# Semidirectly extensible implies linearly pushforwardable for representation over prime field

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

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

## Facts used

1. Automorphism group equals general linear group for vector space over prime field
2. Automorphism group action lemma: Suppose $H$ is a group, and $N,G \le H$ are subgroups such that $G \le N_H(N)$. Suppose $\sigma'$ is an automorphism of $H$ such that the restriction of $\sigma'$ to $N$ gives an automorphism $\alpha$ of $N$, and such that $\sigma'$ also restricts to an automorphism of $G$, say $\sigma$. 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 $c_\alpha$ denotes conjugation by $\alpha$ 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 $F$. $\sigma$ is an automorphism of $G$ that extends to an automorphism $\sigma'$ of $H$, such that $\sigma'$ also restricts to an automorphism $\alpha$ of $V$.

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 $G = G, H = H, N = V, \sigma' = \sigma', \alpha = \alpha, \sigma = \sigma$, gives the desired result.