2 Notation and preliminary results

In this section, we will present the main strategy for the proof of Theorem 1 and prove some preliminary results which will also establish the main theorem for some families of almost simple groups. We fix the notation we will use throughout all the paper.

As usual, if X is a subgroup of group Y, we will denote by $C_X(Y)$ and $N_X(Y)$ the centraliser and the normaliser of Y in $X,$ respectively. Moreover, if $x_1,x_2\in X,$ then $[x_1,x_2]=x_1^{-1}x_2^{-1}x_1x_2.$

If X is a matrix, we denote by ${}^t\!X$ the transpose of X.

For a finite group H, let

$$\begin{align*}\upsilon:\operatorname{\mathrm{Aut}}(H)\to\operatorname{\mathrm{Out}}(H)\cong \operatorname{\mathrm{Aut}}(H)/\mathrm{Inn}(H)\end{align*}$$

be the canonical projection. If $G_0$ is a finite non-abelian simple group, we identify $G_0\cong \mathrm {Inn}(G_0)$ , and from now on, $\upsilon $ will usually denote the above map for $H=G_0$ .

Recall that a subgroup H of a finite group G is said to be a supplement for a normal subgroup N of G if $HN=G.$ The following definition will provide the language we will use in the proof of our main result.

Notice in particular that if T and $\tilde T$ are as in the previous definition and $\upsilon (G)=T$ , then $G=\tilde TG_0$ with $\tilde T$ abelian. Therefore, proving Theorem 1 is equivalent to proving that for every non-abelian simple group $G_0$ and every abelian $T\leqslant \operatorname {\mathrm {Out}}(G_0)$ , there exists a T-abelian supplement. The strategy of the proof of Theorem 1 is in fact the following: given $G_0$ , we analyse all the abelian subgroups T of $\operatorname {\mathrm {Out}}(G_0)$ and, by the classification of the finite simple groups, prove that there exists a T-abelian supplement in a case-by-case inspection. Actually, it is not necessary to check each abelian subgroup of $\operatorname {\mathrm {Out}}(G_0)$ , but only the maximal abelian ones, as it is shown by the following lemma.

In particular, whenever $\operatorname {\mathrm {Out}}(G_0)$ is abelian, to prove Theorem 1, it is enough to check that there exists an $\operatorname {\mathrm {Out}}(G_0)$ -abelian supplement.

Now we will establish some results that give sufficient conditions on $\operatorname {\mathrm {Out}}(G_0)$ and an abelian subgroup T for the existence of a T-abelian supplement.

The previous lemma, together with Lemma 5, shows the following.

Noticing that $\mathrm {Alt}_6\cong \mathrm {PSL}_{2}{(9)}$ , this corollary reduces our investigation to the groups of Lie type.

In what follows, $G_0={{}^{s}L}(q)$ is a simple group of Lie type and $q=p^m$ , where p is a prime. The list of finite simple groups of Lie type and a full explanation of the notation $^sL(q)$ may be found in Section 4. We denote by d the index of $G_0$ in $\mathrm {Inndiag}(G_0)$ , the subgroup of $\operatorname {\mathrm {Aut}}(G_0)$ generated by the inner and diagonal automorphisms of $G_0$ (see Section 4 or [Reference Carter7] for further details). We give the values of d in Table 2 to provide a quick reference to look up, since such values play a central role in the proofs.

Table 2

Index d of $G_0$ in $\mathrm {Inndiag}(G_0)$ .

The Tits group ${}^2F_4(2)'$ , also considered as a group of Lie type, does not appear in Table 2. It is well known that $\operatorname {\mathrm {Aut}}({}^2F_4(2)')={}^2F_4(2)$ , and the extension does not split. We are now able to state another fundamental ingredient for the proof of Theorem 1.

In [Reference Lucchini, Menegazzo and Morigi17], Lucchini, Menegazzo and Morigi gave a complete classification of all simple groups of Lie type $G_0$ for which $\operatorname {\mathrm {Aut}}(G_0)$ splits over $G_0$ . Their main result is the following.

We are now ready to begin the investigation of the various types of almost simple groups, starting with the ones with linear socle.

3 Linear amd unitary groups

In this section, we prove Theorem 1 in the linear case. We begin with the easiest case $n=2$ , which is better understood on its own and gives us an explicit model for the more general setting. Then we deal with the case $n\geqslant 3$ . More specifically, we prove some technical lemmas and analyse all the different types of maximal abelian subgroups T of the outer automorphism group, showing the existence of T-abelian supplements in each case. Finally, the main result of this section is contained in Theorem 19.

We start by recalling the structure of the automorphism group of $\mathrm {PSL}_{n}{(q)}$ (a more detailed description can be found in [Reference Wilson23, 3.3.4]). The group $\operatorname {\mathrm {PGL}}_n(q)$ acts as a group of automorphisms of $\mathrm {PSL}_{n}{(q)},$ and the corresponding quotient group $\operatorname {\mathrm {PGL}}_n(q)/\mathrm {PSL}_{n}{(q)}$ is a cyclic group of order $d = (n, q-1),$ called the group of diagonal outer automorphisms. This group is generated by the element $\delta $ corresponding to the automorphism of $\mathrm {{GL}}_n(q)$ induced by the conjugation with the diagonal matrix $\text {diag}(\lambda ,1\dots ,1),$ being $\lambda $ a generator of the multiplicative group $\mathbb F_q^{\times }.$ The automorphism group of the field $\mathbb F_q$ of order $q = p^m$ is a cyclic group of order m generated by the Frobenius automorphism $x \mapsto x^p.$ This induces an automorphism $\phi $ of $\mathrm {{GL}}_n(q)$ by mapping each matrix entry to its p-th power. We denote by $\mathrm {{\Gamma L}}_n(q)$ the semidirect product of $\mathrm {{GL}}_n(q)$ with this group of field automorphisms, and correspondingly the extension of $\mathrm {{PGL}}_n(q)$ by the induced group of field automorphisms is denoted by $\mathrm {{P\Gamma L}}_n(q).$ The duality automorphism of $\mathrm {{GL}}_n(q)$ is the map that takes a matrix to the transpose of its inverse. For $n = 2,$ this duality map is an inner automorphism of $\mathrm {{SL}}_2(q).$ For $n> 2$ , the duality automorphism induces an automorphism $\gamma $ of $\mathrm {PSL}_{n}{(q)}$ of order $2$ that spans $\operatorname {\mathrm {Aut}}(\mathrm {PSL}_{n}{(q)})/\operatorname {{P\Gamma L}}_n(q)$ . We shall identify field and graph automorphisms with their corresponding images in $\operatorname {\mathrm {Out}} (\mathrm {PSL}_{n}{(q)})$ . They generate a subgroup $\langle \phi ,\gamma \rangle $ which is isomorphic to the direct product of a cyclic group of order m with a cyclic group of order $2.$ It can be easily seen that $\delta ^\phi =\delta ^p$ and $\delta ^\gamma =\delta ^{-1}.$

We identify the general unitary group ${\mathrm {{GU}}}_n(q)$ as the subgroup of the unitary matrices of ${\mathrm {{GL}}}_n(q^2).$ Let $G_0:=\mathord {\mathrm {PSU}_{n}(q)}$ . The quotient ${\mathrm {{PGU}}}_n(q)/\mathord {\mathrm {PSU}_{n}(q)}$ is cyclic of order $d=(n,q+1),$ generated by the automorphism $\delta $ induced by the conjugation with the diagonal matrix $\text {diag}(\lambda ,1,\dots ,1),$ denoting by $\lambda $ and element of the field $\mathbb F_{q^2}$ of order $q+1.$ The outer automorphism group of $G_0$ is described in [Reference Wilson23, 3.6.3]. We have

$$\begin{align*}\operatorname{\mathrm{Out}}(G_0)=\left\langle\delta\right\rangle\rtimes\left\langle\phi\right\rangle, \end{align*}$$

where $\phi $ is the field automorphism which raises the coefficients of every matrix to the power p. In particular, we have $|\phi |=2m$ and $\delta ^\phi =\delta ^p$ .

Proof. Let $Z:=\operatorname {\mathrm {Z}}(\mathord {\mathrm {GL}_{2}(q)})$ . We can suppose that q is odd; otherwise, $d=1$ and $\operatorname {\mathrm {Aut}}(G_0)$ splits over $G_0$ . In this case, $ \operatorname {\mathrm {Out}}(G_0)=\left \langle \delta \right \rangle \times \left \langle \phi \right \rangle , $ with $|\delta |=2$ , $|\phi |=m$ and $[\delta ,\phi ]=1$ . So $\operatorname {\mathrm {Out}}(G_0)$ is abelian, and by Lemma 5, it is enough to prove the case $G = \operatorname {\mathrm {Aut}}(G_0)$ .

Let

$$\begin{align*}A:=\begin{pmatrix} 0 & -\lambda\\ 1 & 0 \end{pmatrix}.\qquad B:=\begin{pmatrix} \lambda^{\frac{p-1}{2}} & 0\\ 0 & 1 \end{pmatrix}. \end{align*}$$

We have

$$\begin{align*}A^{\phi B}=\begin{pmatrix} \lambda^{\frac{1-p}{2}} & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & -\lambda^p\\ 1 & 0 \end{pmatrix}\begin{pmatrix} \lambda^{\frac{p-1}{2}} & 0\\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 0 & -\lambda^{\frac{p+1}{2}}\\ \lambda^{\frac{p-1}{2}} & 0 \end{pmatrix}=\lambda^{\frac{p-1}{2}}A. \end{align*}$$

Therefore,

$$\begin{align*}[A,\phi B]\in Z, \end{align*}$$

$\upsilon (AZ)=\delta $ , and $\upsilon (\phi BZ)$ can be $\phi \delta $ or $\phi $ . In any case,

$$\begin{align*}\upsilon(\left\langle AZ,\phi BZ\right\rangle)=\operatorname{\mathrm{Out}}(G_0), \end{align*}$$

and therefore,

$$\begin{align*}\left\langle A,\phi B\right\rangle Z/Z \end{align*}$$

is an $\operatorname {\mathrm {Out}}(G_0)$ -abelian supplement.

From now on, $G_0\in \{\mathrm {PSL}_{n}{(q)},\mathrm {{PSU}}_n(q)\}$ with $n\geqslant 3$ , so $d=(n,q+e)$ , with $e=-1$ in the linear case, $e=1$ in the unitary case. We write also Z for the center of $\mathord {\mathrm {GL}_{n}(q)}$ or ${\mathrm {{GU}}}_n(q)$ , respectively. If $G_0 = \mathrm {PSL}_{n}{(q)},$ then $\operatorname {\mathrm {Out}}(G_0)=\left \langle \delta \right \rangle \rtimes \left \langle \phi ,\gamma \right \rangle $ , with $|\phi |=m$ , $|\gamma |=2$ , $[\phi ,\gamma ]=1$ , $\delta ^\phi =\delta ^p$ and $\delta ^\gamma =\delta ^{-1}$ . If $G_0 = \mathrm {{PSU}}_n(q),$ then $\operatorname {\mathrm {Out}}(G_0)=\left \langle \delta \right \rangle \rtimes \left \langle \phi \right \rangle $ , with $|\phi |=2m$ and $\delta ^\phi =\delta ^p.$

The following lemma shows that in order to prove Theorem 1 in the linear and unitary cases when $n\geqslant 3$ , it is sufficient to investigate only two cases, which we will deal with in Propositions 15 and 18, respectively.

In the sequel, we will use a lot the following special matrices defined from some integers $w,l,c \in \mathbb {Z}$ with $w\geqslant 2$ (we choose to define such matrices only for $w\geqslant 2$ to avoid ambiguity in the definition and behaviour when $w=1$ ; this choice is irrelevant in the proofs). Let $\mathbb F$ be the algebraic closure of the field with p elements and $\lambda $ be an element of $\mathbb F$ of order $q+e$ , with $e\in \{-1,1\}.$ Then we define

$$\begin{align*}A_{w,l}:=\begin{pmatrix} 0 &0 &\dots & 0&(-1)^{w-1}\lambda^{l}\\ 1 & 0 & \dots &0& 0\\ 0 & 1 & \dots &0& 0\\ \vdots& \vdots &\ddots &\vdots &\vdots\\ 0 & 0 & \dots&1 & 0 \end{pmatrix}\in\mathord{\mathrm{GL}_{w}(\mathbb F)} \end{align*}$$

and

$$\begin{align*}X_{w,c}:=\begin{pmatrix} \lambda^{c(w-1)} &0 &\cdots & 0&0\\ 0 & \lambda^{c(w-2)} & \cdots &0& 0\\ \vdots & \vdots & \ddots &\vdots& \vdots\\ 0& 0 &\cdots &\lambda^{c} &0\\ 0 & 0 & \cdots&0 & 1 \end{pmatrix}\in\mathord{\mathrm{GL}_{w}(\mathbb F)}. \end{align*}$$

Notice that

$$\begin{align*}\det{A_{w,l}}=\lambda^l. \end{align*}$$

We now introduce a technical lemma which is the key ingredient of the proofs in this section.

