Why is $PGL(2,4)$ isomorphic to $A_5$ In the tradition of this question,

why is $\operatorname{PGL}(2,4)\cong A_5$?

 A: A starting point is that a 2-dimensional vector space over the field of 4 elements has 5 1-dimensional subspaces (=the projective line over $F_4$). The group $PGL(2,4)$ then acts on the set of those lines (well, $GL(2,4)$ acts naturally, but the scalar matrices keep the lines fixed setwise). Thus we have a homomorphism $f:PGL(2,4)\rightarrow S_5$. Moreover, if a linear mapping keeps all those 1-dimensional subspaces fixed (setwise), then that linear mapping has to be a scalar multiplication. So $f$ is injective. Thus the image of $f$ is of order 60, and hence must be equal to $A_5$.
A: A minor comment: questions such as these maybe should be phrased slightly differently, in my view. In this case, the question is perhaps "what is a good explanation of this isomorphism, or an easy way to see it?" 
An algebraic way to see the isomorphism is something like this. Note that ${\rm GL}(2,4)$
has a center of order $3$, consisting of scalar matrices. The group of scalar matrices of order dividing $3$ is a normal subgroup $N$ which intersects ${\rm SL}(2,4)$ in the identity, because no non-identity element of $N$ has determinant $1$. Hence we see that 
${\rm PGL}(2,4)$ is isomorphic to ${\rm SL}(2,4)$. Also, it has order $60$. A Sylow $5$
subgroup $S$ of ${\rm SL}(2,4)$ permutes the Sylow $2$-subgroups of ${\rm SL}(2,4)$ by conjugation. None of these Sylow $2$-subgroups is stabilised by $S$, since they each have order $4$, and admit no automorphism of order $5$. Hence there are at least $5$ Sylow $2$-subgroups of ${\rm SL}(2,4).$ By Sylow's theorem, there are no more than that.
${\rm SL}(2,4)$ has no subgroup $M$ of index $2$, for such a subgroup would be normal,
and would either have one or six Sylow $5$-subgroups. If there were one, it would be normal in ${\rm SL}(2,4)$, which is impossible (for example, an element of order $5$ in ${\rm SL}(2,4)$ has centralizer of order $5$, so the normalizer of a Sylow $5$-subgroup has order at most $20$). If there were $6$ subgroups of order $5$ in $M$, then there would only be $6$ elements in $M$ not of order $5$, and then $M$ would have a normal Sylow $3$-subgroup. But that would be normal in ${\rm SL}(2,4)$, while an element of order $3$ in ${\rm SL}(2,4)$ has centralizer of order $3$, so the normalizer of a Sylow $3$-subgroup in ${\rm SL}(2,4)$ has order at most $6$.
The permutation action of ${\rm SL}(2,4)$ on its Sylow $2$-subgroup yields a homomorphism
from ${\rm SL}(2,4)$ into $A_5$ (since ${\rm SL}(2,4)$ has no subgroup of index $2$).
We have already noted that no Sylow $2$-subgroup of ${\rm SL}(2,4)$ is normalized by an element of order $5$. ${\rm SL}(2,4)$ has no non-trivial normal $2$-subgroup (such a subgroup could not be a Sylow $2$-subgroup, as we have seen. On the other hand, if it had order $2$, it would be a central subgroup, and ${\rm SL}(2,4)$ has trivial center).
We have also seen that a Sylow $3$-subgroup of ${\rm SL}(2,4)$ is not normal. Thus 
no non-identity element of ${\rm SL}(2,4)$ normalizes every Sylow $2$-subgroup. Hence
${\rm SL}(2,4)$ is isomorphic to a subgroup of $A_5$ and must be all of $A_5$ by order considerations.
