Proof. Assume $x\in U={\mathbf {O}}_p(B)$ can be written as a product of root elements not involving elements from $U_\alpha $ where $\alpha \in \Delta $ . Let $s\in W$ be the corresponding simple reflection and $\dot s\in {\mathbf {N}}_G({\mathbf {T}})$ a preimage. Then $x^{\dot s}\in B$ , that is, x lies in the Borel subgroups B and $B^{\dot s}$ which are distinct, as $B^{\dot s}$ does not contain root elements for the root $\alpha $ . Hence x lies in two different Sylow p-subgroups of G and thus is not picky.

For the converse, assume x lies in two Sylow p-subgroups, so (up to conjugation) $x\in B$ and $x\in B^g$ for some $g\in G$ . Writing g in the Bruhat decomposition $g=u_1t\dot {w}u_2$ with $u_i\in U$ , $t\in T$ and $w\in W$ we find $x\in (B\cap B^{w})^{u_2}$ , so up to conjugation $x\in B\cap B^{w}$ for some $1\ne w\in W$ . Now by [Reference Carter4, Prop. 2.5.9] then $x\in U_{w_0w}$ , with $w_0\in W$ the longest element, and $U_{w_0w}$ is the product of the root subgroups $U_\alpha $ for $\alpha \in \Phi ^+\cap w(\Phi ^+)$ by [Reference Carter4, Prop. 2.5.16]. Since $w\ne 1$ there is some simple root $\alpha \in \Delta $ made negative by w and so the corresponding root element cannot occur in the unique expression of x as a product of root elements.

Proof. By [Reference Carter4, Prop. 5.1.3], a unipotent element $x\in {\mathbf {G}}$ is regular if and only if it lies in a unique Borel subgroup of ${\mathbf {G}}$ . Assume $x\in G$ lies in two distinct Sylow p-subgroups, so in two Borel subgroups $B_1\ne B_2$ of G, say $B_i={\mathbf {B}}_i^F$ for F-stable Borel subgroups ${\mathbf {B}}_i$ of ${\mathbf {G}}$ , $i=1,2$ . Then x lies in ${\mathbf {B}}_1\ne {\mathbf {B}}_2$ and hence is not regular unipotent. So regular unipotent elements are picky.

For the converse, first assume G is untwisted, so F acts trivially on the Weyl group of ${\mathbf {G}}$ and hence all root subgroups of ${\mathbf {G}}$ in ${\mathbf {B}}$ are F-stable. Let $x\in U$ be not regular unipotent. Then, again by [Reference Carter4, Prop. 5.1.3], it can be written as a product of root elements in which at least one simple root $\alpha $ does not occur, so it cannot be picky by Proposition 3.1.

Now assume F acts nontrivially on W and hence on $\Delta $ . Let $x\in U$ be picky. Then by Proposition 3.1 it is a product involving root elements from all root subgroups for simple roots $\alpha \in \Delta $ . Now by [Reference Malle and Testerman27, Ex. 23.10], for example, the root subgroup $U_\alpha $ consists of elements which are products of root elements of ${\mathbf {G}}$ lying in an F-orbit of simple roots of ${\mathbf {G}}$ , unless $\alpha $ is the image under the orbit map of a pair of roots of ${\mathbf {G}}$ forming a diagram of type $A_2,B_2$ or $G_2$ . Thus, if we are not in one of the latter cases, x is a product of root elements of ${\mathbf {G}}$ involving all simple roots and hence is regular.

It remains to discuss the cases for which the Dynkin diagram has a subgraph of type $A_2,B_2$ or $G_2$ with nontrivial F-action. Thus G is one of ${\operatorname {SU}}_{2n+1}(q)$ , ${}^{2}\!B_2(q^2)$ , $^2G_2(q^2)$ or ${}^{2}\!F_4(q^2)$ . The rank 1 groups ${\operatorname {SU}}_3(q)$ , ${}^{2}\!B_2(q^2)$ and $^2G_2(q^2)$ have TI Sylow p-subgroups [Reference Berger, Landrock and Michler1, Prop. 2.3], so any of their nontrivial p-elements is picky. For $G={}^{2}\!F_4(q^2)$ the list of unipotent class representatives in [Reference Shinoda33] reveals that in addition to the regular classes, there are two further cases, $u_{13},u_{14}$ , involving root elements for both types of simple roots, with centraliser as given in the statement, while all other elements cannot be picky, by Proposition 3.1. (Note that here the elements from the two simple root subgroups are denoted $\alpha _1(t_1)\alpha _2(t_2)$ and $\alpha _3(t)$ respectively.) Finally, for $G={\operatorname {SU}}_{2n+1}(q)$ it remains to identify picky elements x that involve elements from all simple root subgroups of G, but not from all simple root subgroups of ${\mathbf {G}}$ . Thus, expressed as a product of root elements for ${\mathbf {G}}$ , x involves root elements from ${\mathbf {U}}_\alpha $ for all $\alpha \in \{\alpha _1,\ldots ,\alpha _{n-1},\alpha _n+\alpha _{n+1},\alpha _{n+2},\ldots ,\alpha _{2n}\}$ , if the simple roots $\alpha _1,\ldots ,\alpha _{2n}$ of ${\mathbf {G}}$ are numbered along the Dynkin diagram of ${\mathbf {G}}$ , and hence has Jordan block sizes $(2n,1)$ in the natural representation.

Proof. Since x is subnormal in $P(x)$ by definition, it suffices to see that any $H\le G$ such that $\langle x\rangle {\,\triangleleft \triangleleft \,} H$ lies in $P(x)$ . Let H be such a subgroup. In particular $x\in {\mathbf {O}}_p(H)$ by Lemma 2.5. By the Borel–Tits theorem [Reference Malle and Testerman27, Thm 26.5] there is a parabolic subgroup $H'\ge H$ such that $x\in {\mathbf {O}}_p(H')$ , so we may assume H is parabolic. Furthermore, we may replace H by any overgroup $H'$ with $x\in {\mathbf {O}}_p(H')$ . As any such overgroup is again parabolic, we may assume H is parabolic and maximal with respect to $x\in {\mathbf {O}}_p(H)$ .