Lemma 13. Let $w, l, c\in \mathbb {Z}$ be integers with $w\geqslant 2$ and

$$\begin{align*}A:=A_{w,l}(\lambda)\qquad X:=X_{w,c}(\lambda). \end{align*}$$

If

$$\begin{align*}cw\equiv lp^s(-1)^\varepsilon -l\quad \mod q+e,\end{align*}$$

then we have

$$\begin{align*}A^{\phi^s\gamma^\varepsilon X}=\lambda^{c}A. \end{align*}$$

Proof. First, notice that

$$\begin{align*}A^{\gamma}=\begin{pmatrix} 0 &0 &\dots & 0&(-1)^{w-1}\lambda^{-l}\\ 1 & 0 & \dots &0& 0\\ 0 & 1 & \dots &0& 0\\ \vdots& \vdots &\ddots &\vdots &\vdots\\ 0 & 0 & \dots&1 & 0 \end{pmatrix}, \end{align*}$$

and therefore,

$$\begin{align*}A^{\phi^s\gamma^\varepsilon}=\begin{pmatrix} 0 &0 &\dots & 0&(-1)^{w-1}\lambda^{lp^s(-1)^\varepsilon}\\ 1 & 0 & \dots &0& 0\\ 0 & 1 & \dots &0& 0\\ \vdots& \vdots &\ddots &\vdots &\vdots\\ 0 & 0 & \dots&1 & 0 \end{pmatrix}. \end{align*}$$

The statement follows by computing the action of $A^{\phi ^s\gamma ^\varepsilon X}$ and $\lambda ^{c}A$ on the canonical basis.

We show now the existence of T-abelian supplements for T of type $(1)$ . We start with the following lemma.

Proof. Since $\left (\frac {n}{d},\frac {m}{d}\right )=1$ , there exists $\overline {y}\in \mathbb {Z}$ such that $\overline {y}\frac {n}{d}\equiv 1\ \mod \frac {m}{d}$ . Now let

$$\begin{align*}d=p_1^{\alpha_1}\dots p_{\tilde l}^{\alpha_{\tilde l}}p_{\tilde l+1}^{\alpha_{\tilde l+1}}\dots p_l^{\alpha_l} \end{align*}$$

be its prime factorisation, where we have ordered the primes in a way such that $p_i$ divides $\overline {y}$ if and only if $1\leqslant i\leqslant \tilde l$ .

Let

$$\begin{align*}y=\overline{y}+p_{\tilde l+1}\cdots p_l\frac{m}{d}. \end{align*}$$

For every $p_i$ , we have that $p_i$ does not divide y because if $1\leqslant i\leqslant \tilde l$ , then $\overline {y}$ is divisible by $p_i$ while $p_{\tilde l+1}\cdots p_l\frac {m}{d}$ is not (since $(\overline {y},\frac {m}{d})=1$ ), and if $\tilde l<i\leqslant l$ , $p_i$ divides $p_{\tilde l+1}\cdots p_l\frac {m}{d}$ but not $\overline {y}$ . Therefore, $(y,d)=1$ ; moreover, $y\equiv \overline {y}\ \mod \frac {m}{d}$ and $yn\equiv d\ \mod {m}$ .

Proof. By Lemma 14, there exists $y\in \mathbb Z$ such that $yn \equiv d\ \mod q+e$ and $(y,d)=1.$ Since T is abelian,

$$\begin{align*}\delta^k=(\delta^k)^{\phi^s\gamma^\varepsilon\delta^j}=\delta^{kp^s(-1)^\varepsilon},\end{align*}$$

which means $d\mid k((-1)^\varepsilon p^s-1)$ or, equivalently,

$$\begin{align*}t\mid (-1)^\varepsilon p^s-1, \qquad t:=d/k.\end{align*}$$

First, suppose $t=n$ , so $t=d=n$ and $k=1$ . In this case, $T=\left \langle \delta ,\phi ^s\gamma ^\varepsilon \right \rangle $ and

$$\begin{align*}\tilde T:=\left\langle A_{n,1},\phi^s\gamma^\varepsilon X_{n,\frac{(-1)^\varepsilon p^s-1}{n}}\right\rangle Z/Z \end{align*}$$

is a T-abelian supplement since $\upsilon (A_{n,1}Z)=\delta $ and by applying Lemma 13 with

$$ \begin{align*}(w,l,c)=\left(n,1,\frac{(-1)^\varepsilon p^s-1}{n}\right),\end{align*} $$

we have

$$\begin{align*}\left[A_{n,1},\phi^s\gamma^\varepsilon X_{n,\frac{(-1)^\varepsilon p^s-1}{n}}\right]\in Z. \end{align*}$$

So in the sequel, we can suppose $t\neq n$ .

Step 1. We construct matrices $A,X$ such that $\det {A}=\lambda ^k$ and $[A,\phi ^s\gamma ^\varepsilon X]\in Z$ .

Since $t\mid n$ and $t\neq 1,n$ , we have that both $t\geqslant 2$ and $n-t\geqslant 2$ , and so we can define

$$\begin{align*}A:=\begin{pmatrix} A_{t,y} & 0\\ 0 & A_{n-t,k-y} \end{pmatrix}. \end{align*}$$

First, notice that

$$\begin{align*}\det{A}=\det{A_{t,y}}\det{A_{n-t,k-y}}=\lambda^y\lambda^{k-y}=\lambda^k; \end{align*}$$

therefore $\upsilon (AZ)=\delta ^k$ .

Let $r:=((-1)^\varepsilon p^s-1)/t$ and define

$$\begin{align*}X:=\begin{pmatrix} X_{t,yr} & 0\\ 0 & X_{n-t,yr} \end{pmatrix}. \end{align*}$$

We have that

$$\begin{align*}y(p^s(-1)^\varepsilon-1)=yrt\quad \mod{q+e}, \end{align*}$$

so applying Lemma 13 with $(w,l,c)=(t,y,yr)$ , we get

$$\begin{align*}A_{t,y}^{\phi^s\gamma^\varepsilon X_{t,yr}}=\lambda^{yr}A_{t,y}. \end{align*}$$

Moreover, recalling that $kt=d\equiv ny\ \mod {q+e}$ , we have that

$$\begin{align*}(k-y)(p^s(-1)^\varepsilon-1)=(k-y)rt=krt-yrt=yr(n-t)\quad \mod{q+e}, \end{align*}$$

so applying Lemma 13 with $(w,l,c)=(n-t,k-y,yr)$ , we get

$$\begin{align*}A_{n-t,k-y}^{\phi^s\gamma^\varepsilon X_{n-t,yr}}=\lambda^{yr}A_{n-t,k-y}. \end{align*}$$

Therefore, we have

$$\begin{align*}A^{\phi^s\gamma^\varepsilon X}=\lambda^{yr}A \end{align*}$$

or, equivalently,

$$\begin{align*}[A,\phi^s\gamma^\varepsilon X]\in Z. \end{align*}$$

Step 2. We find a matrix C such that $[A,C]=1$ with $\det {C}=\lambda $ . Moreover, $C\in \mathord {\mathrm {GL}_{n}(q)}$ if $e=-1$ , $C\in {\mathrm {{GU}}}_n(q)$ if $e=1.$

Recall that $(y,t)=1$ , so there exist $a,b\in \mathbb {Z}$ such that $ay+bt=1$ . Let

$$\begin{align*}C_0:=\lambda^b A_{t,y}^a. \end{align*}$$

We have that $[A_{t,y},C_0]=1$ and

$$\begin{align*}\det{C_0}=\det{A_{t,y}}^a\lambda^{bt}=\lambda^{ay+bt}=\lambda. \end{align*}$$

Let

$$\begin{align*}C:=\begin{pmatrix} C_0 & 0\\ 0 & 1 \end{pmatrix}. \end{align*}$$

We have $[A,C]=1$ and $\det {C}=\det {C_0}=\lambda $ .

Step 3. We complete the proof by constructing a T-abelian supplement.

Let $u\in \mathbb {Z}$ such that $\upsilon (XZ)=\delta ^u$ . Combining Steps 1 and 2, we get

$$\begin{align*}[A,\phi^s\gamma^\varepsilon XC^{j-u}]\in Z, \end{align*}$$

with $\upsilon (AZ)=\delta ^k$ and $\upsilon (XC^{j-u}Z)=\delta ^u\delta ^{j-u}=\delta ^j $ . Therefore,

$$\begin{align*}\tilde T:=\left\langle A,\phi^s\gamma^\varepsilon XC^{j-u}\right\rangle Z/Z \end{align*}$$

is a T-abelian supplement.

To continue our investigation, we need a few more lemmas.

Lemma 16. Let $A,B\in \mathord {\mathrm {GL}_{n}(q)}$ . Then

$$\begin{align*}[\phi^sA,\gamma B]=1 \end{align*}$$

if and only if

$$\begin{align*}B=A^TB^{\phi^s}A. \end{align*}$$

Lemma 17. Assume that we are in linear case, so $e=-1$ and $\lambda $ is an element of order $q-1$ in $\mathbb F_q.$ If $\alpha ,\beta \in \mathbb {Z}$ are such that

$$\begin{align*}\beta\equiv 2\alpha+p^s\beta\quad \mod{q-1}, \end{align*}$$

then

$$\begin{align*}[\phi^sX_{w,\alpha},\gamma X_{w,\beta}]=1. \end{align*}$$

Proof. Since $X_{w,\alpha },X_{w,\beta }$ are diagonal, for Lemma 16, we just need to check that

$$\begin{align*}X_{w,\beta}=X_{w,\alpha}X_{w,\beta}^{p^s}X_{w,\alpha}.\end{align*}$$

By inspecting the coefficients on the diagonal, for every $1\leqslant i\leqslant w$ , we have

$$\begin{align*}\lambda^{\beta(w-i)}=\lambda^{(2\alpha+p^s\beta)(w-i)}=\lambda^{\alpha(w-i)}\lambda^{p^s\beta(w-i)}\lambda^{\alpha(w-i)}, \end{align*}$$

by the hypothesis on $\alpha $ and $\beta $ .

We show now the existence of T-abelian supplements for T of type $(2)$ . Remember that this case occurs only when T is linear.

Proof. As in the previous case, this proof is articulated in different steps.

Step 1. We find an integer $y\in \mathbb {Z}$ such that $yn\equiv d\ \mod {q-1}$ and y is odd.

This follows from Lemma 14 and the fact that $d=(n,q-1)$ is assumed to be even.

Step 2. We construct matrices $A,X_\phi ,X_\gamma \in \mathord {\mathrm {GL}_{n}(q)}$ such that $\det {A}=\lambda ^{d/2}$ and $\hat {T_1}:=\left \langle A,\phi ^sX_\phi ,\gamma X_\gamma \right \rangle Z/Z$ is abelian.

Since $n-2\geqslant 2$ , we can define

$$\begin{align*}A:=\begin{pmatrix} A_{2,y} & 0\\ 0 & A_{n-2,d/2-y} \end{pmatrix}, \end{align*}$$

so that $\upsilon (AZ)=\delta ^{d/2}$ since

$$\begin{align*}\det{A}=\det{A_{2,y}}\det{A_{n-2,d/2-y}}=\lambda^y\lambda^{d/2-y}=\lambda^{d/2}. \end{align*}$$

Let $r:=(p^s-1)/2$ . Considering the automorphisms $\phi ^s$ and $\gamma $ , let us now argue as in Step 1 of Proposition 15 and construct

$$\begin{align*}X_\phi:=\begin{pmatrix} X_{2,yr} & 0\\ 0 & X_{n-2,yr} \end{pmatrix}, \end{align*}$$

so that

$$\begin{align*}A^{\phi^sX_\phi}=\lambda^{yr}A, \end{align*}$$

and

$$\begin{align*}X_\gamma:=\begin{pmatrix} X_{2,-y} & 0\\ 0 & X_{n-2,-y} \end{pmatrix}, \end{align*}$$

so that

$$\begin{align*}A^{\gamma X_\gamma}=\lambda^{-y}A. \end{align*}$$

Since

$$\begin{align*}-y\equiv 2yr-yp^s\quad \mod{q-1}, \end{align*}$$

by Lemma 17, we have

$$\begin{align*}\left[\phi^sX_{2,yr},\gamma X_{2,-y}\right]=\left[\phi^sX_{n-2,yr},\gamma X_{n-2,-y}\right]=1,\end{align*}$$

and therefore,

$$\begin{align*}\left[\phi^sX_\phi,\gamma X_\gamma\right]=1. \end{align*}$$

From this, we obtain that

$$\begin{align*}\hat{T_1}:=\left\langle A,\phi^sX_\phi,\gamma X_\gamma\right\rangle Z/Z \end{align*}$$

is abelian.

