^W$. In particular, $H^W$ is a subgroup of $\GammaO^-_{2m-2}(q)$ that is transitive on totally singular $(m-2)$-subspaces and does not contain $\Omega_{2m-2}^-(q)$. Since $m\geq 5$, Theorem <ref> implies that no such group exists.

Case $\cU=\cN_2^+$. Here $G=\GammaO_{2m}^+(q)$. We follow the case distinction in Table <ref>, and we note that in each case $q>2$.

(f${}^{+}$1) In this case, $m$ is even and  $q=4$. Furthermore, $G$ factorises as $G=G_U\,B$ where $G_U=N_2^{+}$ and $B=N_G(\SU_m(4))$. The case $\SU_m(4)\lhdeq H$ is recorded in Table <ref> so suppose that $\SU_m(4)\not\le H$.
Then $B=B_U\,H$.
Let $K=\F_{16}$ and $k=\F_4$. It follows from <cit.>*p. 70 that the $K$-subspace $KU$ spanned by $U\in\cN_2^{+}$ has $K$-dimension 2. Hence $KU$ lies
in $\cN_2(\SU_m(4))$ and

is preserved by $B_U$. Therefore $H$ is a core-free subgroup of $B=N_G(\SU_m(4))$ which is transitive on $\cN_2(\SU_m(4))$, but there is no such subgroup by Theorem <ref>.

(f${}^{+}$2) In this case $m=4$ and $q>2$.  We consider the factorisation $G=G_U\,B$ where $G_U=N_2^{+}$ and $B=N_G({}^\wedge\Omega_7(q)^I)$.
Applying $\tau$ to the factorisation $\oG=N_2^{+}\,\oB$ and recalling that $\oG=\PGammaO_8^+(q)$, gives $\oG=(N_2^{+})^\tau\,(\oB)^\tau$ where $(N_2^{+})^\tau=M_2$ and
$(\oB)^\tau=N_{\oG}(\Omega_7(q)^I)^\tau=N_{\oG}(\Omega_7(q)^R)$. Thus we have a factorisation $(\oB)^\tau=(M_2\cap (\oB)^\tau)(\oH)^\tau$. Now $(\oB)^\tau=\oG_{\langle v\rangle}$ for some nonsingular vector $v$ and we let $W=v^\perp$. When $q$ is odd, $(\oB)^\tau$ fixes the decomposition $V=\langle v\rangle\perp W$ while when $q$ is even, $(\oB)^\tau$ preserves the chain $\langle v\rangle <W<V$.
Following <cit.>*3.6.1(b), pp. 63–64 we see that $M_2\cap (\oB)^\tau$ fixes a pair $\{U_1,U_2\}$ of complementary totally singular $3$-subspaces of $W$.  Suppose first that $q$ is even. Then $\Omega_7(q)\cong\Sp_6(q)$ and $M_2\cap (\oB)^\tau$ is contained in the stabiliser in $\GammaSp_6(q)$ of a pair of maximal totally singular subspaces. However, <cit.> then implies that $\Sp_6(q)\lhdeq (\oH)^\tau$ and so $\Omega_7(q)^I\lhdeq H$ as listed in Table <ref>. Next suppose that $q$ is odd.  Then $M_2\cap (\oB)^\tau$ fixes the nonsingular $1$-subspace $W\cap U_1^\perp\cap U_2^\perp$. Thus $(\oH)^\tau$ acts transitively on the set $\cN_1^+$ for $W$ and so is given by Theorem <ref>(c).
As $m=4$, we have $|\cN_2^{+}|=\frac{q^6(q^4-1)(q^3+1)}{2(q-1)}$ by (<ref>).  Letting $q=p^f$ we see that $|\oH^\tau|$ must be divisible by a primitive prime divisor $r$ of $p^{4f}-1$ (one always exists). Note that $r>4f$ and so does not divide the order of the first three groups listed in Table <ref>.
Further, since $3^6$ divides $|\cN_2^{+}|$ when $q=3$, a similar argument
eliminates the last two cases given by Table <ref>. Thus we must have  $\Omega_7(q)^R\lhdeq (\oH)^\tau$ and hence ${}^\wedge\Omega_7(q)^I\lhdeq H$ as listed in Table <ref>  along with the example ${}^\wedge\Omega_7(2)^I\lhdeq H$ already obtained when $q=2$.

Case $\cU=\cN_2^-$. We again follow the case distinction in Table <ref>.

(f${}^{-}$1) In this case $m\ge4$ and $q=p^f\ge2$. We consider the factorisation $G=G_U\,P_m$ where $G_U=N_2^{-}$. Following <cit.>*(3.6.2b), p. 68, we write $U\in\cN_2^-$ as
$U=\langle e_1+f_1,e_1+e_2+\kappa f_2\rangle$ where $x^2+x+\kappa$ is an irreducible quadratic over $\F_q$. Also, we may write $P_m=G_E$ where $E=\langle e_1,\dots,e_m\rangle$.
A direct calculation shows that $Y:=E\cap U^\perp=\langle e_3,\dots,e_m\rangle$. Clearly $G_E\cap G_U\le G_Y$ so $G_E\cap G_U\le G_E\cap G_Y$.
As $H\le G_E$ and $H$ is transitive on $\cN_2^-$, we have $G_E=(G_E\cap G_U)H$, that is to say, $P_m=(P_m\cap G_Y)H$.
Now $G_E=P_m$ is an extension of a $p$-group by a subgroup of $\GammaL_m(q)$. The homomorphism $\pi\colon G_E\to\GammaL_m(q)$ such that $\SL_m(q)\lhdeq\pi(G_E)$ maps the factorisation $P_m=(P_m\cap G_Y)H$
to $\GammaL_m(q)=P_{m-2}(\GammaL_{m}(q))\pi(H)$. One possibility is that $\SL_m(q)\le\pi(H)$. Here $H\le P_m$ and modulo its unipotent radical (the aforementioned $p$-group),
$\pi(H)$ contains $\SL_m(q)$ as stated in Theorem <ref> (f). Suppose now that $\SL_m(q)\not\le\pi(H)$. Then by <cit.>*pp. 10, 13 we have $(m,q)=(5,2)$ and $\pi(H)\le 31\ldotp5$.
However, there are $155=31\cdot5$ subspaces of $(\F_2)^5$ of codimension $2$. Thus $\pi(H)=31\ldotp5$, and $\GL_5(2)=P_3(\GL_5(2))\pi(H)$ is an exact factorisation.
A computation with Magma shows that if $H$ satisfies
$\GO_{10}^+(2)=N_2^-\,H$ and $\pi(H)=31\ldotp5$, then the unipotent radical of $H$ has order $2^{10}$, and we have $H=[2^{10}]\ldotp31\ldotp5$ as in Table <ref>.

