Quotient-pullbackable implies linearly pushforwardable for representation over prime field

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Statement

Suppose \sigma is an automorphism of a group G. Suppose \rho:G \to GL(V) is a finite-dimensional linear representation of G over a prime field (i.e., either the field is of prime order or is the field of rational numbers). Consider the corresponding semidirect product H = V \rtimes G. Suppose, further, that there exists an automorphism \sigma' of H such that \sigma' restricts to an automorphism \alpha of V, and the automorphism induced on G as a quotient equals \sigma. Then, we have:

\rho \circ \sigma = c_\alpha \circ \rho

where c_\alpha denotes conjugation by \alpha.

Note that we need the field to be a prime field in order to ensure that any automorphism of the vector space as an abelian group is also a linear automorphism.

Related facts

Applications

Facts used

  1. Automorphism group equals general linear group for vector space over prime field
  2. Automorphism group action lemma for quotients: Suppose A is an Abelian normal subgroup (?) of a group H, and \sigma' is an automorphism of H that restricts to an automorphism \alpha of A. Note that this also shows that \sigma' descends to an automorphism, say \sigma, of H/A.

Since A is abelian, we have an action of the quotient group on it by conjugation (see quotient group acts on Abelian normal subgroup), giving a homomorphism:

\rho:H/A \to \operatorname{Aut}(A).

The claim is that:

\rho \circ \sigma = c_\alpha \circ \rho.

where c_\alpha denotes conjugation by \alpha as an element of \operatorname{Aut}(A).

Proof

Given: A group G, an automorphism \sigma of G, a representation \rho:G \to GL(V) for a vector space V over a prime field. There is an automorphism \sigma' of H = V \rtimes G such that \sigma' restricts to an automorphism \alpha of V, and it induces the automorphism \sigma on the quotient G.

To prove: \rho \circ \sigma = c_\alpha \circ \rho.

Proof: Since F is a prime field, GL(V) is the whole automorphism group of V by fact (1) (in general, it is a proper subgroup). Thus, the element \alpha, which is a group automorphism of V, is actually in GL(V). Thus, fact (2), setting H = H, A = V, \sigma' = \sigma', \alpha = \alpha, \sigma = \sigma, gives the desired result.