Step 3. We construct matrices $X_{\phi }^{\prime },X_{\gamma }^{\prime }\in \mathord {\mathrm {GL}_{n}(q)}$ such that $\det {X_{\gamma }^{\prime }}=\lambda ^y\det {X_\gamma }$ and $\hat {T_2}:=\left \langle A,\phi ^sX_{\phi }^{\prime },\gamma X_{\gamma }^{\prime }\right \rangle Z/Z$ is abelian.

Let us define

$$\begin{align*}C_\phi:=\begin{pmatrix} A_{2,y}^{-r} & 0\\ 0 & 1 \end{pmatrix}\in\mathord{\mathrm{GL}_{n}(q)}, \quad C_\gamma:=\begin{pmatrix} A_{2,y} & 0\\ 0 & 1 \end{pmatrix}\in\mathord{\mathrm{GL}_{n}(q)}, \end{align*}$$

and

$$\begin{align*}X_{\phi}^{\prime}:=X_\phi C_\phi , \quad X_{\gamma}^{\prime}:=X_\gamma C_\gamma. \end{align*}$$

Since $C_\phi ,C_\gamma \in C_{\mathord {\mathrm {GL}_{n}(q)}}(A)$ , we have

$$\begin{align*}A^{\phi^sX_{\phi}^{\prime}}=\lambda^{yr}A,\qquad A^{\gamma X_{\gamma}^{\prime}}=\lambda^{-y}A. \end{align*}$$

A straightforward computation shows also that

$$\begin{align*}\left[\phi^sX_{\phi}^{\prime},\gamma X_{\gamma}^{\prime}\right]=1. \end{align*}$$

Therefore,

$$\begin{align*}\det{X_{\gamma}^{\prime}}=\det{X_\gamma}\det{C_\gamma}=\det{X_\gamma}\det{A_{2,y}}=\lambda^y\det{X_\gamma}, \end{align*}$$

and

$$\begin{align*}\hat{T_2}:=\left\langle A,\phi^sX_{\phi}^{\prime},\gamma X_{\gamma}^{\prime}\right\rangle Z/Z \end{align*}$$

is abelian.

Step 4. We complete the proof by constructing a T-abelian supplement.

Let $\upsilon (X_\gamma Z)=\delta ^u$ for some $u\in \mathbb {Z}$ , so $\upsilon (X_{\gamma }^{\prime } Z)=\delta ^{y+u}$ . Given that y is odd, one of $u-k$ or $y+u-k$ is even. Since $\gamma \delta ^{x}$ is conjugate to $\gamma \delta ^{y}$ in $\left \langle \delta ,\gamma \right \rangle $ if $y-x$ is even, one of $\gamma \delta ^{u}$ or $\gamma \delta ^{u+y}$ is conjugate to $\gamma \delta ^k$ .

Let

$$\begin{align*}\hat{T}:=\begin{cases} \hat{T_1}\qquad \text{if } u-k \text{ is even}\\ \hat{T_2}\qquad \text{if } u+y-k \text{ is even} \end{cases}, \end{align*}$$

so that there exists a matrix $R\in \mathord {\mathrm {GL}_{n}(q)}$ such that $\upsilon (\hat {T}^R)=\left \langle \delta ^{d/2},\phi ^s\delta ^l,\gamma \delta ^k\right \rangle $ for some $l\in \mathbb {Z}$ . Notice that this group being abelian means $2l\equiv -k(p^s-1)\ \mod {d}$ . In the same way, since $T=\left \langle \delta ^{d/2},\phi ^s\delta ^j,\gamma \delta ^k\right \rangle $ is abelian, it means that $2j\equiv -k(p^s-1)\ \mod {d}$ , but then $2l\equiv 2j\ \mod {d}$ , which means $l\equiv j\ \mod {d/2}$ and $\upsilon (\hat {T}^R)=T$ . Therefore, $\tilde {T}:=\hat {T}^R$ is a T-abelian supplement.

We have therefore proved Theorem 1 in the linear and unitary cases.

4 Notation for groups of Lie type

By Theorem 9 and Table 2, to prove Theorem 1, we are left to deal with the following cases:

$$ \begin{align*}B_n(q),C_n(q) , D_n(q), E_7(q),\ q=p^m, p\not=2, \end{align*} $$

$$ \begin{align*}{}^2D_n(q),\ q=p^m, p\not=2, n\ \text{even}, \end{align*} $$

$$ \begin{align*}E_6(q),\ q=p^m, q\equiv 1\ \mod 3, \end{align*} $$

and

$$ \begin{align*}{}^2\!E_6(q),\ q=p^m, q\equiv -1\ \mod 3. \end{align*} $$

We give a brief introduction of the tools that we are going to use.

For the definitions and automorphisms of simple groups of Lie type, we refer to [Reference Carter7] (see also [Reference Steinberg20]). The automorphism groups of the finite simple groups of Lie type (untwisted and twisted) have been determined by Steinberg in [Reference Steinberg19] and by Griess, Lyons in [Reference Griess and Lyons12]. We denote by $\mathbb {F}_q$ the field with $q=p^m$ elements, where p is a prime. We briefly recall that the Chevalley group (or untwisted group of Lie type) $L(q)$ , viewed as a group of automorphisms of a Lie algebra $L_{\mathbb {F}_q}$ over $\mathbb {F}_q$ , obtained from a complex finite dimensional simple Lie algebra L, is the group generated by certain automorphisms $x_\alpha (t)$ , where t runs over $\mathbb {F}_q$ and $\alpha $ runs over the root system $\Phi $ associated to L. The finite untwisted groups of Lie type $L(q)$ are

$$ \begin{align*}A_n(q), n\geqslant 1, B_n(q), C_n(q), n\geqslant 2, D_n(q), n\geqslant 4, E_6(q), E_7(q), E_8(q), F_4(q), G_2(q). \end{align*} $$

It is well known that $L(q)$ is simple, except in the case $L(q)=A_1(2)$ , $A_1(3)$ , $B_2(2)$ , $G_2(2)$ ([Reference Carter7, Theorem 11.1.2]). The groups $A_1(2)$ , $A_1(3)$ are soluble. The group $B_2(2)$ is isomorphic to $S_6$ . The derived group of $G_2(2)$ is isomorphic to $\mathrm {PSU}_3(3)$ .

For every $\alpha \in \Phi $ , $t\in \mathbb {F}_q^\times $ , one defines $n_\alpha (t)=x_\alpha (t)x_{-\alpha }(-t^{-1})x_\alpha (t)$ , $n_\alpha =n_\alpha (1)$ and the subgroup $N={\langle {n_\alpha (t)\mid \alpha \in \Phi , t\in \mathbb {F}_q^\times }\rangle }$ of $L(q)$ .

Let $\Delta =\{\alpha _1,\ldots ,\alpha _n\}$ be a system of simple roots of $\Phi $ . We shall use the numbering and the description of the simple roots in terms of the canonical basis $(e_1,\ldots , e_r)$ of an appropriate $\mathbb {R}^r$ as in [Reference Bourbaki1], Planches I-IX. We denote by Q the root lattice, by P the weight lattice and by W the Weyl group; $s_i$ is the simple reflection associated to $\alpha _i$ , $\{\omega _1,\ldots ,\omega _n\}$ are the fundamental weights, $w_0$ is the longest element of W, and $A = (a_{ij})$ is the Cartan matrix (hence, $\alpha _i=\sum _j a_{ij}\omega _j$ ).

Let $\operatorname {\mathrm {Hom}}(Q,\mathbb {F}_q^\times )$ be the group of $\mathbb {F}_q$ -characters of Q (i.e., group homomorphisms from Q to $\mathbb {F}_q^\times $ ). For any $\chi \in \operatorname {\mathrm {Hom}}(Q,\mathbb {F}_q^\times )$ , one defines the automorphism $h(\chi )$ of $L_{\mathbb {F}_q}$ ([Reference Carter7, p. 98]). Let $\hat H=\{h(\chi )\mid \chi \in \operatorname {\mathrm {Hom}}(Q,\mathbb {F}_q^\times )\}$ . The map $\chi \mapsto h(\chi )$ is an isomorphism of $\operatorname {\mathrm {Hom}}(Q,\mathbb {F}_q^\times )$ onto $\hat H$ . We have $\hat H\leqslant N_{\mathrm {Aut} L_{\mathbb {F}_q}}(L(q))$ . The automorphism of $L(q)$ induced by $h(\chi )$ maps $x_\alpha (t)$ to $h(\chi )x_\alpha (t)h(\chi )^{-1}=x_\alpha (\chi (\alpha )t)$ ([Reference Carter7, p. 100]). Let $H=\hat H\cap L(q)$ . Then $h(\chi )$ lies in H if and only if $\chi $ can be extended to an $\mathbb {F}_q$ -character of P. The number d in Table 2 relative to the untwisted case is the order of $\hat H/H$ . We denote by $\mathrm {Inndiag}(L(q))$ the group $L(q)\hat H$ , the group of inner-diagonal automorphisms of $L(q)$ . Any automorphism $\phi $ of $\mathbb {F}_q$ induces a field automorphism, still denoted by $\phi $ , of $L(q)$ , which is defined by $x_\alpha (t)^\phi =x_\alpha (t^\phi )$ . In particular, the automorphism $x\mapsto x^p$ of ${\mathbb {F}_q}$ induces the field automorphism denoted by $g\mapsto g^{[p]}$ of $L(q)$ . Note that $h(\chi )^{[p]}=h(\chi )^p=h(p\chi )$ for every $\mathbb {F}_q$ -character $\chi $ of Q.

We recall that a symmetry of the Dynkin diagram of $L(q)$ is a permutation $\rho $ of the nodes of the diagram, such that the number of bonds joining nodes i, j is the same as the number of bonds joining nodes $\rho (i)$ , $\rho (j)$ , for any $i\not = j$ . A nontrivial symmetry $\rho $ of the Dynkin diagram can be extended to a map of the space $E=\mathbb {R} P$ into itself (an isometry if $L=A_n$ , $D_n$ , $E_6$ ), still denoted by $\rho $ . This map yields an outer automorphism, again denoted by $\rho $ , of $L(q)$ ; $\rho $ is said to be a graph automorphism of $L(q)$ (see [Reference Carter7, p. 200–210] for the detailed description). It is defined as follows.

1. (a) If $L(q)= A_n(q)$ , $n\geqslant 2$ , $D_n(q)$ , $E_6(q)$

   $$ \begin{align*} x_\alpha(t)^\rho=x_{\rho(\alpha)}(\gamma_\alpha t), \end{align*} $$
   where $\alpha \in \Phi $ , $t\in \mathbb {F}_q$ , $\gamma _\alpha \in \mathbb {Z}$ ; the $\gamma _\alpha $ can be chosen so that $\gamma _\alpha =1$ if $\alpha \in \pm \Delta $ .

2. (b) If $L(q)= B_2(q)$ , $F_4(q)$ and $q=2^m$

   $$ \begin{align*} x_\alpha(t)^\rho=x_{\rho(\alpha)}(t^{\lambda(\rho(\alpha))}), \end{align*} $$
   where $\alpha \in \Phi $ , $t\in \mathbb {F}_q$ , $\lambda (\alpha )=1$ if $\alpha $ is short, $\lambda (\alpha )=2$ if $\alpha $ is long. Note that $\rho ^2$ is the field automorphism $x_\alpha (t)\mapsto x_\alpha (t^2)$ , so $\rho $ has order $2m$ .

3. (c) If $L(q)= G_2(q)$ and $q=3^m$

   $$ \begin{align*} x_\alpha(t)^\rho=x_{\rho(\alpha)}(t^{\lambda(\rho(\alpha))}), \end{align*} $$
   where $\alpha \in \Phi $ , $t\in \mathbb {F}_q$ , $\lambda (\alpha )=1$ if $\alpha $ is short, $\lambda (\alpha )=3$ if $\alpha $ is long. Note that $\rho ^2$ is the field automorphism $x_\alpha (t)\mapsto x_\alpha (t^3)$ , so $\rho $ has order $2m$ .

Graph and field automorphisms commute; the subgroup R they generate (denoted by $\Phi _K\Gamma _K$ in [Reference Gorenstein, Lyons and Solomon11, Theorem 2.5.12]) normalises $\mathrm {Inndiag}(L(q))$ . We have

$$ \begin{align*}L(q)\trianglelefteq L(q)\hat H=\mathrm{Inndiag}(L(q))\trianglelefteq \mathrm{Inndiag}(L(q)):R= \operatorname{\mathrm{Aut}} (L(q)). \end{align*} $$

We shall identify field and graph automorphisms with their corresponding images in $\operatorname {\mathrm {Out}} (L(q))$ . The action of $\operatorname {\mathrm {Out}}(L(q))/\operatorname {\mathrm {Outdiag}}(L(q))$ on $\operatorname {\mathrm {Outdiag}}(L(q))$ is described in [Reference Bray, Holt and Roney-Dougal2, §1.7.2] and [Reference Gorenstein, Lyons and Solomon11, Theorem 2.5.12].