(f${}^{-}$2) In this case $m\ge4$ and $q=2$ or $4$, with $(m,q)\neq (4,2)$. We consider the factorisation $G=G_U\,B$ where $G_U=N_2^{+}$ and $B=N_G(\GL_m(q))$.  Write $B=G_{\{E,F\}}$ where $E=\langle e_1,\dots,e_m\rangle$ and $F=\langle f_1,\dots,f_m\rangle$.
We view $B$ as $\GammaL(E)\ldotp2$ acting on $E$ where the $2$ acts as the inverse-transpose automorphism.
If $\SL_m(q)\le H$, we obtain the example $\SL_m(q)\lhdeq H$ listed in Table <ref>. Suppose now that $\SL_m(q)\not\le H$. We will argue that
this possibility does not arise.

Since $H\le B$ is transitive, we have a factorisation $B=B_U\,H$. Following <cit.>*(3.6.2b), p. 68, we write $U\in\cN_2^-$ as
$U=\langle e_1+f_1,e_1+e_2+\kappa f_2\rangle$ where $x^2+x+\kappa$ is an irreducible quadratic over $\F_q$.
A direct calculation shows that $F\cap U^\perp=\langle f_3,\dots,f_m\rangle$ and $E_{m-2}:=E\cap U^\perp=\langle e_3,\dots,e_m\rangle$. Since $E':=(F\cap U^\perp)^\perp\cap E=\langle e_1,e_2\rangle$ and $F':=(E\cap U^\perp)^\perp\cap F=\langle f_1,f_2\rangle$,
it follows that $B_U=N_2^-\cap B=G_U\cap G_{\{E,F\}}$ preserves the set $\{E',F'\}$.
Therefore $B_U$ preserves the decomposition $E=E'\oplus E_{m-2}$. Since $B=B_U\,H$ and $B_U\le\textup{Stab}(E'\oplus E_{m-2})$, we have
$\GammaL(E)\ldotp2=\textup{Stab}(E'\oplus E_{m-2})\,H$. However, no such factorisation exists by <cit.>*pp. 10, 12, 13. In summary, the case $\SL_m(q)\not\le H$ does not arise.

(f${}^{-}$3) In this case $m=4$ and $H\leqslant B=N_G({}^\wedge\Omega_7(q)^I)$. The possibility ${}^\wedge\Omega_7(q)^I\lhdeq H$ is listed in Table <ref>. As noted in Remark <ref>, a triality automorphism $\tau$ of $\POmega_8^+(q)$ maps $\oB$ to $N_1$, and $N_2^-$ to an $\SU_4(q)$ subgroup. Thus applying $\tau$ to the factorisation $\oG= \oG_U\,\oB$ we get a factorisation $(\oG)^\tau=(\oG_U)^\tau\, N_1$, where $(\oG)^\tau$ contains $\POmega^+_8(q)$ as a normal subgroup. Moreover, $(\oH)^\tau\leqslant N_1$. Suppose that ${}^\wedge\Omega_7(q)^I\not\leqslant H$. Then by Lemma <ref>, $\Omega_7(q)^R\not\leqslant (\oH)^\tau$.  Hence we obtain a core-free factorisation $N_1=(\oH)^\tau (N_1\cap (\oG_U)^\tau)$.  Let $v$ be the vector such that $N_1=(\oG)_{\langle v\rangle}$.  Let $K=\F_{q^2}$ and $k=\F_q$. Then following  <cit.>*p. 60, 64 we see that $N_1\cap (\oG_U)^\tau$ fixes the $K$-span of $v$, which is a $1$-subspace over $K$ that is nondegenerate with respect to the hermitian form preserved by $(\oG_U)^\tau$. Thus $N_1\cap (\oG_U)^\tau$ also fixes the 3-dimensional nondegenerate $K$-subspace $(Kv)^\perp$ and hence fixes a 6-dimensional $k$-subspace of $v^\perp$ of minus type. Thus $(\oH)^\tau$ acts transitively on the $\cN_6^-$ subspaces of the 7-dimensional $k$-subspace $v^\perp$. When $q$ is odd this also means that $(\oH)^\tau$ acts transitively on the $\cN_1^-$ subspaces of $v^\perp$. Thus when $q$ is even, $(\oH)^\tau$ is given by Theorem <ref>(e) and when $q$ is odd it is given by Theorem <ref>(b). By (<ref>),

and so $|H|$ must be divisible by a primitive prime divisor $r$ of $p^{4f}-1$ where $q=p^f$. Note that $r$ always exists and $r>4f$. Thus none of the possibilities given by Theorem <ref>(b) for $q$ odd has order divisible by $r$. Hence $q$ is even. However, the only possibilities given by Theorem <ref>(e) that have order divisible by $r$ satisfy $\Omega_6^+(q)\lhdeq H^\tau$ when $q=2$ or 4. Thus applying triality we deduce that $\SL_4(q)\lhdeq H$ as listed in Table <ref>.

(f${}^{-}$4) In this case $(m,q)=(4,2)$ and $H\leqslant N_G(\Alt_9)=\Alt_9$. A Magma calculation shows that $H=\Alt_9$ or $\Alt_8$ as recorded in Table  <ref>.  We also note that the $\Alt_8$ provided here is different from the subgroup $\SL_4(2)$ already listed as the $\SL_4(2)$ stabilises two 4- subspaces while $\Alt_8$ only fixes one.

(f${}^{-}$5)  In this case $m=4$, $q=4$ or $16$, and $H\leqslant N_G(\Omega_8^-(q^{1/2}))$ where $\Omega_8^-(q^{1/2})$ is the image under triality of a subfield subgroup. We refer to Remark <ref> for details about this case. In particular,  $\Omega_8^-(q^{1/2})$ is not transitive on $\cN_2^-$, the group $\GO_8^-(2)$ is transitive on $\cN_2^-$ when $q=2$, and the group $\GO_8^-(4).2$ is possibly transitive on $\cN_2^-$ when $q=4$. We record both these as possibilities in Table <ref>. Suppose now that $\Omega_8^-(q)\not\leqslant H$. Then we have a factorisation $N_G(\Omega_8^-(q^{1/2}))=HX$, where $X$ is the stabiliser in $N_G(\GO_8^-(q^{1/2}))$ of a subspace in $\cN_2^-$. Moreover, $X$ is contained in the stabiliser $Y$ of an elliptic 4-dimensional subspace of the natural module for $\GO_8^-(q^{1/2})$. Thus we have a factorisation $N_G(\GO_8^-(q^{1/2}))=HY$. However, by <cit.>, no such factorisation exists, a contradiction.

