Exercise 13.3.9

Proof.

Let

If $\psi (g)=\psi (g^{\prime })$, then $g\cdot {\nu }_i=g^{\prime }\cdot {\nu }_i$ for all $i\in \{1,...,7\}$, then particularly $g\cdot e_1=g^{\prime }\cdot e_1,g\cdot e_2=g^{\prime }\cdot e_2,g\cdot e_3=g^{\prime }\cdot e_3$, where $e_1=(1,0,0),e_2=(0,1,0),e_3=(0,0,1)$ is the standard basis of ${𝔽}_2^3$. Since a linear map (or its matrix) is determined by the images of the vectors of a base, $g=g^{\prime }$

$\psi :G\to S_m$ is a one-to-one group homomorphism. □