We have $H\triangleleft N$ and $N/H\cong W$ . For $w\in W$ , we denote by $\dot w$ a representative of w in N; for each $i=1,\ldots ,n$ , $n_{\alpha _i}$ is a representative of $s_i$ in N. For short, we denote $n_{\alpha _i}$ by $n_i$ . Note that $n_i$ lies in $L(p)$ , so that it is fixed by field automorphisms of $L(q)$ .

Next, we consider the finite twisted groups. These are defined as certain subgroups of appropriate untwisted groups $L(q^s)$ over the field ${\mathbb {F}_{q^s}}$ with $q^s$ elements, $q=p^m$ as usual (the list may be found in [Reference Carter7, p. 251]):

$$ \begin{align*}{}^2\!A_n(q), n\geqslant 2, {}^2\!D_n(q), n\geqslant 4, {}^3\!D_4(q), {}^2\!E_6(q), {}^2\!B_2(2^{2m+1}), {}^2\!F_4(2^{2m+1}), {}^2G_2(3^{2m+1}). \end{align*} $$

Note that for the types $A_n$ , $D_n$ , $E_6$ , we have used the notation ${}^s\!L(q)$ instead of ${}^s\!L(q^s)$ (used in [Reference Carter7, p. 251]) to stick with the notation in [Reference Lucchini, Menegazzo and Morigi17]. They are all simple, except for the groups ${}^2\!A_2(2)$ , ${}^2\!B_2(2)$ , ${}^2\!F_4(2)$ and ${}^2G_2(3)$ ([Reference Carter7, Theorem 14.4.1]). The groups ${}^2\!A_2(2)$ , ${}^2\!B_2(2)$ are soluble. The derived subgroup of ${}^2G_2(3)$ is isomorphic to the simple group $\mathrm {PSL}_{2}{(8)}$ . The derived subgroup ${}^2\!F_4(2)'$ of ${}^2\!F_4(2)$ has index 2 in ${}^2\!F_4(2)$ , and it is a simple group called the Tits group. For the simple groups $G_0$ of type ${}^3\!D_4(q)$ , ${}^2\!B_2(2^{2m+1})$ , ${}^2\!F_4(2^{2m+1})$ , ${}^2G_2(3^{2m+1})$ , ${}^2\!F_4(2)'$ , the group $\operatorname {\mathrm {Out}} (G_0)$ is cyclic ([Reference Gorenstein, Lyons and Solomon11, Theorem 2.5.12, 12]). Therefore, every almost simple group G with socle $G_0$ is abelian supplemented (Corollary 7). Moreover, we have ${}^2\!A_n(q)\cong \mathord {\mathrm {PSU}_{n}(q)}$ , so we are left to deal with ${}^2\!D_n(q)$ and ${}^2\!E_6(q)$ . We observe that if n is odd or $p = 2$ , then $\operatorname {\mathrm {Aut}}({}^2\!D_n(q))$ splits over ${}^2\!D_n(q)$ (Theorem 9), so that there exists a T-abelian supplement for every abelian $T\leqslant \operatorname {\mathrm {Out}}({}^2\!D_n(q))$ (Lemma 8). In view of this discussion, we shall deal with the remaining cases. Below, we give a short description of these groups

So, let us assume that L is of type $D_n$ or $E_6$ , and $\tau $ is an order $2$ symmetry of the Dynkin diagram. The twisted group ${}^2\!L(q)$ is a certain subgroup of the Chevalley group $L(q^2)$ ([Reference Carter7, Definition 13.4.2]). Let E be the real vector space spanned by the roots (or the weights). Then $\tau $ induces an automorphism (in fact an isometry), still denoted by $\tau $ , of E fixing both Q and P. Let $\chi $ be an $\mathbb {F}_{q^2}$ -character of Q (or P). We say that $\chi $ is self-conjugate if $\chi (\tau (x))=\chi (x)^q$ for every x in Q (or P). Let $\hat H^1=\{h(\chi )\mid \chi ~:~Q~\to ~\mathbb {F}_{q^2}^\times \ \text {is a self-conjugate character of} \ Q\}$ . We have $\hat H^1\leqslant N_{\mathrm {Aut} L_{\mathbb {F}_{q^2}}}({}^2\!L(q))$ . Let $H^1=\hat H^1\cap {}^2\!L(q)$ . Then $h(\chi )$ lies in $H^1$ if and only if $\chi $ can be extended to a self-conjugate $\mathbb {F}_{q^2}$ -character of P. The number d in Table 2 relative to the twisted case is the order of $\hat H^1/H^1$ . We denote by $\mathrm {Inndiag}({}^2\!L(q))$ the group ${}^2\!L(q)\hat H^1$ of inner-diagonal automorphisms of ${}^2\!L(q)$ . Any automorphism $\phi $ of $\mathbb {F}_{q^2}$ induces the field automorphism $\phi $ of $L(q^2)$ , which leaves ${}^2\!L(q)$ invariant and therefore induces an automorphism of ${}^2\!L(q)$ (also called a field automorphism). If $R^1$ is the group of field automorphisms of ${}^2\!L(q)$ , we have

$$ \begin{align*}{}^2\!L(q)\trianglelefteq {}^2\!L(q)\hat H^1=\mathrm{Inndiag}({}^2\!L(q))\trianglelefteq \mathrm{Inndiag}({}^2\!L(q)):R^1= \operatorname{\mathrm{Aut}} ({}^2\!L(q)). \end{align*} $$

In general, we have $(1-z)P\leqslant Q$ for every $z\in W$ . For Coxeter elements, equality holds:

Lemma 20. Let $\alpha _1,\ldots ,\alpha _n$ be the simple roots (in any fixed order), $\omega _1,\ldots ,\omega _n$ the corresponding fundamental weights. Then

$$ \begin{align*}(1-s_1\cdots s_n)\omega_i =\alpha_i+z_1\alpha_1+\cdots +z_{i-1}\alpha_{i-1} \end{align*} $$

with $z_1,\ldots ,z_{i-1}\in \mathbb {Z}$ . In particular, $(1-s_1\cdots s_n)P=Q$ .

Proof. We have $s_i(\omega _j)=\omega _j-\delta _{ij}\alpha _i$ for every i, j. For $i=1$ , we have $s_1\cdots s_n\omega _1=s_1\omega _1=\omega _1-\alpha _1$ ; hence, $(1-s_1\cdots s_n)\omega _1=\alpha _1$ . Let $1<i\leqslant n$ . Then $s_1\cdots s_{i-1}(\alpha _i)=\alpha _i+z_1\alpha _1+\cdots +z_{i-1}\alpha _{i-1}$ , with $z_u\in \mathbb {Z}$ for $u=1,\ldots , i-1$ . Then

$$ \begin{align*}(1-s_1\cdots s_n)\omega_i &= \omega_i-s_1\cdots s_i\omega_i= \omega_i-s_1\cdots s_{i-1}(\omega_i-\alpha_i) \\ & =\omega_i-(\omega_i-s_1\cdots s_{i-1}\alpha_i)=s_1\cdots s_{i-1}\alpha_i=\alpha_i+z_1\alpha_1+\cdots+z_{i-1}\alpha_{i-1}. \end{align*} $$

Let $\chi $ be a character of Q, w in W. We define the character $w\chi $ in the following way. For $x\in Q$ , we put $w\chi (x):=\chi (w^{-1}x)$ (i.e., $w\chi =\chi \circ w^{-1}$ ). We also define $\tau \chi $ , where $\tau $ is a graph automorphism, by $(\tau \chi )x:=\chi (\tau ^{-1}x)$ for $x\in Q$ (hence, $\tau \chi =\chi \circ \tau ^{-1}$ ). Note that for $w\in W$ , we have ([Reference Carter7, Theorem 7.2.2])

$$ \begin{align*}\dot w h(\chi)\dot w^{-1}=h(w\chi). \end{align*} $$

Since we are assuming $\Phi $ of type $D_n$ or $E_6$ , there is a Coxeter element w in W fixed by $\tau .$ We may choose a representative $\dot w$ of w in N over the prime field and fixed by $\tau .$ Let $F=\phi ^j$ or $\phi ^j\tau $ , for some integer j. Then F fixes $\dot w$ and acts on $\hat H$ ; hence, it induces an automorphism g of $\operatorname {\mathrm {Hom}}(Q,\mathbb {F}_q^\times )$ given by $F(h(\chi ))=h(g(\chi ))$ . Let $\chi :Q\to \mathbb {F}_q^\times $ be a fixed character, $x=\dot w h(\chi )$ . We shall look for an element $y=h(\chi ')\in \hat H$ such that $[x,Fy]=1$ ; that is,

$$ \begin{align*}xF y=F yx\iff y^{-1}F^{-1}xFy=x\iff y^{-1}F(x)y=x \end{align*} $$

so that

$$ \begin{align*}h(\chi')^{-1}\dot w h(g(\chi))h(\chi')=\dot w h(\chi). \end{align*} $$

We have $h(\chi ')^{-1}\dot w =\dot w\dot w^{-1}h(\chi ')^{-1}\dot w=\dot wh(w^{-1}\chi ')^{-1}=\dot wh(-w^{-1}\chi ')$ ; hence,

$$ \begin{align*}\dot wh(-w^{-1}\chi') h(g(\chi))h(\chi')=\dot w h(\chi), \end{align*} $$

$$ \begin{align*}h(-w^{-1}\chi') h(g(\chi))h(\chi')=h(\chi), \end{align*} $$

and finally,

$$ \begin{align*}h((1-w^{-1})\chi')=h((1-g)\chi),\quad (1-w^{-1})\chi'=(1-g)\chi, \end{align*} $$

$$ \begin{align*}\chi'\circ (1-w)=(1-g)\chi. \end{align*} $$

We shall be interested in the following cases:

$F:s\mapsto s^{[p^i]}$ , then

$$ \begin{align*}\chi'\circ (1-w)=(1-p^i)\chi, \end{align*} $$

$F:s\mapsto s^{[p]\tau }$ , then

$$ \begin{align*}\chi'\circ (1-w)=(1-p\tau)\chi. \end{align*} $$

By Lemma 20, we have $(1-w)^{-1}Q=P$ . Let $\Delta =|\ \! P/Q\!\ |=\det A$ . Then $\Delta P\leqslant Q$ . Note that if $\Phi =D_n$ with n even, then $2 P\leqslant Q$ since $P/Q\cong C_2\times C_2$ (the inverses of the Cartan matrices may be explicitly found in [Reference Wei and Zou22]). We put $\Delta _1=|\ \! P/Q\!\ |$ unless $\Phi =D_n$ , n even, in which case we put $\Delta _1=2$ . Then

$$ \begin{align*}\Delta_1(1-w)^{-1}Q\leqslant Q, \end{align*} $$

and we may define the character

$$ \begin{align*}\zeta_\chi=\chi\circ \Delta_1(1-w)^{-1}:Q\to \mathbb{F}_q^\times \end{align*} $$

and $h(\zeta _\chi )\in \hat H$ .

We start with the cases $B_n(q),C_n(q), E_7(q)$ .

5 $C_n(q)$ , $B_n(q)$ , $n\geqslant 2$ , $E_7(q)$

Here, L is of type $C_n$ , $B_n$ or $E_7$ , $G_0=L(q)$ , $q=p^m$ , $d=(q-1,2)$ , and we assume that $\operatorname {\mathrm {Aut}}(G_0)$ does not split over $G_0$ , so $(\frac {q-1}d,d,m)\not =1$ . Therefore, $d=2$ and p is odd:

$$ \begin{align*}\operatorname{\mathrm{Out}}(G_0)=\left\langle\delta\right\rangle\times\left\langle\phi\right\rangle, \end{align*} $$

$|\ \! \delta \!\ |=2$ , $|\ \! \phi \!\ |=m$ . We fix an $\mathbb {F}_q$ -character $\chi $ of Q which cannot be extended to a character of P, so that $h(\chi )$ induces $\delta $ in $\operatorname {\mathrm {Out}}(G_0)$ . We look for an $\mathbb {F}_q$ -character $\chi '$ so that $[\dot w h(\chi ),\phi h(\chi ')]=1$ ; that is,

$$ \begin{align*}\chi'\circ (1-w)=(1-p)\chi. \end{align*} $$

We have $\Delta _1=2$ , so $\zeta _\chi =\chi \circ 2(1-w)^{-1}$ . We take

$$ \begin{align*}\chi'=\frac{1-p}2\ \zeta_\chi \end{align*} $$

so $h(\chi ')=h(\zeta _\chi )^{\frac {1-p}2}$ . Therefore,

$$ \begin{align*}\tilde T=\left\langle\dot w h(\chi),\phi h(\chi')\right\rangle \end{align*} $$

is an $\operatorname {\mathrm {Out}}(G_0)$ -abelian supplement (arguing as in the $\mathrm {PSL}_{2}{(q)}$ case).

