# Quotient-pullbackable implies linearly pushforwardable for representation over prime field

## Statement

Suppose is an automorphism of a group . Suppose is a finite-dimensional linear representation of 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 . Suppose, further, that there exists an automorphism of such that restricts to an automorphism of , and the automorphism induced on *as a quotient* equals . Then, we have:

where denotes conjugation by .

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

- Finite-quotient-pullbackable implies class-preserving
- Conjugacy-separable implies every quotient-pullbackable automorphism is class-preserving

## Facts used

- Automorphism group equals general linear group for vector space over prime field
- Automorphism group action lemma for quotients: Suppose is an Abelian normal subgroup (?) of a group , and is an automorphism of that restricts to an automorphism of . Note that this also shows that descends to an automorphism, say , of .

Since 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:

.

The claim is that:

.

where denotes conjugation by as an element of .

## Proof

**Given**: A group , an automorphism of , a representation for a vector space over a prime field. There is an automorphism of such that restricts to an automorphism of , and it induces the automorphism on the quotient .

**To prove**: .

**Proof**: Since is a prime field, is the whole automorphism group of by fact (1) (in general, it is a proper subgroup). Thus, the element , which is a group automorphism of , is actually in . Thus, fact (2), setting , gives the desired result.