General affine group of degree one on field of order 4

Question

I want to show $AGL(1,4)$ is isomorphic to $A_4$. I know that $AGL1(k)≅k⋊k^×$, where the semi-direct product is given by the natural action of $k^×$ on $k$ by multiplication. But I dont know how to use it.

Answer 1

In addition to the other answers. You may also remark that $AGL(1,4)$ acts faithfully on $\mathbb{F}_4$ which is a set of cardinal $4$. In particular, the structural morphism of this action allows you to identify $AGL(1,4)$ to a subgroup of $S_4$.

Now $AGL(1,4)$ is a subgroup of index $2$ in $S_4$ and thus $AGL(1,4)=A_4$.

Answer 2

The group $AGL(1,q)\cong \Bbb{F}_q\rtimes \Bbb{F}_q^{\times}$ hays order $q(q-1)$, and its derived groups is isomorphic to $\Bbb{F}_q$ for $q>2$. For $q=4$ this means, $G=AGL(1,4)$ has order $12$ with derived subgroup of order $4$. So $G$ is metabelian, but not abelian and not nilpotent. Furthermore it is not supersolvable, since $4$ is not prime. I suppose, together with the classification list of groups of order $12$ we can now see that $G\cong A_4$:

Classification of groups of order 12

Answer 3

Denote by $k$ the field with $4$ elements. The group of transformations $x\mapsto ax+ b$ imbeds into the group of permutations of $k$. Each translation by an element $b\ne 0$ of $k$ is product of two disjoint transpositions if $a\ne 0$ since $b+b=0$ so $x\mapsto x+b \mapsto x$. Each multiplication by an element $a\ne 1$ is a cycle of length $3$ since $a^3=1$ so $x\mapsto a x \mapsto a^2 x \mapsto x$ ($0$ is fixed). So all the transformation $x\mapsto ax+b$ are even permutation of $k$. Looking at the orders, we conclude that they give all the even permutations of $k$.