We have proved the following.

6 $E_6(q)$

Here, L is of type $E_6$ , $G_0=L(q)$ , $q=p^m$ , $d=(q-1,3)$ , and we assume that $\operatorname {\mathrm {Aut}}(G_0)$ does not split over $G_0$ , so $(\frac {q-1}d,d,m)\not =1$ . Therefore, $d=3$ and $p\not =3$ :

$$ \begin{align*}\operatorname{\mathrm{Out}}(G_0)=\left\langle\delta\right\rangle\rtimes\left\langle\phi,\tau\right\rangle, \end{align*} $$

where $|\ \! \delta \!\ |=3$ , $|\ \! \phi \!\ |=m$ , $\delta ^\phi =\delta ^p$ , $\delta ^\tau =\delta ^{-1}$ and $[\phi ,\tau ]=1$ . We fix an $\mathbb {F}_q$ -character $\chi $ of Q which can not be extended to a character of P.

Let $\pi \colon \operatorname {\mathrm {Out}}(G_0)\to \operatorname {\mathrm {Out}}(G_0)/\left \langle \delta \right \rangle =\left \langle \phi ,\tau \right \rangle $ . Let T be a noncyclic abelian subgroup of $\operatorname {\mathrm {Out}}(G_0)$ . If $\pi (T)$ is not cyclic, then $\pi (T)={\langle {\phi ^s,\tau }\rangle }$ . Therefore, $T= {\langle {\phi ^s\delta ^i,\tau \delta ^k}\rangle }$ . But $\tau \delta ^k$ is conjugate to $\tau $ under ${\langle \delta \rangle }$ ; hence, we may assume $T= {\langle {\phi ^s\delta ^i,\tau }\rangle }$ , so $T= {\langle {\phi ^s,\tau }\rangle } \leqslant {\langle {\phi ,\tau }\rangle }$ and $\tilde T={\langle {\phi ^s,\tau }\rangle }$ is a T-abelian supplement.

We are left with the case where $\pi (T)$ is cyclic; that is, $ \pi (T)={\langle {\phi ^s\tau ^\epsilon }\rangle }$ . Then $T={\langle {\delta ,\phi ^s\tau ^\varepsilon }\rangle }$ . Since $p\not =3$ , we have $p\equiv 1$ or $-1\ \mod 3$ .

Let $p\equiv 1\ \mod 3$ . Then $[\delta ,\phi ]=1$ , and we get $\varepsilon =0$ , $T\leqslant {\langle {\delta ,\phi }\rangle }$ , so by Lemma 5, it is enough to consider the case

$$ \begin{align*}p\equiv 1\quad \mod 3\ ,\ T= {\langle {\delta,\phi}\rangle} \quad\quad\text{(case 1)}. \end{align*} $$

Let $p\equiv -1\ \mod 3$ . Then $\delta ^\phi =\delta ^{-1}$ . If $\varepsilon =1$ , $T={\langle {\delta ,\phi ^s\tau }\rangle }$ . Since $[\delta ,\phi ^s\tau ]=1$ , s must be odd. Therefore, $T\leqslant {\langle {\delta ,\phi \tau ,\phi ^2}\rangle } ={\langle {\delta ,\phi \tau }\rangle }$ . If $\varepsilon =0$ , $T={\langle {\delta ,\phi ^s}\rangle }$ , so s is even, and again, $T\leqslant {\langle {\delta ,\phi ^2}\rangle } < {\langle {\delta ,\phi \tau }\rangle }$ . Therefore, it is enough to consider

$$ \begin{align*}p\equiv -1\quad \mod 3\ ,\ T={\langle {\delta,\phi\tau}\rangle} \quad\quad\text{(case 2)}. \end{align*} $$

Summarising, we only have to deal with cases 1, 2.

We consider the Coxeter element $w=s_1s_4s_6s_3s_2s_5$ , fixed by the graph automorphism $\tau $ . In fact, we have

$$ \begin{align*}\tau(\alpha_1)=\alpha_6, \tau(\alpha_2)=\alpha_2, \tau(\alpha_3)=\alpha_5, \tau(\alpha_4)=\alpha_4, \tau(\alpha_5)=\alpha_3,\tau(\alpha_6)=\alpha_1. \end{align*} $$

We choose a representative $\dot w$ of w in N over the prime field and fixed by $\tau $ , $\dot w=n_1n_4n_6n_3n_2n_5$ , for instance. Hence, $\tau \dot w=\dot w\tau $ , $\dot w\phi =\phi \dot w$ . Here, $\phi $ is the field automorphism of $G_0$ sending x to $x^{[p]}$ . We use the notation $\phi ^{-1}x\phi =x^{[p]}$ . We have $\Delta _1=3$ , so $\zeta _\chi =\chi \circ 3(1-w)^{-1}$ .

Case 1: $p\equiv 1\ \mod 3$ , $T={\langle {\delta ,\phi }\rangle }$ .

We take

$$ \begin{align*}\chi'=\frac{1-p}3\ \zeta_\chi \end{align*} $$

so $h(\chi ')=h(\zeta _\chi )^{\frac {1-p}3}$ . Therefore,

$$ \begin{align*}\tilde T={\langle {\dot w h(\chi),\phi h(\chi')}\rangle}\end{align*} $$

is a T-abelian supplement.

Case 2: $p\equiv -1\ \mod 3$ , $T={\langle {\delta , \phi \tau }\rangle }$ .

Since $\tau w_0=-1$ , we have

$$ \begin{align*}(1+\tau)P=(1+\tau)w_0P=(w_0+\tau w_0)P=(w_0-1)P=(1-w_0)P\leqslant Q. \end{align*} $$

Hence, by Lemma 20,

$$ \begin{align*}(1+\tau)(1-w)^{-1}Q=(1+\tau)P\leqslant Q, \end{align*} $$

so $\chi \circ (1+\tau )(1-w)^{-1}$ is an $\mathbb {F}_q$ -character of Q. We look for an $\mathbb {F}_q$ -character $\chi '$ so that $[\dot w h(\chi ),\phi \tau h(\chi ')]=1$ ; that is,

$$ \begin{align*}\chi'\circ (1-w)=(1-p\tau)\chi. \end{align*} $$

We have $1-p\tau =1+p-p-p\tau =1+p-p(1+\tau )$ , and we may define

$$ \begin{align*}\chi'=\frac{1+p}3\ \zeta_\chi-p\,\chi\circ (1+\tau)(1-w)^{-1}, \end{align*} $$

obtaining a character which satisfies $\chi '\circ (1-w)=(1-p\tau )\chi $ . Therefore,

$$ \begin{align*}\tilde T={\langle {\dot w h(\chi),\phi\tau h(\chi')}\rangle} \end{align*} $$

is a T-abelian supplement.

We have proved the following.

7 ${}^2\!E_6(q)$

Here, L is of type $E_6$ , $G_0={}^2\!E_6(q)\leqslant E_6(q^2)$ , $q=p^m$ , $d=(q+1,3)$ , and we assume that $\operatorname {\mathrm {Aut}}(G_0)$ does not split over $G_0$ ; that is, $(\frac {q+1}d,d,m)\not =1$ . Therefore, $d=3$ and $q\equiv -1\ \mod 3$ , so $p\equiv -1\ \mod 3$ and m is odd:

$$ \begin{align*}\operatorname{\mathrm{Out}}(G_0)=\left\langle\delta\right\rangle\rtimes\left\langle\phi\right\rangle, \end{align*} $$

where $|\ \! \delta \!\ |=3$ , $|\ \! \phi \!\ |=2m$ , $\delta ^\phi =\delta ^{-1}$ .

It is enough to consider the case $T={\langle {\delta , \phi ^{2}}\rangle }$ . We fix a self-conjugate $\mathbb {F}_{q^2}$ -character $\chi $ of Q which can not be extended to a self-conjugate $\mathbb {F}_{q^2}$ -character of P (so that $h(\chi )\in \hat H^1\setminus H^1$ ).

We consider the same Coxeter element $w=s_1s_4s_6s_3s_2s_5$ as in the previuos section, and the same representative $\dot w=n_1n_4n_6n_3n_2n_5$ , which lies in $G_0$ .

We look for an element $h(\chi ')\in \hat H^1$ so that $[\dot w h(\chi ),\phi ^2h(\chi ')]=1$ ; that is,

$$ \begin{align*}\chi'\circ (1-w)=(1-p^2)\chi. \end{align*} $$

We have $\Delta _1=3$ , so $\zeta _\chi =\chi \circ 3(1-w)^{-1}$ . We take

$$ \begin{align*}\chi'=\frac{1-p^2}3\ \zeta_\chi, \end{align*} $$

so $h(\chi ')=h(\zeta _\chi )^{\frac {1-p^2}3}$ .

Note that since $\chi $ is self-conjugate and $\tau w=w\tau $ , $\zeta _\chi $ and $\chi '$ are self-conjugate, so $h(\chi ')$ lies in $\hat H^1$ . Therefore,

$$ \begin{align*}\tilde T=\left\langle\dot w h(\chi),\phi^2 h(\chi')\right\rangle \end{align*} $$

is a T-abelian supplement.

We have proved the following.

8 ${}^2\!D_n(q)$ , n even

Here, L is of type $D_n$ , n even, $G_0={}^2\!D_n(q)\leqslant D_n(q^2)$ , $q=p^m$ , $d=(q+1,2)$ , and we assume that $\operatorname {\mathrm {Aut}}(G_0)$ does not split over $G_0$ . Therefore, $d=2$ , $p\not =2$ , and

$$ \begin{align*}\operatorname{\mathrm{Out}}(G_0)=\left\langle\delta\right\rangle\times\left\langle\phi\right\rangle, \end{align*} $$

where $|\ \! \delta \!\ |=2$ , $|\ \! \phi \!\ |=2m$ .

It is enough to consider the case $T=\operatorname {\mathrm {Out}}(G_0)$ . We fix a self-conjugate $\mathbb {F}_{q^2}$ -character $\chi $ of Q which can not be extended to a self-conjugate $\mathbb {F}_{q^2}$ -character of P (so that $h(\chi )\in \hat H^1\setminus H^1$ ).

We consider the Coxeter element $w=s_1s_2\cdots s_{n-1}s_n$ , fixed by $\tau $ (which exchanges $\alpha _{n-1}$ and $\alpha _n$ ), and the representative $\dot w=n_1n_2\cdots n_{n-1}n_n$ , which lies in $G_0$ . We look for an element $h(\chi ')\in \hat H^1$ so that $[\dot w h(\chi ),\phi h(\chi ')]=1$ ; that is,

$$ \begin{align*}\chi'\circ (1-w)=(1-p)\chi. \end{align*} $$

We have $\Delta _1=2$ (since n is even), so $\zeta _\chi =\chi \circ 2(1-w)^{-1}$ . We take

$$ \begin{align*}\chi'=\frac{1-p}2\ \zeta_\chi, \end{align*} $$

so $h(\chi ')=h(\zeta _\chi )^{\frac {1-p}2}$ .

Since $\chi $ is self-conjugate and $\tau w=w\tau $ , $\zeta _\chi $ and $\chi '$ are self-conjugate, so $h(\chi ')$ lies in $\hat H^1$ . Therefore,

$$ \begin{align*}\tilde T=\left\langle\dot w h(\chi),\phi h(\chi')\right\rangle \end{align*} $$

is an $\operatorname {\mathrm {Out}}(G_0)$ -abelian supplement.

We have proved the following.

9 The remaining case

In the next sections, we shall deal with the remaining case: $D_n(q)$ , $q=p^m$ . We shall use the identifications with classical groups as in [Reference Carter7, Theorem 11.3.2] and [Reference Carter8, 1.11, 1.19]. Here, $\lambda $ is a generator of $\mathbb F_q^\times $ .

We have $G_0=P\Omega _{2n}^+(q)$ , $\mathrm {Inndiag}(G_0)=P(CO_{2n}(q)^\circ )$ , where $CO_{2n}(q)$ if the conformal orthogonal group; that is, the group of orthogonal similitudes of $\mathbb {F}_q^{2n}$ ; $CO_{2n}(q)^\circ $ is the subgroup of index 2 of $CO_{2n}(q)$ of elements which do not interchange the two families of maximal isotropic subspaces of $\mathbb {F}_q^{2n}$ . If $(e_1,\ldots ,e_n,f_1,\ldots ,f_n)$ is the canonical basis of $\mathbb {F}_q^{2n}$ , the bilinear form on $\mathbb {F}_q^{2n}$ corresponds to the matrix

$$ \begin{align*}K_n= \left( \begin{array}{cccccc} 0_{n}&I_{n}\\ I_{n}&0_{n}\\ \end{array} \right). \end{align*} $$

We define the homomorphism $\eta :CO_{2n}(q)^\circ \to \mathbb {F}_q^\times $ by

$$ \begin{align*}\eta(X)=\mu \quad \text{if}\quad {}^t\!XK_nX=\mu K_n. \end{align*} $$

For $\mu \in \mathbb {F}_q^\times $ , let $o_\mu =\left ( \begin {smallmatrix} I_n&0\\ 0&\mu I_n \end {smallmatrix} \right )$ , so that $\eta (o_\mu )=\mu $ .