Case $\cU=\cN_3^\eps.$ In this case $m=4$, $q$ is odd and  $B=N_G({}^\wedge\Omega_7(4)^I)$. As noted earlier, applying an appropriate triality automorphism $\tau$ maps $\oB$ to $N_1$ and $\oG_U$ to a $\mathcal{C}_4$ subgroup of type $\PSp_2(q)\otimes\PSp_4(q)$. (Note that $\PSp_2(q)\cong\PSL_2(q)\cong \POmega_3(q)$ and $\PSp_4(q)\cong \POmega_5(q)$). This gives us a factorisation $(\oG)^\tau= (\oG_U)^\tau\,N_1$, where $(\oG)^\tau$ contains $\POmega_8^+(q)$ as a normal subgroup. Let $v$ be the vector such that $N_1=(\oG)_{\langle v\rangle}$.
By <cit.>*p. 65–66 we see that $N_1\cap (\oG_U)^\tau$ fixes a nondegenerate 4-subspace $W$ containing $v$ and also fixes the 3-subspace $v^\perp\cap W$. Now $H\leqslant B$ and so $(\oH)^\tau\leqslant N_1$ and acts transitively on the set of nondegenerate 3-subspaces of $v^\perp$. Thus by Theorem <ref>, $\Omega_7(q)^R\lhdeq (\oH)^\tau$. Applying triality and using Lemma <ref> we deduce that ${}^\wedge\Omega_7(q)^I\lhdeq H$, as in part (g).

§.§ Actions on  and

First we treat the subspace family $\cP_m=\cP_m^+\cup \cP_m^-$,  the set of all maximal totally singular subspace of $(\F_q)^{2m}$.  Recall that $\cP_m^+, \cP_m^-$ are the two $\Omega_{2m}^+(q)$-orbits in $\cP_m$  and that $\GammaO_{2m}^{+}(q)$ fuses these orbits. A totally singular $(m-1)$-subspace is contained in exactly two totally singular $m$-subspaces – one from each $\Omega_{2m}^+(q)$-orbit.

Suppose that  $H\le\GammaO_{2m}^{+}(q)$, where $m\geq 3$, such that  $\Omega_{2m}^{+}(q)\not\le H$. Let
$V=(\F_q)^{2m}$ and suppose that $H$ acts transitively on $\cP_m$. Then
either $m=3$ and
$H\le N_{\GammaO_6^+(q)}(\Omega_2^-(q^2))\leqslant N_4^-$,  or $H_1\lhdeq H$ where $(H_1,m,q)$ is given in
Table $\ref{T:OPlusBoth}$

Proposition <ref> subgroups $H_1$ with  $H_1\lhdeq H\le\GammaO^+_{2m}(q)$ where $H$ is transitive on $\cP_m$. We exclude all   $H\le N_{\GammaO_6^+(q)}(\Omega_2^-(q^2))\leqslant N_4^-$ with $m=3$.

$H_1$       $\Omega_{2m-1}(q)^R$    $\Omega^{-}_{2m-2}(q)^R$    $2^4\colon \C_5$   $2^4\colon\Alt_5$   $\Alt_5$   $\Sym_7$

$\ge3$   $\ge3$   $3$   $3$   $3$   $3$

$3$   $3$   $3$   $2$

$H_1$       $2^6\colon\Alt_7$   $2^7\colon\Alt_8$    $\nonsplit{2}{\GO_8^+(2)}$   $\nonsplit{2}{\GO_7(2)}$   $\nonsplit{4}{\PSL_3(4)}$    $\nonsplit{2}{\PSL_3(4)}$

$4$   $4$   $4$   $4$   $4$   $4$

$3$   $3$   $3$   $3$   $3$   $3$

$H_1$       $\Alt_8$   $\Alt_9$    $\Sym_9$

$4$   $4$   $4$

$3$   $3$   $2$

Let $\cU=\cP_m$,  and let $G=\GammaO^+_{2m}(q)$, so $G$ fuses the two
$\Omega_{2m}^{+}(q)$-orbits $\cP_m^+$ and $\cP_m^-$ in $\cP_m$, and
the setwise stabiliser $G^+$ of $\cP_m^+$  has index 2 in $G$.
Clearly $\{\cP_m^{-},\cP_m^{+}\}$ is a system of imprimitivity for $G$,
and hence for $H$.
As $H$ is transitive on $\cP_m$, $H^+:=H\cap G^+$ is an index 2 subgroup of $H$ that fixes each of
$\cP_m^+$ and $\cP_m^-$ setwise, and is transitive on both.
Now $\Omega_{2m}^+(q)\not\leqslant H^+$, by our assumptions on $H$, and so $H^+$ is given by Theorem <ref>(d).
One possibility is that $m=3$ and $H^{+}\le N_A(\Omega_2^-(q^2))\leqslant N_4^-$ where $A$ is the stabiliser in $\GammaO_6^+(q)$ of $\cP_3^+$ and so this becomes $N_{\GammaO_6^+(q)}(\Omega_2^-(q^2))\leqslant N_4^-$. Suppose now that this is not the case. Then  by
Theorem <ref>(d), $H^+$
contains a normal subgroup $H_0$ given by Table <ref>.

In all but one case $H_0$ is perfect and is equal to the
soluble residual $(H^+)^{(\infty)}$ of $H^{+}$. In particular $H_0$ is a characteristic subgroup of $H^+$ and hence is normal in $H$.
In the exceptional case  $H_0=2^4\colon 5$ in $\Omega_6^+(3)$, and again
$H_0$ is a characteristic subgroup of $H^+$ and is normal in $H$.
Note that $H_0$ in this proof refers to the subgroup $H_0$ listed
in Table <ref>. The subgroup $H_1$ listed in
Table <ref> contains $H_0$ but may be larger,
such as $H_1=\langle H_0,h\rangle$ where $h$ is described below.