Let $g\in G$ such that $H^g$ is a standard parabolic subgroup, say $H^g=P_\Gamma $ for $\Gamma \subseteq \Delta $ . Then $x^g\in C\cap {\mathbf {O}}_p(P_\Gamma )\subseteq C\cap U$ and hence $P_\Gamma \le P(x)$ by assumption. In fact, by our choice of H we have $P_\Gamma =P(x^g)$ . Using the Bruhat decomposition we can write $g=u\dot {w}b$ with $u\in U$ , $b\in B\le P_\Gamma $ , $w\in W$ , so $H^{uw}=P_\Gamma $ and $x^{uw}\in {\mathbf {O}}_p(P_\Gamma )$ . Since also $y:=x^u\in {\mathbf {O}}_p(P_\Gamma )$ we have $y,y^w\in {\mathbf {O}}_p(P_\Gamma )$ . Let $W_\Gamma \le W$ denote the Weyl group of the standard Levi subgroup of $P_\Gamma $ . Any $w\notin W_\Gamma $ sends at least one simple root in $\Delta \setminus \Gamma $ into a negative root. On the other hand, since $P_\Gamma =P(x^g)$ all simple roots from $\Delta \setminus \Gamma $ occur in y, whence we can’t have ${}^{w}\!y\in P_\Gamma $ . Thus, $w\in W_\Gamma $ , giving $g\in P_\Gamma $ and so $H={}^{g}\!P_\Gamma =P_\Gamma \le P(x)$ . This proves our claim.

Proof. By Proposition 2.6, ${\operatorname {Sub}}_G(x)$ contains a Sylow p-normaliser of G, hence a Borel subgroup. Since all overgroups of B are parabolic subgroups, so is ${\operatorname {Sub}}_G(x)$ .

Proof. Let $\alpha \in \Delta _1$ be a simple root and $P_\alpha :=P_{\{\alpha \}}$ the corresponding standard parabolic subgroup of G. By assumption x lies in the unipotent radical of $P_\alpha $ , so $\langle x\rangle $ is subnormal in $P_\alpha $ . The claim follows since $\langle P_\alpha \mid \alpha \in \Delta _1\rangle =P_{\Delta _1}$ (see [Reference Malle and Testerman27, Prop. 12.2]). The final assertion is now obvious.

Proof. For regular elements this is Theorem 3.2. For nonregular elements, we first consider $G={\operatorname {SL}}_n(q)$ . For $n=2$ the only nonregular unipotent element is the identity, for $n=3$ the nontrivial nonregular unipotent elements have Jordan type $(2,1)$ and thus are conjugate to a root element for the nonsimple positive root, so ${\operatorname {Sub}}_G(x)=G$ by Proposition 3.5. Now assume $n\ge 4$ . Let $x\in G$ be a nonregular unipotent element. Then x has at least two Jordan blocks, so up to conjugation, x can be written as a product of simple root elements in which at least one simple root $\alpha $ is missing. If this $\alpha $ lies at an end of the Dynkin diagram, then $x^{s_\alpha }$ is a conjugate in B in which the simple root next to the end node is missing. So we may assume $\alpha \ne \alpha _1,\alpha _{n-1}$ . Let $P_1,P_2$ be the end-node standard parabolic subgroups of G, corresponding to $\Delta \setminus \{\alpha _1\}$ , $\Delta \setminus \{\alpha _{n-1}\}$ respectively. Let $L_i$ be the corresponding standard Levi subgroups, so $L_i\cong P_i/U_i$ , with $U_i={\mathbf {O}}_p(P_i)$ the unipotent radical of $P_i$ . Then for $i\in \{1,2\}$ , the image $\bar x$ of x in $L_i$ is a product of root elements not involving the simple root $\alpha $ of $L_i$ , hence not regular, so by induction ${\operatorname {Sub}}_{L_i}(\bar x)=L_i$ . (Note that $L_i\cong {\operatorname {GL}}_{n-1}(q)$ , containing ${\operatorname {SL}}_{n-1}(q)$ as a normal subgroup of $p'$ -index.) As $U_i$ is a p-group, ${\operatorname {Sub}}_{P_i}(x)=P_i$ for $i=1,2$ by Lemma 2.15. Since $\langle P_1,P_2\rangle =G$ by [Reference Malle and Testerman27, Prop. 12.2] this achieves the proof.

Now assume G is of one of the other listed types. Let $x\in G$ be nonregular unipotent. By [Reference Carter4, Prop. 5.1.3], up to conjugation, x can be written as a product of root elements in which at least one simple root $\alpha $ is missing. Since the Dynkin diagram of G has three end nodes, there are at least two end node simple roots $\alpha _1,\alpha _2\in \Delta $ such that $\alpha \in \Delta \setminus \{\alpha _i\}$ , $i\in \{1,2\}$ . Let $P_i$ be the corresponding standard parabolic subgroups of G. Since their Levi subgroups are of type $D_{n-1}$ , $E_{n-1}$ or $A_3$ , for which we know the result by induction, respectively by the first part, we can argue exactly as before to see that ${\operatorname {Sub}}_{P_i}(x)=P_i$ for $i=1,2$ and then ${\operatorname {Sub}}_G(x)\ge \langle P_1,P_2\rangle =G$ .

Proof. For all of these groups we know parametrisations of unipotent classes in parabolic subgroups and expressions for class representatives in terms of root elements. The picky elements have subnormaliser conjugate to ${\mathbf {N}}_G(U)=B$ , by Corollary 2.7. We consider the remaining classes.

Let us start with $G=B_2(q)$ . Here representatives for the unipotent conjugacy classes are given in [Reference Enomoto9, Reference Shinoda34]. With the criterion in Proposition 3.5 we see that for $p=2$ there cannot be a class (apart from the regular ones) with ${\operatorname {Sub}}_G(x)<G$ , and for $p>2$ only the class with representative denoted $A_{22}$ in [Reference Shinoda34] could possibly have that property. Representatives for the unipotent conjugacy classes of $G_2(q)$ are given in [Reference Chang and Ree6, Reference Enomoto10, Reference Enomoto and Yamada11], and again by Proposition 3.5 it follows that only the class of $G_2(2^f)$ denoted $A_4$ in [Reference Enomoto and Yamada11] and the class of $G_2(p^f)$ , $p\ge 5$ , with representative $u_4$ in [Reference Chang and Ree6] could have ${\operatorname {Sub}}_G(x)<G$ . For $G={}^{3}\!D_4(q)$ , consulting the tables of class representatives in [Reference Geck12, Reference Himstedt16], we see that we only need to consider the classes with representatives $u_\beta (1)u_{2\alpha +\beta }(a)$ and $u_\alpha (1)u_{\alpha +\beta }(1)$ . Now note that conjugating $u_\beta (1)u_{2\alpha +\beta }(a)$ with the simple reflection $s_\alpha $ gives an element not involving any simple root element. So we are left with the class $D_4(a_1)$ with representative $u_\alpha (1)u_{\alpha +\beta }(1)$ . Finally, for $G={}^{2}\!F_4(q^2)$ by [Reference Shinoda33, Tab. II] apart from the picky cases identified in Theorem 3.2 only the representatives $u_9,u_{10},u_{11},u_{12}$ involve one of the simple root elements.

