Finite Fields and Finite Vector Spaces

Definition An Affine transformation of $\mathbb F_q$ is a map $f_{a,b}$ taking $\lambda$ to $a \lambda + b$ where $a \in \mathbb F_q^\#$ is nonzero. The group of such maps is called $A(\mathbb F_q)$.

Proposition $A(\mathbb F_q)$ is sharply 2-transitive of order $q(q-1)$.
proof: Let $\alpha,\beta$ distinct, the system of equations $\alpha = a 0 + b$, $\beta = a 1 + b$ has a unique solution.

Corollary By a general lemma about sharply 2-transitive groups this group must have a regular characteristic subgroup, this group is $\{f_1,b|b\in\mathbb F_q\}$.

Proposition $A(\mathbb F_q)$ has a one point extension which is sharply $3$-transitive of degree $q+1$.
proof: This is a straightforward application of the one point extension theorem, adjoin $\infty$ to $\mathbb F_q$ and define $x$ on $\mathbb F_q \cup \{\infty\}$ to swap $0$ and $\infty$ and invert all other elements $\lambda x = \lambda^{-1}$. Clearly $x^2=1$. Let $G_0 = \{f_{a,0}\mid a \in \mathbb F_q^\# \}$ and note $f_{a,0}^x$ fixes $0$ and $\infty$ while for $\lambda \in \mathbb F_q^\#$ we $f_{a,0}^x = f_{a^{-1},0}$ so $G_0^x = G_0$. Finally a system for the double cosets is given by just $1$ and any other element e.g. $f = f_{-1,1}$ will do (so $\lambda f = 1 - \lambda$ and $f^2=1$). We see that $xf$ acts on the $\infty, 1, 0$ by cycling them and for the remaining elements $\lambda (xf)^3 = 1 - \frac{1}{1-\frac{1}{\lambda}}=\lambda$ and (apparently...) $x^f = f^x \in GxG$ so we have a one point extension.

Definition $V_n(q)$ is the $n$-dimensional vector space over $\mathbb F_q$, clearly $|V_n(q)|=q^n$.

Definition If $V$  is a vector space then a linear automorphism of $V$ is a bijective linear map $V \to V$. The group of these is called the general linear group $GL(V)$ or $GL_n(q)$ when $V=V_n(q)$.

Definition The special linear group $SL(V)$ of linear automorphisms of determinant 1.

Lemma If $V$ is a vector space over $\mathbb F_q$ then $SL(V)$ is a normal subgroup of $GL(V)$ and the index is $q-1$: $|GL(V):SL(V)|=q-1$.
proof: SL is just the kernel of the surjective determinant map from GL to $\mathbb F_q^\times$. As a consequence $GL/SL \simeq \mathbb F_q^\#$ so $|GL:SL| = q-1$.

Lemma The group $GL(V)$ acts transitively on $V^\#$ and if the dimension of $V$ is $> 1$ the same is true of $SL(V)$.
proof: Take two nonzero vectors $e_1,f_1$ then to get a map between them choose bases $e_1,\ldots,e_n$ and $f_1,\ldots,f_n$ this gives $g \in GL(V)$ mapping between them. If $n > 1$, since we don't necessarily have $\det(g)=1$ let $\det(g)=\mu$ and replace $e_n$ by $\mu^{-1} e_n$. Now $g'$ mapping between these bases has determinant $1$. (did I get this right?)

Proposition $$|GL_n(q)| = q^{n(n-1)/2} \prod_{i=1}^n (q^i-1)$$ and $$|SL_n(q)| = q^{n(n-1)/2} \prod_{i=2}^n (q^i-1).$$
proof: $GL_n(q)$ acts regularly on ordered bases of $V_n(q)$, so the size of $GL_n(q)$ is equal to the number of ordered bases: $q^n-1$ choices for the first element, and having chosen $e_1,\ldots,e_i$ (which spans a $q^i$ sized space) already there are $q^n-q^i$ choices for the next. The size of $SL$ comes from the lemma before the previous.