Past exam of the options I course of the University of Oxford / 2012 / B / B1

A left group action of $G$ on a set $S$ is a map $G\times S\to S$, written $(g,s)\mapsto g\cdot s$, satisfyingfor every $s\in S$ and $g,h\in G$, where $e$ is the identity element of $G$. In particular each $g$ acts by a bijection, whose inverse is the action of $g^{-1}$.

Given a group action, define $\rho (g)(j)=g\cdot j$. Each $\rho (g)$ is a permutation, since $\rho (g^{-1})$ is its inverse. The action law gives, for every $j$,Thus $\boxed{\rho :G\to S_n\text{ is a group homomorphism}}$.

Conversely, given a group homomorphism $\rho :G\to S_n$, put $g\cdot j=\rho (g)(j)$. Since a homomorphism sends the identity to the identity permutation, $e\cdot j=j$. AlsoThese are exactly the two group action axioms, so the construction defines the required action.

The three nonzero vectors in ${\mathbb{F}}_2^2$ are $e_1$, $e_2$ and $e_1+e_2$. An invertible linear map permutes this three-element set, so part (b) gives a group homomorphismIf a matrix acts trivially on all three vectors, it fixes both basis vectors $e_1,e_2$, so it is the identity matrix. Thus the kernel of a group homomorphism is trivial and $\rho$ is injective.

To count the domain, its first column may be any of the three nonzero vectors. The second column must be outside the two-element span of the first, leaving two choices. HenceAn injection between these finite sets of equal size is surjective, proving