Bryant-Kovacs theorem

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Suppose p is a prime number, and V is a vector space over the prime field \mathbb{F}_p of dimension greater than 1. In other words, V is an elementary Abelian p-group that is not cyclic.

Then, if G is a subgroup of GL(V) (the general linear group on V), there exists a finite p-group P such that P/\Phi(P) \cong V, and under the natural homomorphism:

\operatorname{Aut}(P) \to \operatorname{Aut}(P/\Phi(P)) = GL(V)

the image of \operatorname{Aut}(P) is precisely G.

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