It remains to determine the subnormaliser for the elements in (2)–(5). By the tables in the cited literature, respectively in [Reference Himstedt15, Reference Himstedt16] for ${}^{3}\!D_4(q)$ and [Reference Himstedt and Huang18] for ${}^{2}\!F_4(q^2)$ , in all four groups we have the following situation: exactly one of the unipotent radicals of the two maximal standard parabolic subgroups contains an element x as considered. Moreover, the centraliser of x in that parabolic subgroup is the same as in G. By counting, in fact each such x lies inside exactly one such maximal parabolic subgroup, say P. Since any parabolic subgroup contains a Borel subgroup, hence a Sylow normaliser, we conclude by Lemma 2.14 that ${\operatorname {Sub}}_G(x)=P$ .

Proof. Let P be the standard maximal parabolic subgroup of G of type $A_{n-1}$ , that is, the stabiliser of a maximal isotropic subspace in the natural representation, and Q the standard maximal parabolic subgroup of type $B_{n-1}$ . We claim that if (up to conjugation) ${\operatorname {Sub}}_P(x)=P$ then ${\operatorname {Sub}}_G(x)=G$ . Indeed, assume ${\operatorname {Sub}}_P(x)=P$ and let $\bar x$ be the image of x in the standard Levi subgroup $L\cong {\operatorname {GL}}_n(q)$ of P. By Proposition 3.6, $\bar x$ is not regular, so (up to conjugation) its image y in the Levi factor of a standard parabolic subgroup $L_1$ of L of type ${\operatorname {GL}}_{n-1}(q)$ is not regular either. Now, $L_1$ is a standard Levi subgroup of the standard Levi subgroup M of Q of type ${\operatorname {SO}}_{2n-1}(q)$ , and hence by induction ${\operatorname {Sub}}_M(y)=M$ . (The induction base is given by the case $2n-1=5$ from Proposition 3.7.) But then ${\operatorname {Sub}}_Q(x)=Q$ , and since P is maximal in G and Q is distinct from P, we have ${\operatorname {Sub}}_G(x)=G$ .

Now from the explicit class representatives given in [Reference De Franceschi, Liebeck and O’Brien7, §4.1.2] it can be checked that unless x is regular or has type $V_\beta (2n-1)\oplus V_\beta (1)\oplus V(1)$ (in the notation of loc. cit.), there is a conjugate $y\in P$ of x whose image in L is not regular, so has ${\operatorname {Sub}}_P(y)=P$ and then ${\operatorname {Sub}}_G(y)=G$ by the first part. Finally assume x has type $V_\beta (2n-1)\oplus V_\beta (1)\oplus V(1)$ . Then x has Jordan type $(2n-1,1^2)$ . In particular, writing $x\in U$ as a product of root elements, all $u_\alpha $ , $\alpha \in \Delta \setminus \{\alpha _n\}$ , must be nonzero. But then the only standard parabolic subgroups of G that contain x in their radical are B and $P_n$ (of type $B_1$ ). The claim now follows with Lemma 2.14.

Proof. Representatives for the unipotent conjugacy classes of G as well as for a Levi subgroup L of G of type $B_3$ were determined in [Reference Shinoda32, Reference Shoji35]. First assume $p=2$ . We claim that only the regular unipotent elements in L have subnormaliser strictly smaller than L. Indeed, seven of the class representatives given in [Reference Shinoda32, Prop. 2.1] do not involve any simple root element, and the other three nonregular ones can be conjugated by a simple reflection to elements x which only involve the 3rd simple root of L. Then x has trivial image in the standard Levi subgroup of L of type $B_2$ , and image involving just one simple root in the standard Levi subgroup of type $A_2$ , so ${\operatorname {Sub}}_L(x)=L$ by Proposition 3.5.

Now we discuss the unipotent class representatives of G from [Reference Shinoda32, Thm 2.1]. The ones labelled $x_0$ through $x_{19}$ do not involve any simple root element, and similarly for $x_{27},x_{28}$ . For elements ${x\in \{x_{22},\ldots ,x_{26}\}}$ only one of the short simple roots occurs, so the images of x in standard Levi subgroups of type $A_2$ and $B_3$ are nonregular and we conclude by Proposition 3.5 and the previous paragraph. Similarly for $x=x_{20},x_{21}$ the image in a Levi of type $C_3$ is trivial, and in a Levi of type $B_3$ is nonregular. By Lemma 3.8 this leaves us with the elements $x_{29}$ and $x_{30}$ , of type $F_4(a_1)$ . The given representatives show that these elements are regular in the subsystem subgroup of type $B_4$ . Conjugating $x=x_{29}$ or $x_{30}$ with the simple reflection not belonging to the $B_3$ -subsystem, we obtain an element which only involves simple root elements for the 2nd and 3rd simple root. By the first paragraph, the subnormaliser of that element contains the standard parabolic subgroup of type $B_3$ , but by Proposition 3.6 also the one of type $A_2$ , hence it is all of G.

If p is odd, Proposition 3.7 shows that apart from the regular classes, only the unipotent class of L with representative $z_8$ (in the notation of [Reference Shoji35, Tab. 3]) has proper subnormaliser. From [Reference Shoji35, Tab. 5 and 6] we conclude that again at most the classes of G of type $F_4(a_1)$ with representatives $x_{23},x_{24}$ might have a proper subnormaliser. Conjugating $x=x_{23},x_{24}$ by the simple reflection not in the $B_3$ -subsystem we obtain elements whose image in L equals $z_7,z_8$ respectively. As observed above, ${\operatorname {Sub}}_L(z_7)=L$ and so ${\operatorname {Sub}}_G(x_{23})=G$ by the usual argument. On the other hand, we have ${\operatorname {Sub}}_L(z_8)$ is the standard parabolic subgroup corresponding to the 3rd node of the diagram of $B_3$ , which is also the 3rd node of the diagram of $F_4$ . Thus the set of standard parabolic subgroups of G containing $x_{24}$ in their radical has the unique maximal element $P_{\{3,4\}}$ , and so ${\operatorname {Sub}}_G(x_{24})=P_{\{3,4\}}$ by Lemma 2.14.