The subgroups $H$ determined by Table <ref> are transitive
on $\cP^{+}_m$ or $\cP^{-}_m$, but we require $H$ to be
transitive on both. This holds if and only if there is an
element  $h\in  G\setminus G^{+}$ such that $h$ normalises $H_0$.
Such an element $h$ interchanges $\cP^{+}_m$ and $\cP^{-}_m$, and
conjugates the stabiliser in $H$ of a subspace of $\cP_m^{+}$
to the stabiliser of a subspace of $\cP_m^{-}$.

Case  $H_0=\Omega_{2m-1}(q)^R$. Here $H_0$ stabilises a nonsingular $1$-subspace. Suppose first that $q$ is even. Then there is a unique $\Omega_{2m}^+(q)$-orbit on nonsingular $1$-subspaces and hence a unique $\Omega_{2m}^+(q)$-conjugacy class of subgroups $H_0$. Hence $H_0$ is normalised by some element $h\in \GO_{2m}^+(q)\backslash \Omega_{2m}^+(q)$. As discussed above, $h$ interchanges $\cP_m^+$ and $\cP_m^-$, and in this case $H_0$ itself is transitive on both $\cP_m^+$ and $\cP_m^-$, so $\langle H_0,h\rangle$ acts transitively on $\cP_m$. When $q$ is odd there are two $\Omega_{2m}^+(q)$-orbits on nonsingular $1$-subspaces and hence two $\Omega_{2m}^+(q)$-conjugacy classes of subgroups $H_0$. These orbits and conjugacy classes are fixed setwise by $\GO_{2m}^+(q)$ but fused in $\CO_{2m}^+(q)$. Hence $H_0$ is normalised by some element $h\in\GO_{2m}^+(q)\backslash\SO_{2m}^+(q)$. Again, $h$ interchanges $\cP_m^+$ and $\cP_m^-$ and $\langle H_0,h\rangle$ is transitive on $\cP_m$, and we record $H_1=H_0$ in Table <ref>.

Case  $H_0=\Omega_{2m-2}^-(q)^R$. Here there is a unique $\Omega_{2m}^+(q)$-orbit on elliptic subspaces of codimension two and hence a unique $\Omega_{2m}^+(q)$-conjugacy class of subgroups $H_0$.   Hence $H_0$ is normalised by an element of $\GO_{2m}^+(q)\backslash \SO_{2m}^+(q)$, which  interchanges $\cP_m^+$ and $\cP_m^-$ and so $N_G(H_0)$ is transitive on $\cP_m$. We record $H_1=H_0$ in Table <ref>.

Case  $m=4$ and $H_0=\Omega_7(q)^I$.  Here $\overline{H_0}$ is the image under a triality automorphism of $\overline{\Omega_7(q)^R}$. We saw in the first above that $\Omega_7(q)^R$ is transitive on both $\cP_4^+$ and $\cP_4^-$. Moreover,  the stabiliser of a nonsingular vector $\langle u\rangle$ is not transitive on $\cP_1$ as there are singular vectors in $u^\perp$ and $V\backslash u^\perp$. Thus $\Omega_7(q)^R$ is not transitive on $\cP_1$.  As explained in Remark <ref>, a triality automorphism induces a 3-cycle $(\cP_1,\cP_4^+,\cP_4^-)$, and it follows that $H_0$ is transitive on $\cP_1$ and is transitive on exactly one of $\cP_4^+, \cP_4^-$. Thus we do not obtain groups transitive on $\cP_4$ in this case.

Case  $H_0=\SU_4(q)$.  Here $\overline{H_0}$ is the image under a triality automorphism of $\overline{\Omega_6^-(q)^R}$. We saw in the second above that $\Omega_6^-(q)^R$ is transitive on both $\cP_4^+$ and $\cP_4^-$. Moreover,  the stabiliser of an elliptic subspace $U$ of codimension two is not transitive on $\cP_1$ as there are singular vectors in $U$ and $V\backslash U$. Thus $\Omega_6^-(q)^R$ is not transitive on $\cP_1$.  Arguing as in the previous case we deduce that there are no groups transitive on $\cP_4$ in this case.

In all other cases $m\in\{3,4\}$ and $q\in\{2,3\}$.
If $(m,q)=(4,3)$ then the groups occurring in Table <ref> are precisely those identified in Lemma <ref> (2). The remaining cases follow from a Magma calculation informed by Theorem <ref>(b).

Now we consider the families $\cN_k$ where both $k$ and $q$ are odd. Recall that in this situation
$\cN_k=\cN_k^+\cup \cN_k^-$, where $\cN_k^+, \cN_k^-$ are the two $\Omega_{2m}^+(q)$-orbits in $\cN_k$, and that
the group $\CO_{2m}^+(q)$ interchanges $\cN_k^+$ and $\cN_k^-$ and hence is transitive on $\cN_k$.

Suppose that  $H\le\GammaO_{2m}^{+}(q)$, where $m\geq 3$ and $q$ is odd, such that  $\Omega_{2m}^{+}(q)\not\le H$. Let
$V=(\F_q)^{2m}$ and suppose that $H$ acts transitively on the set $\cU=\cN_k$ for some odd $k\leq m$. Then

* $\cU=\cN_1$, $H\leqslant P_m$ fixes a totally singular $m$-subspace $W$
and, modulo
the unipotent radical, $H$ induces a subgroup of
$\GammaL_m(q)$ which is transitive  on $1$-subspaces; or

* $\cU=\cN_1$, and $H_1\lhdeq H$ where $(H_1, m, q)$  is given in
Table $\ref{T:OPlusBothQKOdd}$; or

Proposition <ref>(b) subgroups $H_1$ with $H_1\lhdeq H\le\GammaO^+_{2m}(q)$
where $H$ is transitive on $\cN_1$. We exclude cases when
$H\le P_m$.

$H_1$       $\SL_m(q)$    $\SU_m(q)$   $\Sp_m(q)$   $\Omega_9(q)$   $\nonsplit{2}{\Omega_7(q)}$   $\nonsplit{2}{\Omega_8^{-}(q_0)}$

$\ge3$   $2a\ge4$   $2a\ge4$   $8$   $4$   $4$

* $\cU=\cN_3$, $m=4$, and $\nonsplit{2}{\Omega_7(q)^I}\lhdeq H$.

