You are interested in finite Lie algebras, which is the same thing as finite-dimensional Lie algebras over finite fields. The usual way to understand such a problem is first to consider finite-dimensional Lie algebras algebraically closed fields of characteristic $p$.

So let's assume the ground field $k$ is algebraically closed.

Hogeweij determined in this paper the automorphism group of the following Lie algebras:

• ${\mathfrak g}={\rm Lie}(G)$ where $G$ is a simple algebraic group;

• ${\mathfrak f}={\mathfrak g}/{\mathfrak z}({\mathfrak g})$ where ${\mathfrak g}={\rm Lie}(G)$ for a simply-connected simple algebraic group $G$.

Slightly confusingly, he called ${\mathfrak f}$ the classical Lie algebra of type $X_l$ (where $X_l$ is the root system). (Hogeweij remarked that the automorphism groups of most of the classical Lie algebras were earlier determined by Steinberg.) He used the notation: $\widetilde{G}$, $\widetilde{\mathfrak g}$ for the simply-connected case (called universal by Hogeweij) and $\bar{G}$, $\bar{\mathfrak g}$ for the adjoint case. There are no other possibilities except in type $A_l$ with $(l+1)$ composite or type $D_l$ with $p=2$. Note that ${\mathfrak z}({\mathfrak g})$ is trivial and $\widetilde{\mathfrak g}\cong\bar{\mathfrak g}$ except:

• when $p$ divides $(l+1)$ in type $A_l$;

• when $p=2$ in types $B_l$, $C_l$, $D_l$, $E_7$;

• when $p=3$ in type $E_6$.

The appearance of $G_2$ from your calculations is (surely) related to the following fact, well-known to specialists in positive characteristic Lie algebras: in characteristic 3 (and only in characteristic 3) a Lie algebra of type $G_2$ has an ideal, generated by the short root elements, isomorphic to $\mathfrak{psl}(3,k)$. (Note that this is a 7-dimensional Lie algebra which is not the Lie algebra of an algebraic group.) The simple algebraic group of type $G_2$ therefore acts as automorphisms of $\mathfrak{psl}(3,k)$. However, I don't understand quite how it arises in your set-up.

Indeed, Hogeweij's tables show that (in characteristic 3) the automorphism group of $\mathfrak{psl}(3,k)$ is simple of type $G_2$, but both $\mathfrak{sl}(3,k)$ and $\mathfrak{pgl}(3,k)$ have automorphism group equal to a semidirect product of ${\rm PGL}(3,k)$ and the group of diagram automorphisms (which is cyclic of order 2). So unless ${\rm PGL}(3,\overline{{\mathbb F}_3})$ somehow contains something like $G_2({\mathbb F}_3)$ (which seems unlikely to me, though I am not certain) then I can't see how your result can be correct.

This brings me to my second point. Your algorithm is doomed to fail for large $p$, since the exponential of ${\rm ad}\, x$ is only defined if $({\rm ad}\, x)^p = 0$. In general this holds for some but not all nilpotent elements - it holds for all nilpotent elements if and only if $p$ is greater than or equal to the Coxeter number. The usual construction of the Chevalley groups (if I remember correctly) uses only $\exp(\lambda\,{\rm ad}\, e_\alpha)$ where $e_\alpha$ is a simple (positive or negative) root element. The initial choice of a ${\mathbb Z}$-form ensures that these exponentials are well-defined. But I don't see any reason to expect that the group generated by the exponentials of these simple root elements will be equal to the group generated by the exponentials of all nilpotent ($p$-th power zero) elements. This probably (at least in part) explains the discrepancy between your group of order 9285337152 and the group you expected to obtain.

You are interested in finite Lie algebras, which is the same thing as finite-dimensional Lie algebras over finite fields. The usual way to understand such a problem is first to consider finite-dimensional Lie algebras algebraically closed fields of characteristic $p$.

So let's assume the ground field $k$ is algebraically closed.

Hogeweij determined in this paper the automorphism group of the following Lie algebras:

• ${\mathfrak g}={\rm Lie}(G)$ where $G$ is a simple algebraic group;

• ${\mathfrak f}={\mathfrak g}/{\mathfrak z}({\mathfrak g})$ where ${\mathfrak g}={\rm Lie}(G)$ for a simply-connected simple algebraic group $G$.

Slightly confusingly, he called ${\mathfrak f}$ the classical Lie algebra of type $X_l$ (where $X_l$ is the root system). (Hogeweij remarked that the automorphism groups of most of the classical Lie algebras were earlier determined by Steinberg.) He used the notation: $\widetilde{G}$, $\widetilde{\mathfrak g}$ for the simply-connected case (called universal by Hogeweij) and $\bar{G}$, $\bar{\mathfrak g}$ for the adjoint case. There are no other possibilities except in type $A_l$ with $(l+1)$ composite or type $D_l$ with $p=2$. Note that ${\mathfrak z}({\mathfrak g})$ is trivial and $\widetilde{\mathfrak g}\cong\bar{\mathfrak g}$ except:

• when $p$ divides $(l+1)$ in type $A_l$;

• when $p=2$ in types $B_l$, $C_l$, $D_l$, $E_7$;

• when $p=3$ in type $E_6$.