Proof. There is nothing to show for $m=1$ , so assume $m\ge 2$ . Let $x\in G$ be unipotent and denote by $\lambda _1\ge \lambda _2\ge \cdots $ the lengths of its Jordan blocks. According to the normal forms given in [Reference De Franceschi, Liebeck and O’Brien7, 4.1.3], x has a conjugate in B in which the entry at position $(i,i+1)$ is zero for at least $2m-\lambda _1$ indices $1\le i \le m$ . Assume $2m-\lambda _1>1$ . Then the image of x in the standard Levi subgroup ${\operatorname {GL}}_m(q^2)$ is not regular and thus ${\operatorname {Sub}}_G(x)$ contains the corresponding maximal parabolic subgroup by Propositions 3.5 and 3.6. Furthermore, the image of x in the standard Levi subgroup ${\operatorname {GU}}_{2m-2}(q)$ also has at least two zero entries directly above the main diagonal, unless $\lambda _1=\lambda _2$ . In the latter case we must have $2m=4$ , and then the image of x in ${\operatorname {GU}}_{2m-2}(q)={\operatorname {GU}}_2(q)$ is trivial. So in either case, by induction ${\operatorname {Sub}}_G(x)$ also contains this end node parabolic subgroup and thus equals G.

This only leaves the cases when $2m-\lambda _1\le 1$ , that is, x has Jordan type $(2m)$ or $(2m-1,1)$ . The first of these is the class of regular unipotent elements. If $x\in U$ has a Jordan block of length $2m-1$ , then it must involve elements from all simple root subgroups except possibly for the ‘middle’ one. In particular, the only standard parabolic subgroups whose unipotent radical could contain x are B and $P_m$ . It is easy to see that ${\mathbf {O}}_p(P_m)$ contains elements with a Jordan block of length $2m-1$ ; then Lemma 2.14 shows ${\operatorname {Sub}}_G(x)=P_m$ .

Proof. Let $x\in G$ be an $\ell $ -element, and P a Sylow $\ell $ -subgroup of G containing x. Then P is abelian by Lemma 5.1 and lies in a Sylow d-torus ${\mathbf {S}}$ of ${\mathbf {G}}$ . Conversely, if ${\mathbf {S}}$ is a Sylow d-torus containing P, then ${\mathbf {S}}\le {\mathbf {C}}_{\mathbf {G}}(P)$ , hence even ${\mathbf {S}}\le {\mathbf {H}}:={\mathbf {C}}_{\mathbf {G}}^\circ (P)$ . By inspection of the order formulae [Reference Malle and Testerman27, Tab. 24.1] our condition on $\ell $ implies that $\ell $ does not divide $|W|$ . Thus, $\ell $ is not a torsion prime for ${\mathbf {G}}$ and then neither for ${\mathbf {H}}$ , so $P\le {\mathbf {Z}}^\circ ({\mathbf {H}})=:{\mathbf {S}}'$ , a torus. Since $\ell $ divides a unique $\Phi _d(q)$ this implies that ${\mathbf {S}}\le {\mathbf {S}}'$ , that is, ${\mathbf {S}}$ is the Sylow d-torus of ${\mathbf {Z}}^\circ ({\mathbf {C}}_{\mathbf {G}}^\circ (P))$ and thus uniquely determined by P. Since P is characteristic in ${\mathbf {S}}^F$ we have $N_G(P)=N_G({\mathbf {S}}^F)$ , and our claim follows by Proposition 2.6.

Proof. By Lemma 5.1 we are in the situation of Proposition 5.2. Hence $x\in G$ lies in a unique Sylow $\ell $ -subgroup of G if and only if it lies in a unique Sylow d-torus ${\mathbf {S}}$ of ${\mathbf {G}}$ , if and only if ${\mathbf {S}}$ is the unique Sylow d-torus of ${\mathbf {C}}_{\mathbf {G}}(x)$ , if and only if ${\mathbf {S}}\le {\mathbf {Z}}({\mathbf {C}}_{\mathbf {G}}(x))$ , if and only if ${\mathbf {C}}_{\mathbf {G}}(x)\le {\mathbf {C}}_{\mathbf {G}}({\mathbf {S}})$ . As $x\in {\mathbf {S}}$ we also have ${\mathbf {C}}_{\mathbf {G}}({\mathbf {S}})\le {\mathbf {C}}_{\mathbf {G}}(x)$ , so ${\mathbf {C}}_{\mathbf {G}}({\mathbf {S}})={\mathbf {C}}_{\mathbf {G}}(x)$ . This implies of course ${\mathbf {C}}_G({\mathbf {S}})={\mathbf {C}}_G(x)$ . Conversely, assume ${\mathbf {C}}_G({\mathbf {S}})={\mathbf {C}}_G(x)$ but ${\mathbf {C}}_{\mathbf {G}}({\mathbf {S}})<{\mathbf {C}}_{\mathbf {G}}(x)$ . Since ${\mathbf {C}}_{\mathbf {G}}({\mathbf {S}})$ is a Levi subgroup, hence connected, this means that either ${\mathbf {C}}_{\mathbf {G}}(x)$ is disconnected, or of strictly larger dimension than ${\mathbf {C}}_{\mathbf {G}}({\mathbf {S}})$ . The first is not possible as $\ell $ is not a torsion prime. In the second case, as ${\mathbf {C}}_{\mathbf {G}}({\mathbf {S}})$ contains a maximal torus of ${\mathbf {G}}$ , the maximal unipotent subgroups of ${\mathbf {C}}_{\mathbf {G}}(x)$ must have larger dimension than those of ${\mathbf {C}}_{\mathbf {G}}({\mathbf {S}})$ . But then $|{\mathbf {C}}_G(x)|_p>|{\mathbf {C}}_G({\mathbf {S}})|_p$ , a contradiction. In conclusion, $x\in G$ is picky if and only if ${\mathbf {C}}_G({\mathbf {S}})={\mathbf {C}}_G(x)$ , for ${\mathbf {S}}$ the (unique) Sylow d-torus containing x.