The graph automorphism $\tau $ of $D_n$ exchanging $\alpha _{n-1}$ and $\alpha _n$ is induced by conjugation with

$$ \begin{align*}\tau_n= \left( \begin{array}{cccccc} I_{n-1}&0&0_{n-1}&0\\ 0&0&0&1\\ 0_{n-1}&0&I_{n-1}&0\\ 0&1&0&0\\ \end{array} \right)\in O_{2n}(q), \end{align*} $$

$\tau _n^2=1$ , $x^\tau =\tau _n x\tau _n$ .

$$ \begin{align*}x_{\alpha_i}(z)^\tau=x_{\alpha_i}(z), i=1,\ldots, n-2,\quad x_{\alpha_{n-1}}(z)^\tau=x_{\alpha_n}(z), x_{\alpha_n}(z)^\tau=x_{\alpha_{n-1}}(z). \end{align*} $$

We shall deal with the cases n odd and even separately.

10 $D_n(q)$ , $n\geqslant 3$ , n odd

Here, L is of type $D_n$ , n odd, $G_0=D_n(q)$ , $q=p^m$ , $d=(4 ,q-1)$ , and we assume that $\operatorname {\mathrm {Aut}}(G_0)$ does not split over $G_0$ ; hence, $( \frac {q^n-1}d,d,m)\not =1$ . In particular, $d=2$ or $4$ , and p is odd. Moreover, m is even; hence, $4$ divides $q-1$ . Therefore, $d=4$ .

$$ \begin{align*}\operatorname{\mathrm{Out}}(G_0)={\langle {\delta, \tau, \phi\mid \delta^4=\tau^2=1, \delta^\tau=\delta^{-1}, \phi^m=[\tau,\phi]=1, \delta^\phi=\delta^p}\rangle}. \end{align*} $$

In $\Omega ^+_{2n}(q)$ , we choose

$$ \begin{align*}\dot w_0= \left( \begin{array}{cccccc} 0_{n-1}&0&I_{n-1}&0\\ 0&1&0&0\\ I_{n-1}&0&0_{n-1}&0\\ 0&0&0&1\\ \end{array} \right), \end{align*} $$

a representative of the longest element $w_0$ of the Weyl group. We have $\dot w_0^2=1$ , $\dot w_0\tau _n=\tau _n\dot w_0=K_n$ . Let $X\in CO_{2n}(q)^\circ $ , $\eta (X)=\mu $ (i.e., ${}^t\!XK_nX=\mu K_n$ ). Then ${}^t\!X=\mu K_nX^{-1}K_n$ , so that

(10.1) $$ \begin{align} {}^t\!X^{-1}=\eta(X)^{-1}\dot w_0 \tau_n X\tau_n \dot w_0=\eta(X)^{-1}\dot w_0 X^\tau \dot w_0. \end{align} $$

We start with $D_3$ , exploiting the fact that $D_3=A_3$ . Let $V={\mathbb {F}}_q^4$ with canonical basis ${\mathcal B}=(v_1,\ldots ,v_4)$ over ${\mathbb {F}}_q$ , $\overline V=\overline {\mathbb {F}}_q^4$ with the same basis over $\overline {\mathbb {F}}_q$ . Let

$$ \begin{align*}\sigma:GL(\overline V)\to GL(\wedge^2 \overline V),\quad f\mapsto \wedge^2 f. \end{align*} $$

We choose the basis $\mathcal B$ for $\overline V$ , and the basis ${\mathcal C}=(v_{12}, v_{13}, v_{23}, v_{34}, v_{42}, v_{14})$ , where $v_{ij}=v_i\wedge v_j$ , for $\wedge ^2 \overline V$ . We endow $\wedge ^2\overline V$ with the symmetric bilinear form with matrix $K_3$ with respect to ${\mathcal C}$ . Then $\sigma (GL(\overline V))\leqslant CO(\wedge ^2\overline V)^\circ $ , $\sigma (GL(V))\leqslant CO(\wedge ^2V)^\circ $ , and, by considering bases, we obtain the homomorphism $\sigma :GL_4(q) \to CO_6(q)^\circ $ . We have

$$ \begin{align*}\sigma: \left( \begin{array}{cc} I_3&0\\ 0&\mu \end{array} \right)\mapsto \left( \begin{array}{cc} I_3&0_3\\ 0_3&\mu I_3 \end{array} \right)=o_\mu; \end{align*} $$

in particular,

$$ \begin{align*}\det \left( \begin{array}{cc} I_3&0\\ 0&\mu \end{array} \right)=\mu=\eta(o_\mu). \end{align*} $$

Moreover, $ \sigma : \mu I_4 \mapsto \mu ^2 I_6 $ . If $X\in GL_4(q)$ , $\det X=\mu $ , then $ X=Y\left ( \begin {smallmatrix} I_3&0\\ 0&\mu \end {smallmatrix} \right ) $ with $Y\in SL_4(q)$ , $ \sigma (X)=\sigma (Y)o_\mu $ with $\sigma (Y)\in \Omega _6^+(q)$ ([Reference Taylor21, Theorem 12.20]); hence,

(10.2) $$ \begin{align} \eta(\sigma(X))=\mu=\det X. \end{align} $$

From (10.1) and (10.2), we get

(10.3) $$ \begin{align} \sigma({}^t\!X^{-1})={}^t\!(\sigma(X))^{-1}=(\det X)^{-1}\dot w_0\sigma(X)^\tau\dot w_0. \end{align} $$

For X, $Y\in GL_4(q)$ , $z\in \mathbb {F}_q^\times $ , we get

(10.4) $$ \begin{align} \begin{aligned} Y^{-1}X^{[p]} Y = z X \quad &\Rightarrow\quad \sigma(Y)^{-1}\sigma(X)^{[p]} \sigma(Y)= z^2 \sigma(X)\\ Y^{-1}({}^t\!X^{-1} )Y = z X \quad &\Rightarrow\quad Z^{-1} \sigma(X)^\tau Z= z^2\det (X) \sigma(X)\ , \ Z=\dot w_0\sigma(Y ) \\ {}^t\!X^{-1} Y = z Y^{[p]} X \quad &\Rightarrow\quad \sigma(X)^\tau Z= z^2 \det(X) Z^{[p]}\sigma(X)\ ,\ Z=\dot w_0\sigma(Y ) \end{aligned} \end{align} $$

since $\dot w_0^{[p]}=\dot w_0$ .

In Section 3, for a given abelian subgroup T of $\operatorname {\mathrm {Out}}(\mathrm {PSL}_{4}{(q)})$ , we have exhibited a T-abelian supplement $\tilde T$ by giving matrices in $\mathord {\mathrm {GL}_{4}(q)}$ : the map $\sigma $ allows to solve the problem for $G_0=P\Omega ^+_6(q)$ , by giving matrices in $CO_{6}(q)^\circ $ . Now we consider $D_n$ , n odd, $n=1+2m$ , $n>3$ . The space $\mathbb {F}_q^{2n}$ is the orthogonal direct sum $\mathbb {F}_q^{2n}=U\oplus U^\perp $ , where $U={\langle {e_1,\ldots ,e_{n-3},f_1,\ldots ,f_{n-3}}\rangle }$ , $U^\perp ={\langle {e_{n-2},e_{n-1},e_n,f_{n-2},f_{n-1},f_n}\rangle }$ , with $\dim U=2n-6=4(m-1)$ . Moreover, U is the direct orthogonal sum of subspaces of dimension 4:

$$ \begin{align*}U_1={\langle {e_1,e_2,f_1,f_2}\rangle},\ldots, U_{m-1}={\langle {e_{n-4}, e_{n-3},f_{n-4},f_{n-3}}\rangle}. \end{align*} $$

To define an isometry or more generally an orthogonal similitude of $\mathbb {F}_q^{2n}$ , we may give matrices $X_i\in CO_4(q)^\circ $ , $\eta (X_i)=\mu $ , $i=1,\ldots ,m-1$ , $X\in CO_6(q)^\circ $ , $\eta (X)=\mu $ and define Y in $\mathrm {GL}_{2n}(q)$ by

$$ \begin{align*}Y=X_1\oplus \cdots \oplus X_{m-1}\oplus X. \end{align*} $$

Then $Y\in CO_{2n}(q)^\circ $ , with $\eta (Y)=\mu $ . If $Y\in CO_{2n}(q)^\circ $ fixes $U^\perp $ , then it fixes U, and if we write $Y=X\oplus Z$ , with $X\in CO_6(q)^\circ $ , $Z\in CO_{2n-6}(q)^\circ $ , and consider the action of $\phi $ and $\tau $ , we get

$$ \begin{align*}Y^{[p]}=X^{[p]}\oplus Z^{[p]},\quad Y^\tau=X^\tau \oplus Z \end{align*} $$

since $\tau _n$ acts on the basis $(e_1,\ldots ,e_n,f_1,\ldots ,f_n)$ just by switching $e_n$ and $f_n$ (here, $Y^\tau =\tau _n Y\tau _n$ , $X^\tau = \tau _3 X\tau _3$ ).

We shall proceed as follows. Assume T is an abelian subgroup of $\operatorname {\mathrm {Out}} (G_0)$ . We consider the analogous subgroup T of $\operatorname {\mathrm {Out}} (\mathrm {PSL}_{4}{(q)})$ . From the $\mathrm {PSL}_{4}{(q)}$ case, we have an abelian subgroup of $\operatorname {\mathrm {Aut}} (\mathrm {PSL}_{4}{(q)})$ given by explicit matrices in $\mathrm {GL}_4(q)$ . By using $\sigma $ , we obtain corresponding matrices in $CO_6(q)^\circ $ satisfying certain relations. For each such matrix X, we define a matrix $X_1\in CO_4(q)^\circ $ and finally define the matrix $Y=X_1\oplus \cdots \oplus X_1\oplus X$ in $CO_{2n}(q)^\circ $ ( $m-1$ copies of $X_1$ ). We shall then obtain a T-abelian supplement $\tilde T$ in $\mathrm {Aut} (G_0)$ .

Let A, $B\in GL_2(q)$ with

$$ \begin{align*}B^{-1}A^{[p]}B=z A, \quad \det A=\mu, \ z=\mu^{\frac12(p-1)} \end{align*} $$

and let $\nu \in \mathbb {F}_q^\times $ . Our aim is to define orthogonal similitudes of $\mathbb {F}_q^4$ (with respect to the form given by $K_2$ ). We put

$$ \begin{align*}a&=a(A)= \left( \begin{array}{cc} A & 0_2 \\ 0_2 & {}^t\!A^{-1} \\ \end{array} \right)\left( \begin{array}{cc} I_2& 0_2 \\ 0_2 &(\det A) I_2 \\ \end{array} \right)\in CO_4(q)^\circ ,\quad \eta(a)=\det A\\b&=b(B,\nu)= \left( \begin{array}{cc} B & 0_2 \\ 0_2 & {}^t\!B^{-1} \\ \end{array} \right)\left( \begin{array}{cc} I_2& 0_2 \\ 0_2 & \nu I_2 \\ \end{array} \right)\in CO_4(q)^\circ ,\quad \eta(b)=\nu. \end{align*} $$

From $ B^{-1}A^{[p]}B=\mu ^{\frac 12(p-1) }A $ , we get

$$ \begin{align*}b^{-1}a^{[p]}b=\mu^{\frac12(p-1)} a,\quad \eta(a)=\det A=\mu, \eta(b)=\nu. \end{align*} $$

We shall take

$$ \begin{align*}A=\left( \begin{array}{cc} 0 & -\mu \\ 1 & 0 \\ \end{array} \right),\quad B=\left( \begin{array}{cc} \mu^{\frac12(p-1)}& 0 \\ 0 & 1 \\ \end{array} \right),\quad B^{-1}A^{[p]}B=\mu^{\frac12(p-1)} A, \end{align*} $$

(10.5) $$ \begin{align} a=a(A)=a(\mu)=\left( \begin{array}{cccc} 0 & -\mu& 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & \mu & 0 \\ \end{array} \right)\ , \ \eta(a)=\mu, \end{align} $$

(10.6) $$ \begin{align} b=b(B,\nu)=b(\mu,\nu)= \left( \begin{array}{cccc} \mu^{\frac{1}{2}(p-1)} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \mu^{-\frac{1}{2}(p-1)} & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \nu & 0 \\ 0 & 0 & 0 & \nu \\ \end{array} \right) \ ,\ \eta(b)=\nu. \end{align} $$

Then

$$ \begin{align*}b^{-1}a^{[p]}b=\mu^{\frac12(p-1)} a ,\quad \eta(a(\mu))=\mu,\ \eta(b(\mu,\nu))=\nu. \end{align*} $$

Note that for any $i\in {\mathbb Z}$ , we have

$$ \begin{align*}B^{-1}(A^i)^{[p]}B=\mu^{\frac12i(p-1)} A^i,\quad \det A^i=\mu^i, \end{align*} $$

$$ \begin{align*}b(\det A,\nu)^{-1}a(A^i)^{[p]}\,b(\det A,\nu)=\mu^{\frac12i(p-1)} a(A^i),\quad \eta(a(A^i))=\det A^i=\mu^i, \ \eta(b)=\nu. \end{align*} $$

We shall make use of the explicit matrices in $\mathord {\mathrm {GL}_{4}(q)}$ from Section 3.