The appearance of $G_2$ from your calculations is (surely) related to the following fact, well-known to specialists in positive characteristic Lie algebras: in characteristic 3 (and only in characteristic 3) a Lie algebra of type $G_2$ has an ideal, generated by the short root elements, isomorphic to $\mathfrak{psl}(3,k)$. (Note that this is a 7-dimensional Lie algebra which is not the Lie algebra of an algebraic group.) The simple algebraic group of type $G_2$ therefore acts as automorphisms of $\mathfrak{psl}(3,k)$. However, I don't understand quite how it arises in your set-up.

Indeed, Hogeweij's tables show that (in characteristic 3) the automorphism group of $\mathfrak{psl}(3,k)$ is simple of type $G_2$, but both $\mathfrak{sl}(3,k)$ and $\mathfrak{pgl}(3,k)$ have automorphism group equal to a semidirect product of ${\rm PGL}(3,k)$ and the group of diagram automorphisms (which is cyclic of order 2). So unless ${\rm PGL}(3,\overline{{\mathbb F}_3})$ somehow contains something like $G_2({\mathbb F}_3)$ (which seems unlikely to me, though I am not certain) then I can't see how your result can be correct.

This brings me to my second point. Your algorithm is doomed to fail for small $p$, since the exponential of ${\rm ad}\, x$ is only defined if $({\rm ad}\, x)^p = 0$. In general this holds for some but not all nilpotent elements - it holds for all nilpotent elements if and only if $p$ is greater than or equal to the Coxeter number. The usual construction of the Chevalley groups (if I remember correctly) uses only $\exp(\lambda\,{\rm ad}\, e_\alpha)$ where $e_\alpha$ is a simple (positive or negative) root element. The initial choice of a ${\mathbb Z}$-form ensures that these exponentials are well-defined. But I don't see any reason to expect that the group generated by the exponentials of these simple root elements will be equal to the group generated by the exponentials of all nilpotent ($p$-th power zero) elements. This probably (at least in part) explains the discrepancy between your group of order 9285337152 and the group you expected to obtain.

You are interested in finite Lie algebras, which is the same thing as finite-dimensional Lie algebras over finite fields. The usual way to understand such a problem is first to consider finite-dimensional Lie algebras algebraically closed fields of characteristic $p$.

So let's assume the ground field $k$ is algebraically closed.

Hogeweij has determined here the automorphism group of the following Lie algebras:

• ${\mathfrak g}={\rm Lie}(G)$ where $G$ is a simple algebraic group;

• ${\mathfrak f}={\mathfrak g}/{\mathfrak z}({\mathfrak g})$ where ${\mathfrak g}={\rm Lie}(G)$ for a simply-connected simple algebraic group $G$.

Slightly confusingly, he called ${\mathfrak f}$ the classical Lie algebra of type $X_l$ (where $X_l$ is the root system). (Hogeweij remarked that the automorphism groups of most of the classical Lie algebras were earlier determined by Steinberg.) He used the notation: $\widetilde{G}$, $\widetilde{\mathfrak g}$ for the simply-connected case (called universal by Hogeweij) and $\bar{G}$, $\bar{\mathfrak g}$ for the adjoint case. There are no other possibilities except in type $A_l$ with $(l+1)$ composite or type $D_l$ with $p=2$. Note that ${\mathfrak z}({\mathfrak g})$ is trivial and $\widetilde{\mathfrak g}\cong\bar{\mathfrak g}$ except:

• when $p$ divides $(l+1)$ in type $A_l$;

• when $p=2$ in types $B_l$, $C_l$, $D_l$, $E_7$;

• when $p=3$ in type $E_6$.

The appearance of $G_2$ from your calculations is (surely) related to the following fact, well-known to specialists in positive characteristic Lie algebras: in characteristic 3 (and only in characteristic 3) a Lie algebra of type $G_2$ has an ideal, generated by the short root elements, isomorphic to $\mathfrak{psl}(3,k)$. (Note that this is a 7-dimensional Lie algebra which is not the Lie algebra of an algebraic group.) The simple algebraic group of type $G_2$ therefore acts as automorphisms of $\mathfrak{psl}(3,k)$. However, I don't understand quite how it arises in your set-up.

