# What is $G_2(2^m)$, and how is it embedded in $\Gamma L_6(2^m)$?

I am trying to understand the classification of doubly transitive groups, specifically the nonsolvable affine case. Dixon and Mortimer (p.244) says there are three infinite families, one of which is $\mathbb{F}_{2^m}^6 \rtimes G_2(2^m) \leq \mathbb{F}_{2^m}^6 \rtimes \Gamma L_6(2^m)$.

What exactly is the group $G_2(2^m)$, and how is it embedded in $\Gamma L_6(2^m)$?

Ideally, the answer would say something like, "It's the subgroup of $\Gamma L_6(2^m)$ that preserves $X$." I am looking for a reference that provides such a description, written in contemporary English, and preferably at an introductory level.

So far, I have found the following:

• Dixon and Mortimer points me to Hering 1974. Hering points me to "Dickson, 1915", but there is no such entry in the bibliography. MathSciNet lists six research articles by Dickson in 1915. None of them appear relevant.

• This question is relevant and has a long list of references, but they are aimed at proving that the list in Dixon and Mortimer is complete. I want something that just describes the groups in that list.

• This paper by Cooperstein talks about $G_2(2^m)$ as a subgroup of $Sp_6(2^m)$, but it doesn't explicitly describe the embedding. Instead, it points me to this paper by Tits and Borel (in French) and this earlier paper by Cooperstein, but the latter is quite technical and does not obviously contain what I need. I would like something at an introductory level.

• Wikipedia references this paper by Dickson, which first introduced $G_2(2^m)$ in 1905. Maybe Hering was trying to cite this one instead. In any case, the language is extremely outdated. I'm looking for something more understandable.

• I am not sure what $\Gamma L_6(2^m)$ is supposed to be, but $G_2(2^m)$ is the simple Chevalley group of type $G_2$ over the finite field with $2^m$ elements. The embedding into $\operatorname{Sp}_6(2^m)$ is given by the $6$-dimensional irreducible representation in characteristic two, this you can find explicitly in pg. 34 of the first paper by Cooperstein you mention. – spin Dec 6 '17 at 15:31
• Personally I like the definition of $G_2(q)$ which starts from octonions over $\mathbb F_q$. See my question mathoverflow.net/questions/270781/… You can find the basis $1,u,v,uv, w,uw, vw, (uv)w$ for some invertible octonions $u,v,w$ and see that automorphism of octonions preserve $1$, so it can be seen as element of $O_7(q)$. Next we should somehow understand the automorphism $O_7(q) =S_6(q)$ for $q=2^n$. Each element of group $G_2$ preserve some subalgebra of $\mathbb O_q$. – user21230 Dec 7 '17 at 11:05
• The comment of @MarekMitros is fully explained in Wilson's book The finite simple groups. Go to p.118 - 121 and you will see that $G_2(F)$ is precisely the elements inside ${\rm Sp}_6(F)$ that preserve "a quaternion algebra". So this is the sort of characterisation that you are asking for. An alternative route to the same construction (also in that same part of Wilson's book) uses the terminology of trilinear forms, in which case, once again, $G_2(F)$ is the set of elements of ${\rm Sp}_6(F)$ that preserve a particular trilinear form. So that's two characterisations of the form you wanted. – Nick Gill Dec 7 '17 at 16:42
• @spin, $\Gamma L_6(2^m)$ is the set of semilinear automorphisms of a 6-dimensional vector space over a field of order $2^m$. It is perfectly natural to define $G_2(2^m)$ via an action on this subspace, rather than as a Chevalley group -- in the same way that one can choose to define ${\rm Sp}_{2m}(q)$ as a Chevalley group, or else via an action on a vector space that preserves a bilinear form. – Nick Gill Dec 7 '17 at 16:45
• This may not be of interest, but the embeddings $G_2(2^n)\le Sp_6(2^n)\le D_4(2^n)$ can be described as follows. $Aut(D_4(2^n))$ has a subgroup $\Sigma\cong S_3$ of "graph" automorphisms derived from the symmetries of the Dynkin diagram. Here for $\sigma\in\Sigma$, $\sigma(x_{\pm\alpha}(t))=x_{\pm\sigma(\alpha)}(t)$ for all fundamental roots $\alpha$ and all $t\in F_{2^n}$. For any fixed $\sigma\in\Sigma$ of order $2$, $C_{D_4(2^n)}(\sigma)\cong B_3(2^n)\cong Sp_6(2^n)$, while $C_{D_4(2^n)}(\Sigma)\cong G_2(2^n)$. Your embedding is the inclusion $C_{D_4(2^n)}(\Sigma)\le C_{D_4(2^n)}(\sigma)$. – Richard Lyons Dec 7 '17 at 18:59