Proof. The centraliser ${\mathbf {C}}:={\mathbf {C}}_{\mathbf {G}}({\mathbf {S}})$ is an F-stable Levi subgroup of ${\mathbf {G}}$ . Its structure is described in [Reference Geck and Malle14, Exmp. 3.5.14 and 3.5.15] for ${\mathbf {G}}$ of classical type. Note that since ${\mathbf {S}}$ is a Sylow e-torus, it has the form ${\mathbf {T}}{\mathbf {H}}$ with ${\mathbf {T}}$ a torus whose order polynomial only involves factors $\Phi _d$ with $d\le 2e$ , and ${\mathbf {H}}$ a semisimple group of rank less than e. The claim can easily be checked by inspection of the order formulas (e.g., in [Reference Malle and Testerman27, Tab. 24.1]). For groups of exceptional type, Table 3.3 in [Reference Geck and Malle14] shows that ${\mathbf {C}}$ is itself a torus, with the required property, unless either $e=4$ and ${\mathbf {G}}$ has type $E_7$ , but the claim still holds in that case, or ${\mathbf {G}}^F={}^{3}\!D_4(q)$ and $\ell =3$ (where the claim fails).

Proof. The relative Weyl groups are described in [Reference Geck and Malle14, Exmp. 3.5.29] for ${\mathbf {G}}$ of classical type. Namely, they are either symmetric groups ${\mathfrak {S}}_n$ , in which case $e\in \{1,2\}$ , wreath products $C_d\wr {\mathfrak {S}}_n$ with $d\in \{e,2e\}$ , or certain subgroups of index 2 in the latter. Note that $\ell>e=e_\ell (q)$ if $\ell \ne 2$ . The claim follows in this case from the known normal structure of ${\mathfrak {S}}_n$ . For ${\mathbf {G}}$ of exceptional type, the occurring relative Weyl groups are given in [Reference Broué, Malle and Michel3, Tab. 3]. No further examples occur.

Proof. If ${\mathbf {G}}$ is of classical type, then $W_e/{\mathbf {O}}_2(W_e)$ is a symmetric group ${\mathfrak {S}}_n$ (see [Reference Geck and Malle14, 3.5.29]), and in ${\mathfrak {S}}_n$ the Sylow $\ell $ -normalisers are easily seen not to contain elements of prime order $r>\ell $ . For ${\mathbf {G}}$ of exceptional type, it suffices to consider $W(E_8)$ since all relative Weyl groups are subquotients of this. Now the only element order $\ell r$ for primes $2<\ell <r$ occurring in $W(E_8)$ is 15, but here $r{\not \equiv }1\ \pmod \ell $ .

Proof. Let $P\le G$ be a Sylow $\ell $ -subgroup of G and set $e=e_\ell (q)$ . By [Reference Malle25, Thm 5.16] there is a Sylow e-torus ${\mathbf {S}}$ of $({\mathbf {G}},F)$ such that the normaliser ${\mathbf {N}}_G({\mathbf {S}})$ contains the normaliser ${\mathbf {N}}_G(P)$ of P. Now ${\mathbf {L}}:={\mathbf {C}}_{\mathbf {G}}({\mathbf {S}})$ is an F-stable Levi subgroup of ${\mathbf {G}}$ . An application of [Reference Malle25, Prop. 5.3] to $[{\mathbf {L}},{\mathbf {L}}]$ shows that $[{\mathbf {L}},{\mathbf {L}}]^F$ is an $\ell '$ -group. Now ${\mathbf {L}}/[{\mathbf {L}},{\mathbf {L}}]$ is a torus, hence abelian, so as P is nonabelian by assumption, $\ell $ must divide the order of the relative Weyl group $W_e={\mathbf {N}}_G({\mathbf {S}})/{\mathbf {C}}_G({\mathbf {S}})$ .

Let $x\in G$ be an $\ell $ -element. First assume $x\in {\mathbf {S}}_\ell ^F$ . By Proposition 5.7, since $\ell \ge 5$ , $W_e$ has more than one Sylow $\ell $ -subgroup, so x lies in two different Sylow $\ell $ -subgroups of ${\mathbf {N}}_G({\mathbf {S}})$ and hence of G. Now assume x does not lie in any Sylow e-torus, and let ${\mathbf {T}}$ be an F-stable torus of ${\mathbf {G}}$ containing x. Then the order polynomial of ${\mathbf {T}}$ must be divisible by a cyclotomic polynomial $\Phi _{e\ell ^i}$ for some $i\ge 1$ and hence $|{\mathbf {C}}_G(x)|$ is divisible by $\Phi _{e\ell ^i}(q)$ . Let r be a Zsigmondy primitive prime divisor of $\Phi _{e\ell ^i}(q)$ , which exists as $e\ell ^i$ is divisible by a prime at least 5. On the other hand, by Proposition 5.6, the order polynomial of ${\mathbf {C}}_{\mathbf {G}}({\mathbf {S}})$ is not divisible by $\Phi _{e\ell ^i}$ , so $|{\mathbf {C}}_G({\mathbf {S}})|$ is prime to r. Thus $W_e$ must contain an element of order $\ell r$ . Now note that $r\equiv 1\ \pmod \ell $ . Thus Lemma 5.8 together with Lemma 2.1 show that x cannot be picky in ${\mathbf {N}}_G({\mathbf {S}})$ and so neither in G.

Proof. The proof of Theorem 5.9 goes through for the prime $\ell =3$ unless either we are in one of the exceptions of [Reference Malle25, Thm 5.16], in an exception of Proposition 5.7, or if $\Phi _{e3^i}(q)$ has no primitive prime divisor for some $i\ge 1$ , where again $e=e_3(q)$ . We discuss these in turn. The exceptions from [Reference Malle25, Thm 5.16] are $G={\operatorname {SL}}_3(\epsilon q)$ with $\epsilon q\equiv 4,7\ \pmod 9$ and $G=G_2(q)$ with $q\equiv 2,4,5,7\ \pmod 9$ . For $G={\operatorname {SL}}_3(\epsilon q)$ , by Lemma 2.4 we may consider $S={\operatorname {PSL}}_3(\epsilon q)$ instead. The Sylow 3-subgroups of S are elementary abelian of order 9, and the centraliser of an element of order 3 has structure $((q-\epsilon )\times (q-\epsilon )/3).3$ . Since the Sylow 3-normalisers have the form $3^{1+2}.Q_8$ for the appropriate congruences, there cannot exist any picky 3-elements for $q>4$ by Lemma 2.1.

Now let $G=G_2(q)$ with $q\equiv 2,4,5,7\ \pmod 9$ and let $\epsilon \in \{\pm 1\}$ with $q\equiv \epsilon \ \pmod 3$ . The centraliser $C:={\mathbf {C}}_G(t)\cong {\operatorname {SL}}_3(\epsilon q)$ of a 3-central element $t\in G$ contains a Sylow 3-subgroup of G. Thus t lies in several Sylow 3-subgroups of C (and thus of G) unless $q=2$ , hence cannot be picky. There is one further class of nontrivial 3-elements of G, centralised by an $A_1$ -subgroup (see [Reference Chang and Ree6, Reference Enomoto and Yamada11]), but the normaliser in G of a Sylow 3-subgroup (of order 27) does not contain such a subgroup unless again $q=2$ . In the latter case, $G'={\operatorname {SU}}_3(3)$ for which all nontrivial 3-elements are picky by Theorem 3.2, yielding conclusion (3).