Let $\cU=\cN_k$,  and let $G=\GammaO^+_{2m}(q)$, so $G$ fuses the two
$\Omega_{2m}^{+}(q)$-orbits $\cN_k^+$ and $\cN_k^-$ in $\cN_k$, and
the setwise stabiliser $G^+$ of $\cN_k^+$  has index 2 in $G$.
Clearly $\{\cN_k^{-},\cN_k^{+}\}$ is a system of imprimitivity for $G$,
and hence for $H$.
As $H$ is transitive on $\cN_k$, $H^+:=H\cap G^+$ is an index 2 subgroup of $H$ that
fixes $\cN_k^+$ and $\cN_k^-$ setwise and is transitive on both, and $H=\langle H^{+},h\rangle$ for any
$h\in H\setminus H^+$ with $h^2\in H^{+}$.
Now $\Omega_{2m}^+(q)\not\leqslant H^+$, by our assumptions on $H$, and so $H^+$ is given by
Theorem <ref>.
In particular, $k=1$ or $(m,k)=(4,3)$.

Suppose first that $(m,k)=(4,3)$. Then by Theorem <ref> (g) we have $\nonsplit{2}{\Omega_7(q)^I}\lhdeq H^{+}$, and moreover $\nonsplit{2}{\Omega_7(q)^I}$ is transitive on $\cN_3^+$.   By <cit.>*Table 8.50, p. 402, the group $\Omega_8^{+}(q)$ has four conjugacy classes of subgroups isomorphic to
$\nonsplit{2}{\Omega_7(q)}$ and by <cit.> these
are fused by $\GO_8^{+}(q)$.   Hence there is a unique conjugacy
class of subgroups of $\GO_8^{+}(q)$ isomorphic to $\nonsplit{2}{\Omega_7(q)}$. Thus there exists $h$ in $\CO_8^{+}(q)\setminus\GO_8^{+}(q)$
normalising $\nonsplit{2}{\Omega_7(q)^I}$, and since $h\notin\GO_8^+(q)$ it interchanges $\cN_3^+$ and $\cN_3^-$. Thus $\nonsplit{2}{\Omega_7(q)^I}$ is also transitive on $\cN_1^-$ and  $\langle \nonsplit{2}{\Omega_7(q)^I},h\rangle$ is transitive on $\cN_1$. Since $\nonsplit{2}{\Omega_7(q)^I}$ is characteristic in $H^+$,  part  (c) holds.

We assume from now on that $k=1$.  If $(m,q)=(4,3)$ then the transitive subgroups $H$ are those given by Lemma <ref>(3). Thus we may in addition assume that $(m,q)\neq(4,3)$. Then by Theorem <ref>(e) either $H^+$ fixes a totally singular $m$-subspace $W$ and $H^+$ induces a subgroup of $\GammaL_m(q)$ that is transitive on the set of $1$-subspaces of $W$, or  $H^+$ contains a normal subgroup $H_0$ as follows (recall that $q$ is odd):

* $\SL_m(q)$;

* $\Sp_m(q)$ or $\SU_m(q)$ with $m$ even;

* $\nonsplit{2}{\Omega_7(q)}$ with $m=4$;

* $\nonsplit{2}{\Omega_8^-(q^{1/2})}$ with $m=4$;

* $\Omega_9(q)$ with $m=8$.

Suppose first that $H^+$ fixes a totally singular $m$-subspace $W$ and induces a subgroup of $\GammaL_m(q)$ that is transitive on the set of $1$-subspaces of $W$. Thus $H^+$ acts irreducibly on $W$ and hence also on $V/W$. Since the action of $H^+$ on $V/W$ is dual to the action of $H^+$ on $W$, it follows that either $W$ is the only totally singular $m$-subspace fixed by $H^+$ or $H^+$ fixes precisely two totally singular $m$-subspaces, namely $W$ and a complementary space $W'$. Since $N_G(H^+)$ fixes setwise the set of totally singular $m$-subspaces fixed by $H^+$ it follows that either $H$ fixes $W$, or $H$ fixes the decomposition $V=W\oplus W'$. In the first case $H$ is as in part (a) of Proposition <ref>. In the second case $H$, and hence also $H^+$, is contained in the normaliser in $G$ of $\SL_m(q)$. Then as seen in the analysis of case ($\delta$) of the proof of <cit.>*Lemma 4.5 the group $H^+$ has a normal subgroup $H_0$ as listed in cases (i) or (ii) above. We deal with these below.

It remains to consider the cases (i)–(v). In all such instances $H_0$ is characteristic in $H^+$ and hence is normal in $H$. Thus to complete the proof it suffices to show that examples arise in each case.

For case (i), we see from <cit.>*Table 1 that the stabiliser  $K$ in $\GO^+_{2m}(q)$ of a pair of complementary totally singular $m$-subspaces is transitive on $\cN_1^+$. Moreover, $\SL_m(q)$ is characteristic in $K$. The group $\CO_{2m}^+(q)$ fixes the $\GO_{2m}^+(q)$-conjugacy class of such subgroups and so there exists $h\in\CO^+_{2m}(q)\backslash\GO_{2m}^+(q)$ that normalises $K$. Such an element $h$ interchanges $\cN_1^+$ and $\cN_1^-$, and it follows that $K$ is transitive on $\cN_1^-$ and $\langle K,h\rangle$ is transitive on $\cN_1$. Thus examples arise in this case.

For case (ii) we see from <cit.>*Table 1 that the normaliser $K$ of $\SU_m(q)$ in $\GO^+_{2m}(q)$ is transitive on $\cN_1^+$ when $m$ is even. By <cit.>*Proposition 4.3.18 there is only one conjugacy class in $\GO^+_{2m}(q)$ of such subgroups and so arguing as in the previous case we again find an example.  Next suppose that $\Sp_m(q)$ is normal in $H^+$. Then by the proof of <cit.>*Lemma 4.5, the group $\Sp_{m}(q)$ is contained in the stabiliser $K$ of a pair of totally singular $m$-subspaces. As $K$ contains $\GL_m(q)$, there is only one $\GO^+_{2m}(q)$-conjugacy class, and hence only one $\CO_{2m}^+(q)$-conjugacy class, of such subgroups $\Sp_m(q)$.
Thus there exists $h\in\CO^+_{2m}(q)\backslash\GO_{2m}^+(q)$ that normalises the subgroup $H_0=\Sp_m(q)$, and such an element $h$ interchanges $\cN_1^+$ and $\cN_1^-$. So if $H^+$ is transitive on $\cN_1^+$ it is also transitive on $\cN_1^-$ and  $\langle H^+,h\rangle$ is transitive on $\cN_1$,  and we again get examples.

