Need help about $P\Gamma L_2(q)$, $q=4,3$ I am asking kindly,

For which values of $n$ we have $$S_n≅P\Gamma L_2(3),S_n≅P\Gamma L_2(4)$$ This may be correct if we replace $S_n$ by $A_n$.

Any help will be appreciated. :)

Edit (JL): Adding the definition of the group. Let $\Omega$ denote the set $GF(q)\cup\{\infty\}$. Then the group $P\Gamma L_2(q)$ consists of bijections from $\Omega$ to itself the type
$$
P\Gamma L_2(q)=\{f:\Omega\rightarrow\Omega\mid f(z)=\frac{az^\sigma+b}{cz^\sigma+d}; a,b,c,d\in GF(q); ad-bc\neq0; \sigma\in Aut(GF(q))\}.
$$
This group has order $q(q^2-1)\cdot |Aut(GF(q))|$.
 A: By definition $P\Gamma L_2(q)$ is a subgroup of $\mathrm{Sym}(\Omega)\simeq S_{q+1}$. 
The claims would follow immediately from the knowledge of the order of the group $P\Gamma L_2(q)$:
When $q=3$, the group $Aut(GF(3))$ is trivial, and we have $|P\Gamma L_2(3)|=24=|S_4|$, so it follows that $P\Gamma L_2(q)$ must be all of $S_4$.
When $q=4$, the group $Aut(GF(4))$ is cyclic of order tow, and we have 
$$|P\Gamma L_2(4)|=4\cdot(4^2-1)\cdot2=120=|S_5|.$$ Again we see that $P\Gamma L_2(4)$ must be all of $S_5$.
So the real question is to prove the formula for the order of the group in these two cases.
In the case $q=3$ we observe that the transformations $z\mapsto z+1$ and $z\mapsto 2z$
give the permutations $\alpha=(012)(\infty)$ and $\beta=(12)(0)(\infty)$ of $\Omega$ respectively.
Clearly these two generate the group $S_3$ acting on the affine part $GF(3)$.
The transformation $z\mapsto 1/z$ gives the permutation $\gamma=(0\infty)(1)(2)$. Together with $\alpha$ we get a group that acts transitively on $\Omega$, because $\gamma(\infty)=0$,
and $(\alpha^i\gamma)(\infty)=i$ for $i=1,2$. Therefore the group generated by all of
$\alpha,\beta,\gamma$ must have at least $4\cdot|\langle \alpha,\beta\rangle|=4\cdot6=24$
elements, and hence be all of $S_4$.
For the case $q=4$ I write $GF(4)=\{0,1,a,b=a+1=a^2\}$. Here $F:x\mapsto x^2$ is the
only non-trivial automorphism. It corresponds to the permutation $\phi=(0)(1)(ab)(\infty)$.
In addition to that we have $z\mapsto z+1$ giving us $\alpha=(01)(ab)(\infty)$, and
$z\mapsto az$ giving us the permutation $\beta=(0)(1ab)(\infty)$. The elements $\phi$ and $\beta$
fix both $0$ and $\infty$, and we see that they generate a copy of $S_3$ acting on $\{1,a,b\}$. The group of translations $K$ generated by $\alpha$ and $\beta\alpha\beta^{-1}$, the latter
corresponding to $z\mapsto z+a$, gives us a copy of the Klein four group acting transitively on the finite part $GF(4)$. As in the preceding case we thus see that the group
$H=\langle \alpha,\beta,\phi\rangle$ generated by these three transformations must act as $S_4$ on $GF(4)=\Omega\setminus\{\infty\}$ because it has at least $4\cdot6=24$ elements.
It remains to show that adding the generator $\gamma=(0\infty)(1)(ab)$ gives us a group of size at least $120$. This follows from transitivity as in the case $q=3$. We can map $\infty$ to $0$ with $\gamma$, so $\gamma$ followed by an appropriate element of $K$
maps $\infty$ to any other element of $\Omega$. Therefore the order of the group
$G=P\Gamma L_2(4)$ is at least $5\cdot|H|=120$. As $G$ was known to be a subgroup of $S_5$, the isomorphism $G\simeq S_5$ follows.