Next, the exceptions of Proposition 5.7 occur precisely for the groups ${\operatorname {SL}}_3(\epsilon q)$ , $G_2(q)$ and ${}^{3}\!D_4(q)$ , since these are the only groups with a relative Weyl group of a Sylow e-torus isomorphic to ${\mathfrak {S}}_3$ or $W(G_2)$ . If $G={\operatorname {SL}}_3(\epsilon q)$ we have $3|(q-\epsilon )$ since we assume Sylow 3-subgroups of G are nonabelian, and even $9|(q-\epsilon )$ by the previous paragraph. Then, by [Reference Malle25, Thm 5.16] the normaliser of a Sylow 3-subgroup P of G lies inside the normaliser ${\mathbf {N}}_G({\mathbf {S}})$ of a Sylow e-torus ${\mathbf {S}}$ . If x is a 3-element not lying in a conjugate of ${\mathbf {S}}_3^F$ , then we may conclude as in the proof of Theorem 5.9 that x is not picky. If $x\in {\mathbf {S}}_3^F$ is not regular, its centraliser involves an $A_1$ -type group, hence is not contained in ${\mathbf {N}}_G({\mathbf {S}})={\mathbf {S}}^F.{\mathfrak {S}}_3$ and again x is not picky by Lemma 2.1. Now assume $x\in {\mathbf {S}}_3^F$ is regular. If ${\mathbf {N}}_G({\mathbf {S}})$ has more than one Sylow 3-subgroup, again x is not picky. If ${\mathbf {N}}_G({\mathbf {S}})$ has a unique Sylow 3-subgroup, that is, the Sylow 3-subgroup is normal, then the elements of order three in $W_e$ need to centralise the $3'$ -part of ${\mathbf {S}}^F$ which by inspection forces ${\mathbf {S}}^F$ to be a 3-group. Thus $q-\epsilon $ is a power of 3 which together with $q\equiv \epsilon \ \pmod 9$ forces $q=8$ , $\epsilon =-1$ . In the latter case, all regular $x\in {\mathbf {S}}_3^F$ are picky by direct computation, as claimed in (1).

Next, we consider $G=G_2(q)$ with $q\equiv \pm 1\ \pmod 9$ (the other congruences were already discussed above). The argument is now entirely analogous to the one given for ${\operatorname {SL}}_3(\epsilon q)$ , and only $G_2(8)$ gives rise to picky elements, listed in (1).

Next, if $G={}^{3}\!D_4(q)$ with $q\equiv \epsilon \ \pmod 3$ , $\epsilon \in \{\pm 1\}$ , then the normaliser of a Sylow 3-subgroup is contained in a torus normaliser of the form $N=(q^3-\epsilon )(q-\epsilon ).W(G_2)$ (see the discussion in the proof of [Reference Malle25, Thm 5.14]). Now elements of order 3 in $W(G_2)$ act like a field automorphism on the torus of order $q^3-\epsilon $ . In particular, as soon as $q^2+\epsilon q+1$ has a primitive prime divisor (necessarily distinct from 3), a Sylow 3-subgroup is not normal in N and we may conclude as before. This leaves $q=2$ , so $G={}^{3}\!D_4(2)$ . Here, by explicit computation, the elements of order 9 with centraliser $A_1(q).(q^3+1)$ are picky.

Finally, if there exists no primitive prime divisor for $\Phi _{e3^i}(q)$ with $i\ge 1$ , then $i=1$ , $e=q=2$ . If $x\in G$ is a picky 3-element, then it has order at most 9, as elements of order 27 have centraliser order divisible by $\Phi _{18}(2)=3^3 19$ in contradiction to Lemma 5.8. We now discuss the various types. For G of classical type, all elements of order 3 lie in a Sylow $2$ -torus of G and thus in at least two Sylow 3-subgroups, arguing as in the proof of Theorem 5.9. So in particular, ${\mathbf {C}}_{\mathbf {G}}(x)$ must contain a torus of rational type $\Phi _6$ . In ${\operatorname {SL}}_n(q)$ any centraliser order divisible by $q^3+1$ is in fact divisible by $(q^6-1)/(q-1)$ , hence by 7 when $q=2$ , contradicting Lemma 5.8. In the remaining groups of classical type, the primitive 9th roots of unity can occur at most once as an eigenvalue of x, as otherwise the centraliser contains a subgroup ${\operatorname {SU}}_2(8)$ and hence elements of order 7. Let V be the submodule of the natural G-module spanned by the eigenspaces for the eigenvalues of x of order at most 3 and H the induced classical group on V. Then H has the same type as G, respectively type ${\operatorname {SO}}^{-\epsilon }$ if G has type ${\operatorname {SO}}^\epsilon $ (since in that case by what we showed x has exactly all six primitive 9th roots of unity as eigenvalues and the induced orthogonal group on the corresponding 6-dimensional sum of eigenspaces is of minus type). Then ${\operatorname {Sub}}_G(x)$ will contain the subnormaliser of $x|_V$ (of order 3), and by direct computation the latter is all of H if $H={\operatorname {SU}}_3(2),{\operatorname {Sp}}_6(2),{\operatorname {SO}}_8^+(2)$ or ${\operatorname {SO}}_6^-(2)$ . As all of these have order divisible by a prime larger than 3, we see that x cannot be picky in ${\operatorname {SU}}_n(2),{\operatorname {Sp}}_{2n}(2),{\operatorname {SO}}_{2n}^+(2),{\operatorname {SO}}_{2n}^-(2)$ when $n\ge 9,6,6,7$ respectively. By direct computation, there do exist picky elements of order 9 in all remaining cases.