Finally we deal with cases (iii)-(v). If $m=4$ and $H_0=\nonsplit{2}{\Omega_7(q)}^I$ then we have already seen in the case $(m,k)=(4,3)$ that there is a unique conjugacy class of subgroups of $\GO_8^+(q)$ isomorphic to $\nonsplit{2}{\Omega_7(q)}^I$.  Since $\nonsplit{2}{\Omega_7(q)}^I$ is transitive on $\cN_1^+$ then the same argument as before yields an example here as well.

If $m=4$ and $H_0=\nonsplit{2}{\Omega_8^-(q^{1/2})}$  then by <cit.>*Table 8.50, p. 402, the group $\Omega_8^{+}(q)$ has four conjugacy classes of subgroups isomorphic to  $\nonsplit{2}{\Omega_8^-(q^{1/2})}$ and by <cit.> these are all fused in $\GO^+_8(q)$. Hence the argument above yields an example here.

Finally suppose that $m=8$.  Then it is seen in the last line of the the proof of the proposition of <cit.>*Appendix 3 that $N_G(\Omega_9(q))$ is transitive on $\cN_1$.

§ TRANSITIVE GENERALISED QUADRANGLES

A generalised quadrangle $\mathcal{Q}=(\cP,\mathcal{L})$ of
order $(s,t)$ is a point-line incidence
relation satisfying:

(GQ1) each $\ell\in\mathcal{L}$ is incident with $s+1$ points

(GQ2) each $p\in\cP$ is incident with $t+1$ lines
$\ell\in\mathcal{L}$, and

(GQ3) for each point $p$ not on a line $\ell$, there is
a unique point $p'\in\cP$ and a unique line $\ell'\in\mathcal{L}$, such that
$p$ is on $\ell'$, and $p'$ is on $\ell$ and $\ell'$.

To avoid degeneracies we henceforth assume that $\mathcal{Q}$ is thick, that is,  $s,t>1$.
It well known that $|\cP|=(s+1)(st+1)$, $|\mathcal{L}|=(t+1)(st+1)$ and
$\sqrt{t}\le s\le t$, see <cit.>*Lemma 2.1.

A beautiful theorem of Buekenhout and Lefévre <cit.> says that
if the point set $\cP$ and line set
$\mathcal{L}$ of a generalised quadrangle $\mathcal{Q}$ are chosen from the
points and lines of a projective space $\textup{PG}(d,q)$, then $\mathcal{Q}$
is a classical generalised quadrangle. By this we mean that
$(\cP,\mathcal{L})=(\cP_1,\cP_2)$ where $\cP_i$ is the set of
totally singular $i$-subspaces obtained from the natural module
for $\Sp_4(q)$,
$\GU_4(q)$, or
$\GU_5(q)$, and incidence is containment. The five thick classical generalised quadrangles are listed in Table <ref> along with their automorphism groups.

The classical generalised quadrangles $\mathcal{Q}$. Let $q_i=q^i+1$.

$\mathcal{Q}$   $W_3(q)$   $Q_4(q)$   $Q^{-}_5(q)$   $H_3(q^2)$   $H_4(q^2)$

$(s,t)$   $(q,q)$   $(q,q)$   $(q,q^2)$   $(q^2,q)$   $(q^2,q^3)$

$\Aut(\mathcal{Q})$   $\textup{P}\Gamma\Sp_4(q)$   $\textup{P}\Gamma\textup{O}_5(q)$   $\textup{P}\Gamma\textup{O}_6^{-}(q)$   $\textup{P}\Gamma\textup{U}_4(q)$   $\textup{P}\Gamma\textup{U}_5(q)$

We note that the axioms for a generalised quadrangle are symmetric and so we can interchange the role of points and lines, that is, if $\mathcal{Q}=(\cP,\mathcal{L})$ is a generalised quadrangle then $(\mathcal{L},\cP)$ is also a generalised quadrangle and is referred to as the dual of $\mathcal{Q}$.
The dual of $W_3(q)$ is isomorphic to $Q_4(q)$ and the dual of $Q_5^-(q)$ is isomorphic to $H_3(q^2)$. The dual of $H_4(q^2)$ is not usually regarded as being a classical generalised quadrangle.

From our results we are able to determine all the point- or line-transitive
subgroups of $\Aut(\mathcal{Q})$ where $\mathcal{Q}$ is a classical generalised
quadrangle. In each case $\soc(\Aut(\mathcal{Q}))$ is transitive on the
set $\cP$ of points and the set $\mathcal{L}$ of lines of $\mathcal{Q}$, so it suffices to determine the
transitive subgroups not containing $\soc(\Aut(\mathcal{Q}))$.

The groups that act regularly on the set of points of a classical generalised quadrangle were determined in <cit.> and the only examples are $3^{1+2}_{\pm}$ acting on $Q_5^-(2)$ and $\C_{513}\rtimes \C_9$ acting on $Q_5^-(8)$. (Here $3_{\pm}^{1+2}$ denote extraspecial groups of order
$27$ where $3_{+}^{1+2}$ has exponent $3$ and $3_{-}^{1+2}$ has exponent $9$.)

Let $\mathcal{Q}=(\cP,\mathcal{L})$ be a classical generalised quadrangle as
in Table . Suppose $H\le\Aut(\mathcal{Q})$ where
$\soc(\Aut(\mathcal{Q}))\not\leq H$.

* If $H\le\Aut(\mathcal{Q})$ is transitive on the points $\cP$, then
there exists $H_0\lhdeq H$ where $(\mathcal{Q},H_0,q)$ is given in
Table $\ref{tab:transGQP1}$.

* If $H\le\Aut(\mathcal{Q})$ is transitive on the lines $\mathcal{L}$, then
there exists $H_0\lhdeq H$ where $(\mathcal{Q},H_0,q)$ is given in
Table $\ref{tab:transGQP2}$.

Subgroups $H_0\lhdeq H\le\Aut(\mathcal{Q})$ where $H$ is transitive
on the set $\cP$ of points of the classical generalised
quadrangle $\mathcal{Q}$.