Indeed, Hogeweij's tables show that (in characteristic 3) the automorphism group of $\mathfrak{psl}(3,k)$ is simple of type $G_2$, but both $\mathfrak{sl}(3,k)$ and $\mathfrak{pgl}(3,k)$ have automorphism group equal to a semidirect product of ${\rm PGL}(3,k)$ and the group of diagram automorphisms (which is cyclic of order 2). So unless ${\rm PGL}(3,\overline{{\mathbb F}_3})$ somehow contains something like $G_2({\mathbb F}_3)$ (which seems unlikely to me, though I am not certain) then I can't see how your result can be correct.

This brings me to my second point. Your algorithm is doomed to fail for large $p$, since the exponential of ${\rm ad}\, x$ is only defined if $({\rm ad}\, x)^p = 0$. In general this holds for some but not all nilpotent elements - it holds for all nilpotent elements if and only if $p$ is greater than or equal to the Coxeter number. The usual construction of the Chevalley groups (if I remember correctly) uses only $\exp(\lambda\,{\rm ad}\, e_\alpha)$ where $e_\alpha$ is a simple (positive or negative) root element. The initial choice of a ${\mathbb Z}$-form ensures that these exponentials are well-defined. But I don't see any reason to expect that the group generated by the exponentials of these simple root elements will be equal to the group generated by the exponentials of all nilpotent ($p$-th power zero) elements. This probably (at least in part) explains the discrepancy between your group of order 9285337152 and the group you expected to obtain.

You are interested in finite Lie algebras, which is the same thing as finite-dimensional Lie algebras over finite fields. The usual way to understand such a problem is first to consider finite-dimensional Lie algebras algebraically closed fields of characteristic $p$.

So let's assume the ground field $k$ is algebraically closed.

Hogeweij determined in this paper the automorphism group of the following Lie algebras:

• ${\mathfrak g}={\rm Lie}(G)$ where $G$ is a simple algebraic group;

• ${\mathfrak f}={\mathfrak g}/{\mathfrak z}({\mathfrak g})$ where ${\mathfrak g}={\rm Lie}(G)$ for a simply-connected simple algebraic group $G$.

Slightly confusingly, he called ${\mathfrak f}$ the classical Lie algebra of type $X_l$ (where $X_l$ is the root system). (Hogeweij remarked that the automorphism groups of most of the classical Lie algebras were earlier determined by Steinberg.) He used the notation: $\widetilde{G}$, $\widetilde{\mathfrak g}$ for the simply-connected case (called universal by Hogeweij) and $\bar{G}$, $\bar{\mathfrak g}$ for the adjoint case. There are no other possibilities except in type $A_l$ with $(l+1)$ composite or type $D_l$ with $p=2$. Note that ${\mathfrak z}({\mathfrak g})$ is trivial and $\widetilde{\mathfrak g}\cong\bar{\mathfrak g}$ except:

• when $p$ divides $(l+1)$ in type $A_l$;

• when $p=2$ in types $B_l$, $C_l$, $D_l$, $E_7$;

• when $p=3$ in type $E_6$.

The appearance of $G_2$ from your calculations is (surely) related to the following fact, well-known to specialists in positive characteristic Lie algebras: in characteristic 3 (and only in characteristic 3) a Lie algebra of type $G_2$ has an ideal, generated by the short root elements, isomorphic to $\mathfrak{psl}(3,k)$. (Note that this is a 7-dimensional Lie algebra which is not the Lie algebra of an algebraic group.) The simple algebraic group of type $G_2$ therefore acts as automorphisms of $\mathfrak{psl}(3,k)$. However, I don't understand quite how it arises in your set-up.

Indeed, Hogeweij's tables show that (in characteristic 3) the automorphism group of $\mathfrak{psl}(3,k)$ is simple of type $G_2$, but both $\mathfrak{sl}(3,k)$ and $\mathfrak{pgl}(3,k)$ have automorphism group equal to a semidirect product of ${\rm PGL}(3,k)$ and the group of diagram automorphisms (which is cyclic of order 2). So unless ${\rm PGL}(3,\overline{{\mathbb F}_3})$ somehow contains something like $G_2({\mathbb F}_3)$ (which seems unlikely to me, though I am not certain) then I can't see how your result can be correct.

This brings me to my second point. Your algorithm is doomed to fail for large $p$, since the exponential of ${\rm ad}\, x$ is only defined if $({\rm ad}\, x)^p = 0$. In general this holds for some but not all nilpotent elements - it holds for all nilpotent elements if and only if $p$ is greater than or equal to the Coxeter number. The usual construction of the Chevalley groups (if I remember correctly) uses only $\exp(\lambda\,{\rm ad}\, e_\alpha)$ where $e_\alpha$ is a simple (positive or negative) root element. The initial choice of a ${\mathbb Z}$-form ensures that these exponentials are well-defined. But I don't see any reason to expect that the group generated by the exponentials of these simple root elements will be equal to the group generated by the exponentials of all nilpotent ($p$-th power zero) elements. This probably (at least in part) explains the discrepancy between your group of order 9285337152 and the group you expected to obtain.