Similarly, the claim for $G_2(2),{}^{3}\!D_4(2),F_4(2)$ and $E_6(2)$ follows by explicit computation in GAP. The maximal subgroup $H=Fi_{22}$ of $S={}^{2}\!E_6(2)$ contains a Sylow 3-subgroup of S. By direct computation, H possesses one class of picky 3-elements, but this is fused in S with one of the nonpicky classes of H, so S has no picky 3-element. The group $G=E_7(2)$ has six classes of elements of order 9, with centralisers

$$ \begin{align*}\Phi_2^2\Phi_6.{}^{2}\!A_2(2)A_1(2),\ \Phi_2.{}^{2}\!A_2(8),\ \Phi_2\Phi_6.{}^{2}\!A_3(2)A_1(2),\end{align*} $$

$$ \begin{align*}\Phi_2\Phi_6.{}^{2}\!A_4(2),\ \Phi_2\Phi_6.{}^{3}\!D_4(2),\ \Phi_2\Phi_6.A_1(8)A_1(2)\end{align*} $$

(private communication of Frank Lübeck), while all elements of order 3 lie in a Sylow 2-torus by [Reference Lübeck21]. Now the normaliser of a Sylow 3-subgroup of G is contained in the normaliser of a Sylow 2-torus and thus has the form $3^7.N$ where N is the normaliser of a Sylow 3-subgroup of the Weyl group $W(E_7)$ , whose 2-part is $2^3$ . Since all of the above centralisers have order divisible by at least $2^4$ , G cannot have picky 3-elements by Lemma 2.1. Finally, $G=E_8(2)$ has four classes of elements of order 9, with centralisers

$$ \begin{align*}\Phi_2.{}^{2}\!A_2(8)A_1(2),\ \Phi_2\Phi_6.{}^{2}\!A_4(2)A_1(2),\ \Phi_2\Phi_6.{}^{2}\!D_5(2),\ \Phi_2\Phi_6.{}^{3}\!D_4(2)A_1(2)\end{align*} $$

(again computed by Frank Lübeck), and all elements of order 3 have centralisers of semisimple rank at least 7 by [Reference Lübeck21]. Since the 2-part of the normaliser of a Sylow 3-subgroup of G is just $2^4$ , we may argue as before to see that G has no picky 3-elements.

Proof. First assume that $\ell>3$ and x is regular. Then $\ell $ does not divide the order of the Weyl group of ${\mathbf {G}}$ and hence the Sylow $\ell $ -subgroups of G are abelian (see [Reference Broué and Malle2, Cor. 3.13, p. 259]). We may now proceed as in the proof of Theorem 5.3, replacing cyclotomic polynomials over ${\mathbb {Q}}$ by suitable cyclotomic polynomials over ${\mathbb {Q}}(\sqrt {p})$ , with corresponding Sylow tori, for which the Sylow theorems continue to hold, see [Reference Geck and Malle14, 3.5.3, 3.5.4], to conclude that x is picky.

We now discuss the remaining cases. All semisimple elements $x\ne 1$ of ${}^{2}\!B_2(q^2)$ are regular and have order prime to 6, so there is nothing to prove in this case. In $G={}^2G_2(q^2)$ the only $\ell $ -elements $x\ne 1$ that are either nonregular or have $\ell \le 3$ are the involutions, with ${\mathbf {C}}_G(x)=\langle x\rangle \times {\operatorname {PSL}}_2(q^2)$ . Such x are also contained in a Sylow 2-normaliser, of structure $2^3.7.3$ , not lying in the maximal subgroup ${\mathbf {C}}_G(x)$ . Thus ${\operatorname {Sub}}_G(x)=G$ as claimed.

Let $G={}^{2}\!F_4(q^2)$ . The conjugacy classes of semisimple elements are given in [Reference Shinoda33, Tab. IV]. First assume $\ell>3$ . The nonregular representatives $x\ne 1$ are $t_i$ with $i\in \{1,2,4,5,7,8\}$ , where $t_1,t_2,t_5$ and $t_8$ only occur if $q^2\ge 8$ . The noncyclic Sylow $\Phi $ -tori are $T(j)$ for $j\in \{1,6,7,8\}$ in [Reference Shinoda33, (3.1)], with relative Weyl groups of order $16,96,96,48$ respectively. Comparison with the list of maximal subgroups of G in [Reference Malle23, Main Thm] shows that none of them can contain both ${\mathbf {C}}_G(x)$ and ${\mathbf {N}}_G({\mathbf {T}})$ for ${\mathbf {T}}$ a Sylow $\Phi $ -torus of ${\mathbf {G}}$ containing x.

So finally assume $\ell =3$ . Any 3-element $x\in G$ is contained in a Sylow $\Phi _4$ -torus ${\mathbf {T}}$ of ${\mathbf {G}}$ , where ${\mathbf {N}}_G({\mathbf {T}})=(q^2+1)^2.G_{12}$ with the primitive complex reflection group $G_{12}$ of order 48. As ${\mathbf {N}}_G({\mathbf {T}})$ is a maximal subgroup of G by [Reference Malle23, Main Thm], either ${\operatorname {Sub}}_G(x)={\mathbf {N}}_G({\mathbf {T}})$ or ${\operatorname {Sub}}_G(x)=G$ . The structure of a Sylow 3-normaliser N is discussed in [Reference Malle25, Thm 8.4]. If $q^2\equiv 8\ \pmod 9$ then N is contained in the torus normaliser $(q^2+1)^2.G_{12}$ . Now $G_{12}$ has non-normal Sylow 3-subgroups and hence arguing as in the proof of Theorem 5.9 there cannot exist picky 3-elements, whence ${\operatorname {Sub}}_G(x)=G$ by what we said above. On the other hand, if $q^2\equiv 2,5\ \pmod 9$ a Sylow 3-normaliser is isomorphic to ${\operatorname {SU}}_3(2).2$ , not contained in any conjugate of ${\mathbf {N}}_G({\mathbf {T}})$ by [Reference Malle25, Thm 8.4(b)], so again we have ${\operatorname {Sub}}_G(x)=G$ .

Proof. By [Reference Malle and Testerman27, Prop. 2.9] we may embed ${\mathbf {U}}$ into the (unipotent) group of upper uni-triangular matrices of ${\operatorname {GL}}_n$ for a suitable n, so without loss we may assume ${\mathbf {U}}$ is connected. By [Reference Malle and Testerman27, Cor. 2.10] the unipotent group ${\mathbf {U}}$ is nilpotent, so by [Reference Humphreys19, Prop. 17.4], for any proper closed subgroup ${\mathbf {H}}<{\mathbf {U}}$ we have $\dim {\mathbf {H}}<\dim {\mathbf {N}}_{\mathbf {U}}({\mathbf {H}})$ . Thus by induction on the dimension, any closed subgroup is subnormal in ${\mathbf {U}}$ . As $\langle x\rangle $ is normal in its abelian (see [Reference Humphreys19, Lemma 15.1.C]) closure $\overline {\langle x\rangle }$ this achieves the proof.

