The automorphism group of $2^{2n}{:}Sp_{2n}(2)$

Question

Let $G=2^{2n}{:}Sp_{2n}(2)$ be the split extension, where the symplectic group $Sp_{2n}(2)$ acts naturally on the vector space $2^{2n}$. With the aid of GAP it turns out that the automorphism group $\textrm{Aut}(G)\cong G{:}2\cong 2^{2n+1}{:}Sp_{2n}(2)=G_1$ for $n=2, 3, 4$. Furthermore, the group $G_1$ is not isomorphic to the affine subgroup $G_2=2^{2n+1}{:}Sp_{2n}$ of $Sp_{2n+2}(2)$ which has a similar structure. Note that the group $G_3=2^2{:}Sp_2(2)\cong S_4$ is complete and $G_4=2^3{:}Sp_2(2)\cong G_3\times 2$, where $G_4$ is the affine subgroup of $Sp_4(2)$.

My question is how do one prove that $\textrm{Aut}(G)\cong G{:}2\cong 2^{2n+1}{:}Sp_{2n}(2)$ for any $n\geq 2$ (if it is true) or is there already literature available on this topic.

Answer 1

Since ${\rm Sp}(2n,2)$ has trivial outer automorphism group and its natural module $M$ is absolutely irreducible, this follows from the fact that $|H^1({\rm Sp}(2n,2),M)| =2$.

You can find that result, for example, in Table 4.5 of

Cohomology of finite groups of Lie type, I, Edward Cline; Brian Parshall; Leonard Scott, Publications Mathématiques de l'IHÉS (1975) Volume: 45, page 169-191 (see here),

but that might not be the earliest proof.