$\mathcal{Q}$   $W_3(q)$    $W_3(q)$   $W_3(q)$   $W_3(q)$   $Q_4(q)$    $Q_4(q)$   $Q_4(q)$   $H_3(q^2)$

$H_0$    $\PSp_2(q^2)$   $2^4\ldotp\C_5$   $2^4\ldotp\Alt_5$   $\Alt_5$   $\Sp_2(q^2)$   $2^4\ldotp F_{20}$   $2^4\ldotp\Alt_5$   $\PSL_3(4)$

$3$   $3$   $3$      $3$   $3$   $3$
$\mathcal{Q}$   $Q^{-}_5(q)$   $Q^{-}_5(q)$   $Q^{-}_5(q)$   $Q^{-}_5(q)$   $Q^{-}_5(q)$   $Q^{-}_5(q)$   $Q^{-}_5(q)$

$H_0$    $\SU_3(q)$   $3_\pm^{1+2}$   $(\C_3)^3$   $\PSL_2(7)$   $\PSL_3(4)$   $\nonsplit{3}{\Alt_7}$   $\C_{513}\rtimes\C_9$

$2$   $2$   $3$   $3$   $5$   $8$

Subgroups $H_0\lhdeq H\le\Aut(\mathcal{Q})$ where $H$ is transitive
on the set $\mathcal{L}$ of lines of the classical generalised
quadrangle $\mathcal{Q}$.

$\mathcal{Q}$   $W_3(q)$    $W_3(q)$   $W_3(q)$   $Q_4(q)$   $Q_4(q)$   $Q_4(q)$   $Q_4(q)$   $Q^{-}_5(q)$

$H_0$    $\Omega_4^{-}(q)$   $2^4\ldotp F_{20}$   $2^4\ldotp\Alt_5$   $\Omega_4^{-}(q)$   $2^4\ldotp \C_5$   $2^4\ldotp\Alt_5$   $\Alt_5$   $\PSL_3(4)$

$3$   $3$      $3$   $3$   $3$   $3$
$\mathcal{Q}$   $H_3(q^2)$   $H_3(q^2)$   $H_3(q^2)$   $H_3(q^2)$   $H_3(q^2)$   $H_3(q^2)$   $H_3(q^2)$

$H_0$    $\SU_3(q)$   $3_\pm^{1+2}$   $(\C_3)^3$   $\PSL_2(7)$   $\PSL_3(4)$   $\nonsplit{3}{\Alt_7}$   $\C_{513}\rtimes \C_9$

$2$   $2$   $3$   $3$   $5$   $8$

The entries in Tables <ref> and  <ref>  are obtained from
Theorems <ref>, <ref>, <ref>, <ref>, and <ref>,   and by taking $\cU$ to be the appropriate set of totally singular subspaces and taking the projective versions of the matrix groups listed. For $Q_5^-(8)$ and $H_3(8^2)$ we consult the proofs of Theorem <ref> and <ref>, respectively, to obtain the extra information provided for the projective groups involved when $\C_{19}\lhdeq H$. Also for $Q^-_5(2)$ and $H_3(2^2)$ we use Magma to see that the groups $3^{1+2}_{\pm}$ and $(\C_3)^3$ remain in the projective setting and are not `quotiented out' by a subgroup of order three.

We note that since $W_3(q)$ and $Q_4(q)$ are dual, any group that is transitive on the set of points of $W_3(q)$ should also appear as a group that is transitive on the set of lines of $Q_4(q)$ and vice versa. Similarly,  any group that is transitive on the set of lines of $W_3(q)$ will also appear as a group that is transitive on the set of points of $Q_4(q)$ and vice versa. Since $\PSp_2(q^2)\cong\PSL_2(q^2)\cong\Omega_4^-(q)$, and $\Omega_2(q^2)\cong \PSL_2(q^2)\cong\Omega_4^-(q)$, Tables <ref> and <ref> satisfy this condition. The same duality occurs between $H_3(q^2)$ and $Q_5^-(q)$.

We also note that for $q$ even,   the groups $\PSL_2(q^2)$ acting transitively on the set of points of $W_3(q)$ are distinct from the groups $\Omega_4^-(q)$ that are transitive on the set of lines of $W_3(q)$. They are interchanged by a graph automorphism of $\PSp_4(q)$ <cit.>*Section 14. Similarly, the groups $\Sp_2(q^2)$ that are transitive on the set of points of $Q_4(q)$ for $q$ even are distinct from the groups $\Omega_4^-(q)$ that acts transitively on the set of lines.

Recall that $(p,\ell)\in \cP\times\mathcal{L}$ is a flag if $p$ lies on $\ell$, and
an antiflag otherwise.  The sets $\mathcal{F}$ of flags and the set
$\mathcal{A}$ of antiflags have cardinalities:

\begin{align*}
\end{align*}

We note that if $G$ acts flag-transitively or antiflag-transitively on a generalised quadrangle $\mathcal{Q}$ then it also acts flag-transitively or antiflag-transitively on its dual.  It is well known that the actions of $\PSp_4(q)$, $\PSU_4(q)$ and $\PSU_5(q)$ on totally isotropic 1-subspaces have rank 3, see for example KLrank3,L1,L2. Thus if $G$ is one of these three groups and $p$ is a point of the corresponding generalised quadrangle $\mathcal{Q}$ then $G_p$ has two orbits on the remaining points: the points collinear with $p$ and the points not collinear with $p$. Since each pair of points lies on at most one line, it follows that $G_p$ acts transitively on the set of lines of $\mathcal{Q}$ incident with $p$ and so $G$ acts flag-transitively on $\mathcal{Q}$.
Kantor <cit.> conjectured that, up to duality, the only flag-transitive generalised quadrangles are the classical ones and the unique generalised quadrangles of order $(3,5)$ and $(15,17)$. This conjecture is still wide-open but recently, Bamberg, Li and Swartz <cit.>  showed that, up to duality, the only antiflag-transitive generalised quadrangles are the classical ones and the unique generalised quadrangle of order $(3,5)$.

Let $\mathcal{Q}$ be a classical generalised quadrangle as
in Table  and let $H\leqslant \Aut(\mathcal{Q})$ with $\soc(\Aut(\mathcal{Q}))\not\leqslant H$.

* If $H$ is flag-transitive,
then either

* $\mathcal{Q}$ is $W_3(3)$ or $Q_4(3)$, and
$H\in\{2^4\ldotp F_{20}, 2^4\ldotp\Alt_5, 2^4\ldotp\Sym_5\}$; or

* $\mathcal{Q}$ is $H_3(3^2)$ or $Q_5^-(3)$, and $H=\PSL_3(4).2$ (two possibilities) or $\PSL_3(4).2^2$, where in the first case the extension is by a field or graph automorphism.