10.1 $p\equiv 1\ \mod {4}$

By Lemma 11, in the case when $p\equiv 1\ \mod 4$ , the maximal abelian subgroups of $\operatorname {\mathrm {Out}}(D_3(q))\cong \operatorname {\mathrm {Out}}(\mathrm {PSL}_{4}{(q)})$ are $\left \langle \delta ,\phi \right \rangle $ , $\left \langle \delta ^2,\phi ,\tau \right \rangle $ and $\left \langle \delta ^2,\phi ,\tau \delta \right \rangle $ . We are therefore going through such cases.

Case $T=\left \langle \delta ,\phi \right \rangle $ . In the $\mathrm {PSL}_{4}{(q)}$ case, we took

$$ \begin{align*}L=\begin{pmatrix} 0 & 0 & 0 & -\lambda\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{pmatrix}, M=\begin{pmatrix} \lambda^{\frac{3(p-1)}{4}} & 0 & 0 & 0\\ 0 & \lambda^{\frac{2(p-1)}{4}} & 0 & 0\\ 0 & 0 & \lambda^{\frac{p-1}{4}} & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}, \end{align*} $$

$$ \begin{align*}\tilde T={\langle {L,\phi M}\rangle} Z(\mathord{\mathrm{GL}_{4}(q)})/Z(\mathord{\mathrm{GL}_{4}(q)}) \end{align*} $$

We have $ M^{-1}L^{[p]} M=\lambda ^{\frac {p-1}{4}}L $ , hence in $CO_6(q)^\circ $ , with $\ell =\sigma (L)$ , $m=\sigma (M)$ , by (10.2), (10.4),

$$ \begin{align*}m^{-1}\ell^{[p]} m= \lambda^{\frac{p-1}{2}}\ell, \end{align*} $$

$$ \begin{align*}\eta(\ell)=\det L=\lambda,\quad \eta(m)=\det M=\lambda^{\frac{3 (p-1)}{2}}. \end{align*} $$

We look for a, $b\in CO_4(q)^\circ $ satisfying the same relations using the above procedure. We take $\mu =\lambda $ , $\nu =\lambda ^{\frac {3 (p-1)}{2}}$ , $a=a(\lambda )$ , $b=b(\lambda ,\lambda ^{\frac {3 (p-1)}{2}})$ : if we put $A_1=a\oplus \cdots \oplus a \oplus \ell $ , $B_1 =b\oplus \cdots \oplus b\oplus m$ , then

$$ \begin{align*}A_1, B_1\in CO_{2n}(q)^\circ, B_1^{-1}A_1^{[p]}B_1= \lambda^{\frac{p-1}{2}} A_1 \end{align*} $$

and

$$ \begin{align*}\tilde T={\langle {A_1,\phi B_1}\rangle} Z(CO_{2n}(q)^\circ)/Z(CO_{2n}(q)^\circ) \end{align*} $$

is a T-abelian supplement.

Case $T=\left \langle \delta ^2,\phi ,\tau \right \rangle $ .

In the $\mathrm {PSL}_{4}{(q)}$ case for $\left \langle \delta ^2,\phi ,\gamma \right \rangle $ , we took

$$\begin{align*}L=\begin{pmatrix} 0 & -\lambda & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -\lambda\\ 0 & 0 & 1 & 0 \end{pmatrix}, M=\begin{pmatrix} \lambda^{\frac{p-1}{2}} & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & \lambda^{\frac{p-1}{2}} & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}, N=\begin{pmatrix} \lambda^{-1} & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & \lambda^{-1} & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}, \end{align*}$$

$$ \begin{align*}\tilde T=\left\langle L,\phi M,\gamma N\right\rangle Z/Z, \end{align*} $$

with

$$ \begin{align*}M^{-1}L^{[p]}M = z_1 L\ ,\ N^{-1}({}^t\!L^{-1})N = z_2 L\ , \ ({}^t\!M^{-1})N=z_3 N^{[p]} M, \end{align*} $$

$$ \begin{align*}z_1=\lambda^{\frac12(p-1)}, z_2= \lambda^{-1}, z_3=1, \det L=\lambda^2, \det M=\lambda^{p-1}, \det N=\lambda^{-2}. \end{align*} $$

Hence, in $CO_6(q)^\circ $ , with $\ell =\sigma (L)$ , $m=\sigma (M)$ , $n=\dot w_0\sigma (N)$ ,

$$ \begin{align*}m^{-1}\ell^{[p]}m = \lambda^{p-1} \ell\ , \ n^{-1}\ell^\tau n = \ell\ , \ m^\tau n= \lambda^{p-1} n^{[p]} m, \end{align*} $$

$$ \begin{align*}\eta(\ell)=\det L=\lambda^2, \eta(m)=\det M=\lambda^{p-1}, \eta(n)=\eta(\dot w_0)\eta(\sigma(N))=\det N=\lambda^{-2}. \end{align*} $$

Recall that $\tau _n$ acts trivially on U; hence, we have to define matrices a, b, $c\in CO_4(q)^\circ $ such that

$$ \begin{align*}b^{-1}a^{[p]}b = \lambda^{p-1}a\ , \ c^{-1}a c = a\ , \ b c= \lambda^{p-1} c^{[p]} b,\ \text {that is,} \quad b^{-1}c^{[p]} b=\lambda^{-(p-1)} c, \end{align*} $$

$$ \begin{align*}\eta(a)=\lambda^2, \eta(b)=\lambda^{p-1}, \eta(c)=\lambda^{-2}. \end{align*} $$

Once we have solved $b^{-1}a^{[p]}b = \lambda ^{p-1}a$ , we may take $c=a^{-1}$ . We take

$$ \begin{align*}a=a(\lambda^2), b=b(\lambda^2,\lambda^{p-1}), c=a^{-1}. \end{align*} $$

If we put $A_1=a\oplus \cdots \oplus a \oplus \ell $ , $B_1 =b\oplus \cdots \oplus b\oplus m$ , $C_1=c\oplus \cdots \oplus c\oplus n$ , then $ A_1, B_1, C_1\in CO_{2n}(q)^\circ , $ with

$$ \begin{align*}B_1^{-1}A_1^{[p]}B_1 = \lambda^{p-1} A_1 \ ,\ C_1^{-1}A_1^\tau C_1 = A_1 \ , \ B_1^\tau C_1= \lambda^{p-1} C_1^{[p]} B_1, \end{align*} $$

$$ \begin{align*}\eta(A_1)=\lambda^2, \eta(B_1)=\lambda^{p-1}, \eta(C_1)=\lambda^{-2}, \end{align*} $$

so that

$$ \begin{align*}\tilde T={\langle {A_1,\phi B_1,\tau C_1}\rangle} Z/Z \end{align*} $$

is a T-abelian supplement.

Case $T=\left \langle \delta ^2,\phi ,\tau \delta \right \rangle $ . In the $\mathrm {PSL}_{4}{(q)}$ case for $\left \langle \delta ^2,\phi ,\gamma \delta \right \rangle $ , we took

$$\begin{align*}L=\begin{pmatrix} 0 & -\lambda & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -\lambda\\ 0 & 0 & 1 & 0 \end{pmatrix}, M=\begin{pmatrix} \lambda^{\frac{p-1}{2}} & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & \lambda^{\frac{p-1}{2}} & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & -\lambda & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}^{\frac{1-p}{2}},N= \begin{pmatrix} 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & \lambda^{-1} & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}, \end{align*}$$

$$ \begin{align*}\tilde T=\left\langle L,\phi M,\gamma N\right\rangle Z/Z, \end{align*} $$

with

$$ \begin{align*}M^{-1}L^{[p]}M = z_1 L\ ,\ N^{-1}({}^t\!L^{-1})N = z_2 L\ , \ ({}^t\!M^{-1})N=z_3 N^{[p]} M, \end{align*} $$

$$ \begin{align*}z_1=\lambda^{\frac12(p-1)}, z_2= \lambda^{-1}, z_3=1, \det L=\lambda^2, \det M=\lambda^{\frac12(p-1)}, \det N=\lambda^{-1}. \end{align*} $$

Hence, in $CO_6(q)^\circ $ , with $\ell =\sigma (L)$ , $m=\sigma (M)$ , $n=\dot w_0\sigma (N)$ ,

$$ \begin{align*}m^{-1}\ell^{[p]}m = \lambda^{p-1} \ell \ , \ n^{-1}\ell^\tau n = \ell \ , \ m^\tau n= \lambda^{\frac12(p-1)} n^{[p]} m, \end{align*} $$

$$ \begin{align*}\eta(\ell)=\det L=\lambda^2, \eta(m)=\det M=\lambda^{\frac12(p-1)}, \eta(n)=\eta(\dot w_0)\eta(\sigma(N))=\det N=\lambda^{-1}. \end{align*} $$

We have to define matrices a, b, $c\in CO_4(q)^\circ $ such that

$$ \begin{align*}b^{-1}a^{[p]}b = \lambda^{p-1}a \ , \ c^{-1}a c = a \ , \ b c= \lambda^{\frac12(p-1)} c^{[p]} b,\ \text {that is,} \quad b^{-1}c^{[p]} b=\lambda^{-\frac12(p-1)} c, \end{align*} $$

$$ \begin{align*}\eta(a)=\lambda^2, \eta(b)=\lambda^{\frac12(p-1)}, \eta(c)=\lambda^{-1}. \end{align*} $$

Once we have solved $b^{-1}c^{[p]}b = \lambda ^{-\frac 12(p-1)}c$ , we may take $a=c^{-2}$ . We take

$$ \begin{align*}c=a(\lambda^{-1}), b=b(\lambda^{-1},\lambda^{\frac12(p-1)}), a=c^{-2}. \end{align*} $$

If we put $A_1=a\oplus \cdots \oplus a \oplus \ell $ , $B_1 =b\oplus \cdots \oplus b\oplus m$ , $C_1=c\oplus \cdots \oplus c\oplus n$ , then $ A_1, B_1, C_1\in CO_{2n}(q)^\circ , $ with

$$ \begin{align*}B_1^{-1}A_1^{[p]}B_1 = \lambda^{p-1} A_1 \ , \ C_1^{-1}A_1^\tau C_1 = A_1 \ , \ B_1^\tau C_1= \lambda^{\frac12(p-1)} C_1^{[p]} B_1, \end{align*} $$

$$ \begin{align*}\eta(A_1)=\lambda^2, \eta(B_1)=\lambda^{\frac12(p-1)}, \eta(C_1)=\lambda^{-1}, \end{align*} $$

so that

$$ \begin{align*}\tilde T={\langle {A_1,\phi B_1,\tau C_1}\rangle} Z/Z \end{align*} $$

is a T-abelian supplement.

10.2 $p\equiv -1\ \mod {4}$

By Lemma 11, in the case when $p\equiv -1\ \mod 4$ , the maximal abelian subgroups of $\operatorname {\mathrm {Out}}(D_3(q))\cong \operatorname {\mathrm {Out}}(\mathrm {PSL}_{4}{(q)})$ are $\left \langle \delta ,\phi \tau \right \rangle $ , $\left \langle \delta ^2,\phi ,\tau \right \rangle $ and $\left \langle \delta ^2,\phi \delta ,\tau \delta \right \rangle $ . We are therefore going through such cases.