Proof. Let $\langle x\rangle \unlhd H_1\unlhd \cdots \unlhd H_r=H$ be a subnormal series. Set $U:={\widehat {R_u}}(H)$ and $\tilde H_i:=H_iU$ . Since x normalises U, the group $\langle x,U\rangle $ is unipotent. Then $\langle x,U\rangle \unlhd \tilde H_1$ implies $\langle x,U\rangle \le {\widehat {R_u}}(\tilde H_1)=:U_1$ . As $\tilde H_1\unlhd \tilde H_2$ we now have $U_1\unlhd \tilde H_2$ and thus $U_1\le {\widehat {R_u}}(\tilde H_2)=:U_2$ . By induction this yields $\langle x,U\rangle \le U_1\le \cdots \le U_r={\widehat {R_u}}(\tilde H_r)$ , whence $x\in {\widehat {R_u}}(H)$ .

Now the Zariski closure $\overline {{\widehat {R_u}}(H)}$ is unipotent [Reference Malle and Testerman27, Prop. 2.9], giving $\langle x\rangle {\,\triangleleft \triangleleft \,}\overline {{\widehat {R_u}}(H)}$ by Lemma 6.1 and then of course $\langle x\rangle {\,\triangleleft \triangleleft \,}{\widehat {R_u}}(H)$ . Hence, $\langle x\rangle {\,\triangleleft \triangleleft \,}{\mathbf {N}}_{\mathbf {G}}({\widehat {R_u}}(H))\ge H$ and the last assertion follows.

Proof. Let ${\mathbf {U}}\le {\mathbf {G}}$ be a closed unipotent subgroup with ${\mathbf {U}}={\widehat {R_u}}({\mathbf {N}}_{\mathbf {G}}({\mathbf {U}}))$ . By [Reference Malle and Testerman27, Cor. 17.15], ${\mathbf {U}}$ lies in some Borel subgroup of ${\mathbf {G}}$ and then by [Reference Malle and Testerman27, Thm 17.10] there is a parabolic subgroup ${\mathbf {P}}\le {\mathbf {G}}$ with ${\mathbf {U}}\le R_u({\mathbf {P}})$ and ${\mathbf {N}}_{\mathbf {G}}({\mathbf {U}})\le {\mathbf {P}}$ . In particular, ${\mathbf {N}}_{\mathbf {G}}({\mathbf {U}})$ normalises $R_u({\mathbf {P}})$ . Thus, $V:={\mathbf {N}}_{R_u({\mathbf {P}})}({\mathbf {U}})$ is contained in and normalised by ${\mathbf {N}}_{\mathbf {G}}({\mathbf {U}})$ , so contained in ${\widehat {R_u}}({\mathbf {N}}_{\mathbf {G}}({\mathbf {U}}))={\mathbf {U}}$ , whence $V={\mathbf {U}}$ . By [Reference Humphreys19, Prop. 17.4] this forces ${\mathbf {U}}=R_u({\mathbf {P}})$ .

Conversely, if ${\mathbf {P}}\le {\mathbf {G}}$ is parabolic, hence connected, then ${\mathbf {P}}/R_u({\mathbf {P}})$ is connected reductive, and so has no nontrivial normal unipotent subgroups (since any such would have to be finite, hence central, but all central elements of connected reductive groups are semisimple, for example, by [Reference Malle and Testerman27, Cor. 8.13(b)]), whence ${\widehat {R_u}}({\mathbf {P}})=R_u({\mathbf {P}})$ .

Proof. By Lemma 6.1, ${\operatorname {Sub}}_{\mathbf {G}}(x)$ contains all Borel subgroups of ${\mathbf {G}}$ containing x. For the converse, assume $\langle x\rangle {\,\triangleleft \triangleleft \,} H$ for some subgroup $H\le {\mathbf {G}}$ , where without loss of generality H is maximal with respect to inclusion. By Lemma 6.2 then $H={\mathbf {N}}_{\mathbf {G}}({\widehat {R_u}}(H))$ and $x\in Q:={\widehat {R_u}}(H)$ . Now Q lies in the unipotent radical ${\mathbf {U}}$ of some Borel subgroup of ${\mathbf {G}}$ ([Reference Malle and Testerman27, Cor. 17.15]). If $Q<{\mathbf {U}}$ then $Q_1:={\mathbf {N}}_{\mathbf {U}}(Q)>Q$ , and $\langle x\rangle {\,\triangleleft \triangleleft \,} \langle H,Q_1\rangle>H$ , a contradiction, so in fact $Q={\mathbf {U}}$ and $H={\mathbf {N}}_{\mathbf {G}}(Q)={\mathbf {N}}_{\mathbf {G}}({\mathbf {U}})$ is closed (see [Reference Malle and Testerman27, Ex. 10.18]). Then ${\mathbf {N}}_H({\mathbf {U}})^\circ $ is a Borel subgroup of H, and thus $H^\circ $ is generated by the H-conjugates of ${\mathbf {N}}_H({\mathbf {U}})^\circ $ (see [Reference Malle and Testerman27, Thm 6.10]), all of which contain Q and hence x. Thus, H is generated by the normalisers (in H) of the maximal unipotent subgroups (of H and hence of G) it contains. The claim follows.

Proof. Any regular x is picky by [Reference Carter4, Prop. 5.1.3]. Now let $x\in {\mathbf {G}}$ be nonregular unipotent. There is nothing to prove if ${\mathbf {G}}$ is of type $A_1$ . If ${\mathbf {G}}$ is of rank 2, the class representatives for the corresponding finite groups given in [Reference Srinivasan37, Reference Enomoto9, Reference Chang and Ree6, Reference Enomoto10, Reference Enomoto and Yamada11] show that any nonregular unipotent element has a ${\mathbf {G}}$ -conjugate in ${\mathbf {B}}$ not involving any simple root element, and thus ${\operatorname {Sub}}_{\mathbf {G}}(x)={\mathbf {G}}$ by the analogue of Proposition 3.5. Now assume ${\mathbf {G}}$ has rank at least 3. Then a suitable conjugate of x in ${\mathbf {B}}$ can be written as a product of root elements in which at least one simple root $\alpha $ of ${\mathbf {G}}$ does not occur. If $\alpha $ is an end node, then the observation for rank 2 shows that x is conjugate to an element in which neither $\alpha $ nor the simple root connected to it occurs. Thus, we may assume $\alpha $ is not an end node. We can now complete the proof as in Proposition 3.6 using Proposition 6.4.