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Call a finite group $A$ affine if it has a normal, self-centralizing, complemented, elementary abelian subgroup $V$. Such a group $A$ is a semi-direct product $G\ltimes V$ where $V$ is a vector space of dimension $n$ over $\mathbb{F}_p$ and $G$ is a group of matrices in $\operatorname{GL}(n,p)$. The elements of $A$ can be written as matrices $\left(\begin{smallmatrix}g& v \\ 0& 1 \end{smallmatrix}\right)$, where $g \in G$, $v \in V$ and $0,1$ are row vectors of the appropriate length. Conversely, given $G ≤ \operatorname{GL}(V)$, $V$ a vector space over $\mathbb{F}_p$, $A=G\ltimes V$ is an affine group.

An example is the full affine group, $\operatorname{AGL}(n,p)$ where $G$ is $\operatorname{GL}(n,p)$ and $V$ is $\mathbb{F}_p^n$, that is $A$ is all $(n+1)×(n+1)$ matrices of the form $\left(\begin{smallmatrix}g& v \\ 0& 1 \end{smallmatrix}\right)$ where $g$ is $n×n$ invertible, $v$ is anything, $0$ is a zero vector, and $1$ is just a $1×1$ identity matrix.

$V$ becomes a $G$-module and its $G$-module structure has a large influence on the group theoretic structure of $A$. In particular, $V$ contains no "trivial" (central) summand as a $G$-module iff $A$ is centerless.

Supposing $A$ is centerless, what conditions on $V$ ensure $A$ is a complete group, that is, so that $A$ is also "outerless"?

See the previous questions:

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Caution: I think I implicitly assume V is semi-simple in both the question and answer. My V are usually simple anyways, but YMMV. –  Jack Schmidt Dec 7 '10 at 17:06
    
Anyone feel free to add an answer. I'm accepting my own just because the software hassles me to do so. –  Jack Schmidt Dec 15 '10 at 19:50

1 Answer 1

up vote 4 down vote accepted

Derek Holt helped me figure this out.

Suppose $V$ is not just normal, but characteristic in $A$. Then $\operatorname{Out}(A)$ has a normal series with factors $N, S, H$ with:

  • $N ≤ \operatorname{Out}(G)$ consisting of those automorphisms that take $V$ to an isomorphic $G$-module
  • $S$ a quotient of the group $\operatorname{Aut}_G(V)$ of $G$-module automorphisms of $V$ by the normal subgroup of automorphisms induced by the center of $G$
  • $H = H^1(G,V)$, the first cohomology group

For $A$ to be complete, $\operatorname{Out}(A) = 1$, so $N=S=H=1$. In particular,

  • No (non-identity) outer automorphism can take $V$ to an isomorphic $G$-module
  • $V$ has to be multiplicity-free and split, and the center of $G$ needs to include all the (block) scalar matrices
  • The first cohomology group has to vanish

Assuming then that $V$ is irreducible and characteristic, then the second condition is just that $V$ is absolutely irreducible and $G$ contains the scalar matrices $Z(\operatorname{GL}(V))$.

If $V$ is not characteristic (say if $G$ is very small and unipotent) then this analysis fails, but I think if $G$ starts out close to being complete, $V$ is likely to be characteristic. Certainly if $V$ is irreducible.

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I vaguely recall that G including a bunch of scalars tends to make the cohomology vanish, so the second and third conditions might work together. –  Jack Schmidt Dec 7 '10 at 16:56

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