Case $T=\left \langle \delta ,\phi \tau \right \rangle $ . In the $\mathrm {PSL}_{4}{(q)}$ case for $\left \langle \delta ,\phi \gamma \right \rangle $ , we took

$$\begin{align*}L=\begin{pmatrix} 0 & 0 & 0 & -\lambda\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{pmatrix}, M= \begin{pmatrix} \lambda^{\frac{3(-p-1)}{4}} & 0 & 0 & 0\\ 0 & \lambda^{\frac{2(-p-1)}{4}} & 0 & 0\\ 0 & 0 & \lambda^{\frac{-p-1}{4}} & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}, \end{align*}$$

$$ \begin{align*}\tilde T={\langle {L,\phi \gamma M}\rangle} Z/Z, \end{align*} $$

with

$$ \begin{align*}M^{-1}({}^t\!(L^{[p]})^{-1}) M=\lambda^{-\frac{p+1}{4}}L ,\quad \det L=\lambda, \det M=\lambda^{-\frac32(p+1)}. \end{align*} $$

Hence, in $CO_6(q)^\circ $ , with $\ell =\sigma (L)$ , $m=\dot w_0\sigma (M)$ ,

$$ \begin{align*}m^{-1}(\ell^{[p]})^\tau m = \lambda^{\frac12(p-1)} \ell, \end{align*} $$

$$ \begin{align*}\eta(\ell)=\det L=\lambda, \eta(m)=\eta(\dot w_0)\eta(\sigma(M))=\det M=\lambda^{-\frac32(p+1)}. \end{align*} $$

We have to define matrices a, $b\in CO_4(q)^\circ $ such that

$$ \begin{align*}b^{-1}a^{[p]}b = \lambda^{\frac12(p-1)}a ,\quad \eta(a)=\lambda, \eta(b)=\lambda^{-\frac32(p+1)}. \end{align*} $$

We take

$$ \begin{align*}a=a(\lambda), b=b(\lambda,\lambda^{-\frac32(p+1)}). \end{align*} $$

If we put $A_1=a\oplus \cdots \oplus a \oplus \ell $ , $B_1 =b\oplus \cdots \oplus b\oplus m$ , then $ A_1, B_1\in CO_{2n}(q)^\circ , $ with

$$ \begin{align*}B_1^{-1}(A_1^{[p]})^\tau B_1 = \lambda^{\frac12(p-1)} A_1 ,\quad \eta(A_1)=\lambda, \eta(B_1)=\lambda^{-\frac32(p+1)}, \end{align*} $$

so that

$$ \begin{align*}\tilde T={\langle {A_1,\phi \tau B_1}\rangle} Z/Z \end{align*} $$

is a T-abelian supplement.

Case $T=\left \langle \delta ^2,\phi ,\tau \right \rangle $ . In the $\mathrm {PSL}_{4}{(q)}$ case for $\left \langle \delta ^2,\phi , \gamma \right \rangle $ , we took

$$ \begin{align*}\tilde T= {\langle {L,\phi M, \gamma N}\rangle} Z/Z, \end{align*} $$

with the same L, M, N as in the case $p\equiv 1\ \mod 4$ , $T=\left \langle \delta ^2,\phi , \gamma \right \rangle $ . We define $A_1, B_1, C_1\in CO_{2n}(q)^\circ $ as in this case, and $\tilde T={\langle {A_1,\phi B_1,\tau C_1}\rangle } Z/Z$ is a T-abelian supplement.

Case $T=\left \langle \delta ^2,\phi \delta ,\tau \delta \right \rangle $ . In the $\mathrm {PSL}_{4}{(q)}$ case for $\left \langle \delta ^2,\phi \delta , \gamma \delta \right \rangle $ , we took

$$ \begin{align*}\tilde T= {\langle {L,\phi M, \gamma N}\rangle} Z/Z, \end{align*} $$

with the same L, M, N as in the case $p\equiv 1\ \mod 4$ , $T=\left \langle \delta ^2,\phi , \gamma \delta \right \rangle $ . Again, we define $A_1, B_1, C_1\in CO_{2n}(q)^\circ $ in the same way, and $\tilde T={\langle {A_1,\phi B_1,\tau C_1}\rangle } Z/Z$ is a T-abelian supplement.

We have proved the following.

Theorem 25. Let G be an almost simple group with socle $G_0=D_n(q)$ , n odd. If $G/G_0$ is abelian, then there exists an abelian subgroup A such that $G=AG_0$ .

11 $D_n(q)$ , n even

Here, L is of type $D_n$ , n even, $G_0=D_n(q)$ , $q=p^m$ , $d=(2 ,q-1)^2$ , and we assume that $\operatorname {\mathrm {Aut}}(G_0)$ does not split over $G_0$ ; hence, $( \frac {q^n-1}d,d,m)\not =1$ . In particular, $d\not =1$ ; hence, p is odd, m is even and $d=4$ , $\hat H/H\cong C_2\times C_2$ .

If $n=4,$ then

$$ \begin{align*}\operatorname{\mathrm{Out}}(G_0)=({\langle {\delta_1, \delta_2, \delta_3}\rangle} \times {\langle \phi\rangle}) :\ S_3, \end{align*} $$

where $S_3={\langle {\rho , \tau }\rangle }$ , $\tau ^2=1$ , $\rho ^3=1$ , $\delta _1\delta _2=\delta _3$ , $\delta _i^2=\phi ^m=[\rho , \phi ]=[\tau ,\phi ]=1$ , $\delta _1^\tau =\delta _2$ , $\delta _3^\tau =\delta _3$ , $ \delta _1^\rho =\delta _2$ , $\delta _2^\rho =\delta _3$ , $\delta _3^\rho =\delta _1$ .

If $n\not =4,$ then

$$ \begin{align*}\operatorname{\mathrm{Out}}(G_0)=({\langle {\delta_1, \delta_2, \delta_3}\rangle} \times {\langle \phi\rangle} ) :{\langle \tau\rangle}, \end{align*} $$

where $\tau ^2=1$ , $\delta _1\delta _2=\delta _3$ , $\delta _i^2=\phi ^m=[\tau ,\phi ]=1$ , $\delta _1^\tau =\delta _2$ , $\delta _3^\tau =\delta _3$ .

Note that $(\tau \delta _1)^2=\tau \delta _1\tau \delta _1=\delta _2\delta _1=\delta _3$ , $(\tau \delta _2)^2=\delta _3$ ; hence,

$$ \begin{align*}{\langle {\delta_3,\phi,\tau\delta_1}\rangle} = {\langle {\phi,\tau\delta_1}\rangle} ={\langle {\phi,\tau\delta_2}\rangle}. \end{align*} $$

We have to consider the following cases. Assume T is an abelian, noncyclic subgroup of $ {\langle {\delta _1,\delta _2,\phi ,\tau }\rangle }$ (which is $\operatorname {\mathrm {Out}}(G_0)$ if $n\not =4$ ).

Let $D={\langle {\delta _1,\delta _2}\rangle }$ , $\pi :\operatorname {\mathrm {Out}}(G_0)\to \operatorname {\mathrm {Out}}(G_0)/D={\langle {\phi ,\tau }\rangle }$ . If $\pi (T)$ is cyclic, then $ \pi (T)={\langle {\phi ^s\tau ^\epsilon }\rangle } $ . If $\varepsilon =0$ , $T\leqslant {\langle {D,\phi ^s}\rangle }$ , so

$$ \begin{align*}T\leqslant {\langle {\delta_1,\delta_2,\phi}\rangle}. \end{align*} $$

If $\varepsilon =1$ , $T\leqslant {\langle {D,\phi ^s\tau }\rangle }$ , and T contains an element $\alpha =\phi ^s\tau \delta $ , $\delta \in D$ , $\delta \not =1$ , so either $T={\langle {\delta _3,\phi ^s\tau }\rangle }$ or $T={\langle {\delta _3,\phi ^s\tau \delta _1}\rangle } ={\langle {\delta _3,\phi ^s\tau \delta _2}\rangle }$ . In the first case,

$$ \begin{align*}T\leqslant {\langle {\delta_3,\phi,\tau}\rangle}. \end{align*} $$

In the second case,

$$ \begin{align*}T\leqslant {\langle {\phi,\tau\delta_1}\rangle} ={\langle {\phi,\tau\delta_2}\rangle}. \end{align*} $$

If $\pi (T)$ is not cyclic, then $\pi (T)={\langle {\phi ^s,\tau }\rangle }$ . Therefore, either

$$ \begin{align*}T\leqslant {\langle {\delta_3,\phi,\tau}\rangle} \end{align*} $$

or

$$ \begin{align*}T\leqslant {\langle {\delta_3,\phi,\tau\delta_1}\rangle} ={\langle {\phi,\tau\delta_1}\rangle} ={\langle {\phi,\tau\delta_2}\rangle}. \end{align*} $$

Therefore, if $T\leqslant {\langle {\delta _1,\delta _2,\phi ,\tau }\rangle }$ , by Lemma 5, we only have to deal with the following cases:

$$ \begin{align*} \begin{aligned} \text{case 1:}&\quad T= {\langle {\delta_1,\delta_2,\phi}\rangle} ,\\ \text{case 2:}&\quad T= {\langle {\delta_3,\phi,\tau}\rangle} ,\\ \text{case 3:}&\quad T= {\langle {\phi,\tau\delta_1}\rangle} ={\langle {\phi,\tau\delta_2}\rangle}. \end{aligned} \end{align*} $$

Assume $n=4$ . Let $M={\langle {\delta _1,\delta _2,\phi }\rangle } $ , $\zeta :\operatorname {\mathrm {Out}}(G_0)\to \operatorname {\mathrm {Out}}(G_0)/M={\langle {\rho ,\tau }\rangle }$ , and T a noncyclic abelian subgroup of $\operatorname {\mathrm {Out}}(G_0)$ not contained in ${\langle {\phi ,\rho ,\tau }\rangle }$ . Hence, $\zeta (T)=\{1\}$ , ${\langle {\rho ^i\tau }\rangle }$ or ${\langle \rho \rangle }$ . However, $\rho ^i\tau $ is conjugate to $\tau $ ; therefore, we may assume $\zeta (T)=\{1\}$ , ${\langle {\tau }\rangle }$ or ${\langle \rho \rangle }$ .

If $\zeta (T)=\{1\}$ or ${\langle \tau \rangle }$ , we are in the previous case $T\leqslant {\langle {\delta _1,\delta _2,\phi ,\tau }\rangle }$ . We are left with $\zeta (T)={\langle \rho \rangle } $ , $T\leqslant {\langle {\delta _1,\delta _2,\phi ,\rho }\rangle }$ , so $T={\langle {\phi ^s,\phi ^t\rho \delta }\rangle }$ , $\delta \in D$ , $\delta \not =1$ since T is abelian and not contained in ${\langle {\phi ,\rho ,\tau }\rangle }$ . It follows that $T\leqslant {\langle {\phi ,\rho \delta }\rangle }$ . Moreover, since ${\langle \rho \rangle } $ acts transitively on $\{\delta _1,\delta _2,\delta _3\}$ and $[\rho ,\phi ]=1$ , we may assume

$$ \begin{align*}\text{case 4:}\quad T= {\langle {\phi,\rho\delta_2}\rangle} \quad\quad\text{only for } D_4. \end{align*} $$

We use the same procedure used to deal with the odd n case. It is convenient to start with $G_0=D_2(q)=P\Omega _4^+(q)\cong \mathrm {PSL}_{2}{(q)} \times \mathrm {PSL}_{2}{(q)}$ .

We have

$$ \begin{align*}n_1=\left( \begin{smallmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ \end{smallmatrix} \right), n_2= \left( \begin{smallmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ \end{smallmatrix} \right) \text { in } \Omega^+_4(q).\end{align*} $$

Note that $n_2=\tau _2n_1\tau _2$ and $n_1n_2=n_2n_1$ . If

$$ \begin{align*}g= \left( \begin{smallmatrix} f_1& 0 & 0 & 0 \\ 0 & f_2 & 0 & 0 \\ 0 & 0 & \frac{\mu} {f_1}& 0 \\ 0 & 0 & 0 & \frac{\mu}{f_2} \\ \end{smallmatrix} \right) \end{align*} $$

is a diagonal matrix in $CO_4(q)^\circ $ , then $\alpha _1(g)=\frac {f_1}{f_2}$ , $\alpha _2(g)=\frac {f_1f_2}{\mu }$ . We define $\delta _1$ , $\delta _2$ , $\delta _3$ . Let

$$ \begin{align*}h_1= \left( \begin{smallmatrix} 1 & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & 1 \\ \end{smallmatrix} \right) \end{align*} $$

in $CO_4(q)^\circ $ . Then $\alpha _1(h_1)=\lambda ^{-1}$ , $\alpha _2(h_1)=1$ . We write for short $h_1\mapsto h(\chi _1)\in \hat H$ , where $\chi _1=(\lambda ^{-1},1)$ is the $\mathbb {F}_q$ -character of Q with $\chi _1(\alpha _1)=\lambda ^{-1}$ , $\chi _1(\alpha _2)=1$ . We define $\delta _1:=h(\chi _1)G_0$ . Moreover, $\chi _2:=\chi _1\circ \tau =(1,\lambda ^{-1})$ , $h_2:=h_1^\tau $ , $h_2\mapsto h(\chi _2)\in \hat H$ , $\delta _2:=h(\chi _2)G_0$ ; finally, $h_3:=h_1h_2$ , and hence, $ h_3\mapsto h(\chi _3)$ , $\chi _3=\chi _1+\chi _2=(\lambda ^{-1},\lambda ^{-1})$ , $\delta _3:=h(\chi _3)G_0$ , so $\delta _3=\delta _1\delta _2$ .