Semidirectly extensible implies linearly pushforwardable for representation over prime field

From Groupprops

Statement

Suppose $F$ is a prime field (i.e., either a field of prime order or the field of rational numbers), and $G$ is a group. Suppose $V$ is a finite-dimensional vector space over $F$ , and $\rho:G \to GL(V)$ be a linear representation of $G$ . Let $H = V \rtimes G$ with respect to the induced action of $G$ on $V$ .

Suppose, further, that $σ$ is an automorphism of $G$ that can be extended to an automorphism $σ'$ of $H$ such that $σ'$ also restricts to an automorphism $α$ of $V$ . Then, $\rho \circ \sigma = c_\alpha \circ \rho$ where $c α$ is conjugation by $α$ in $G L (V)$ .

Note that we need the field to be a prime field in order that $G L (V)$ is equal to the automorphism group of $V$ as a group.

Related facts

Quotient-pullbackable implies linearly pushforwardable for representation over prime field

Applications

Facts used

Automorphism group equals general linear group for vector space over prime field
Automorphism group action lemma: Suppose $H$ is a group, and $N,G \le H$ are subgroups such that $G \le N_H(N)$ . Suppose $σ'$ is an automorphism of $H$ such that the restriction of $σ'$ to $N$ gives an automorphism $α$ of $N$ , and such that $σ'$ also restricts to an automorphism of $G$ , say $σ$ . Consider the map:

$\rho: G \to \operatorname{Aut}(N)$

that sends an element $g \in G$ to the automorphism of $N$ induced by conjugation by $g$ (note that this is an automorphism since $G \le N_H(N)$ ). Then, we have:

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

where $c α$ denotes conjugation by $α$ in the group $\operatorname{Aut}(N)$ .

Proof

Given: A group $G$ , a homomorphism $\rho:G \to GL(V)$ for a finite-dimensional vector space $V$ over a prime field $F$ . $σ$ is an automorphism of $G$ that extends to an automorphism $σ'$ of $H$ , such that $σ'$ also restricts to an automorphism $α$ of $V$ .

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

Proof: Since $F$ is a prime field, $G L (V)$ is the whole automorphism group of $V$ by fact (1) (in general, it is a proper subgroup). Thus, the element $α$ , which is a group automorphism of $V$ , is actually in $G L (V)$ . Thus, fact (2), setting $G = G, H = H, N = V,σ' = σ',α = α,σ = σ$ , gives the desired result.