Moreover, all six groups listed above are flag-transitive.

* The group $H$ is not antiflag-transitive.

A flag- or antiflag-transitive subgroup
$H\le\Aut(\mathcal{Q})$ must be transitive on both $\cP$ and $\mathcal{L}$,
and hence must occur in both Tables <ref>
and <ref> (for the same generalised quadrangle $\mathcal{Q}$). Thus noting Remark <ref> we see that if $\soc(\Aut(\mathcal{Q}))\not\leqslant H$, then one of the following occurs:

* $\mathcal{Q}=W_3(3)$ and either $2^4\ldotp F_{20}\lhdeq H$ or $2^4\ldotp\Alt_5\lhdeq H$;

* $\mathcal{Q}=Q_4(3)$ and either $2^4\ldotp F_{20}\lhdeq H$ or $2^4\ldotp\Alt_5\lhdeq H$;

* $\mathcal{Q}=H_3(3^2)$ and $\PSL_3(4)\lhdeq H$;

* $\mathcal{Q}=Q_5^-(3)$ and  $\PSL_3(4)\lhdeq H$.

Suppose first that $\mathcal{Q}=W_3(3)$ or $Q_4(3)$.  The possibilities for $H$ are $2^4\ldotp F_{20}$, $2^4\ldotp\Alt_5$ and $2^4\ldotp\Sym_5$. Now $(|\mathcal{F}|,|\mathcal{A}|)=(160,1440)$ and 1440 does not divide $|2^4\ldotp\Sym_5|$ and so none of these groups are antiflag-transitive. A Magma calculation shows that all three groups are flag-transitive, as stated in part (a)(i).

Next suppose that $\mathcal{Q}=H_3(3^2)$ or $Q_5^-(3)$. Then $\PSL_3(4)\lhdeq H\leqslant \PSL_3(4).2^2$ and $(|\mathcal{F}|,|\mathcal{A}|)=(1120,30240)$. Order arguments again eliminate the possibility of an antiflag-transitive group.  A Magma calculation shows  $\PSL_3(4)$ is not transitive on lines of $H_3(3^2)$ but $\PSL_3(4).2$ is, where the extension is by a field or graph automorphism, but not a graph-field automorphism. Moreover, both are flag-transitive.

The proof of Theorem <ref> actually shows something quite remarkable: any group that is both point-transitive and line-transitive is also flag-transitive.

Let $\mathcal{Q}$ be a classical generalised quadrangle as
in Table  and let $H\leqslant \Aut(\mathcal{Q})$ be both point-transitive and
line-transitive. Then $H$ is flag-transitive.

The full automorphism group of a classical generalised quadrangle acts primitively on the set of points. Moreover, there has been considerable recent effort devoted to classifying the generalised quadrangles with a group of automorphisms that acts primitively on the set of points: it is known that, if $H\leqslant\Aut(\mathcal{Q})$ acts primitively on both the set of points and the set of lines then it must be an almost simple group <cit.>. If $H$ acts primitively on points and transitively on lines then it has a unique minimal normal subgroup $N$ <cit.>, and if the group $N$ is elementary abelian then $\mathcal{Q}$ must be one of the unique generalised quadrangles of order $(3,5)$ or $(15,17)$ <cit.>. Detailed information about a subgroup $H$ of automorphisms which is assumed to act primitively on points (without any other assumptions),  is given in <cit.>. We shed light on these questions, for the classical examples, by determining all groups that act primitively on the set of points of a classical generalised quadrangle.

Let $\mathcal{Q}$ be a classical generalised quadrangle as
in Table , and suppose
that $H\leqslant \Aut(\mathcal{Q})$ acts primitively on the set of points of $\mathcal{Q}$ and $\soc(\Aut(\mathcal{Q}))\not\leqslant H$. Then $\mathcal{Q}=H_3(3^2)$ and $\PSL_3(4)\lhdeq H$.

Note that $H$ acts faithfully on  the set  $\cP$ of points of $\mathcal{Q}$. Since $H$ is primitive
on $\cP$, each of its nontrivial normal subgroups is transitive on $\cP$. In particular  $H$ is transitive on $\cP$, and so has a normal subgroup $H_0$ given in Table <ref>.
Since $\soc(H_0)$ is characteristic in $H_0$, it is normal in $H$, and hence is
transitive on $\cP$. Thus also
$\soc(H_0)$ has a normal subgroup, say $H_1$ appearing in Table <ref>.
If $\soc(H_0)$ is abelian then  $\soc(H_0)$ must be   elementary abelian
and regular on $\cP$. However the only abelian group appearing in Table <ref> is $(\C_3)^3$ acting on $Q_5^-(2)$, and by <cit.> this group is not point-regular on $Q_5^-(2)$.
Thus $\soc(H_0)$ is nonabelian, and hence is a direct product of isomorphic nonabelian simple groups.
If $H_0=\PSp_2(q^2)$, $\Omega_3(q^2)$ or $\SU_3(q)$ then $H$ preserves an extension field structure on the underlying vector space and so preserves the partition of $\cP$ given by the 1-dimensional subspaces over $\F_{q^2}$. Hence these examples are not primitive.

This leaves four almost simple examples to consider. Suppose first that $\soc(H_0)=\PSL_3(4)$. If $\mathcal{Q}=Q_5^-(3)$, then the $H$-action on $\cP$ is equivalent to its action on the set of lines of the dual quadrangle $H_3(3^2)$. However, we saw in the proof of Theorem <ref> that $\PSL_3(4)$ is not transitive on the lines of $H_3(3^2)$, so its orbits form a system of imprimitivity for the $H$-action on these lines. Thus $H$ is not primitive on the lines of $H_3(3^2)$, and hence is not primitive on the point set $\cP$, which is a contradiction. Hence in this case $\mathcal{Q}=H_3(3^2)$, and a Magma calculation shows that $\PSL_3(4)$ acts primitively on the set points of $H_3(3^2)$, as in the statement.
If $\mathcal{Q}=Q_5^-(3)$ and $H_0=\PSL_2(7)$, then the number of points is $112$ which does not divide
$|H_0|$,  so $H_0$ is not transitive on points (though $\PGL_2(7)$ is), which is a contradiction. Finally, if $\mathcal{Q}=W_3(3)$ and $H_0=\Alt_5$, then the number of points is $40$, and so $H_0$ is not transitive on points (though $\Sym_5$ is), ruling this case out as well.

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