Why is $PGL(2,4)$ isomorphic to $A_5$

Question

In the tradition of this question,

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

Answer 1

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 $A_5$.

Answer 2

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.

Answer 3

My answer is an elementary and straightforward one. As Jyrki Lahtonen wrote, $PGL_2(\mathbb{F}_4)$ permutes the points of $\mathbb{P}_1(\mathbb{F}_4)$ so there is a homomorphism $\varphi: PGL_2(\mathbb{F}_4) \to S_5$.
It is injective cause $\varphi(A) = id \implies A = \lambda E \equiv E$ in $PGL_2(\mathbb{F}_4).$
Now let's look at cyclic types in $S_5$. These are
$\small \bullet \ (5)$ with $\dfrac{5!}{5}=24$ elements,
$\small \bullet \ (4,1)$ with $\dfrac{5!}{4}=30$ elements,
$\small \bullet \ (3, 2):$ $\dfrac{5!}{3 \cdot 2}=20$,
$\small \bullet \ (3, 1, 1):$ $\dfrac{5!}{3 \cdot 2!}=20$,
$\small \bullet \ (2, 2, 1):$ $\dfrac{5!}{2^2\cdot 2!}=15,$
$\small \bullet \ (2, 1, 1, 1):$ $\dfrac{5!}{2 \cdot 3!}=10,$
$\small \bullet \ (1, 1, 1, 1, 1)$ with only one element, $id$.
$Im(\varphi)$ is a subgroup of $S_5$ so it contains $id$. To make $\varphi$ isomorphic to $PGL_2(\mathbb{F}_4)$ we need $59$ more elements. If it contains a permutation from some conjugacy class, it contains all permutations of this class.
Surely we need $24: (5)$ and $15: (2, 2, 1)$ to make the number end with 9. There are $59 - 24-15 = 20$ left. These are the classes $(3, 2)$ or $(3, 1, 1)$. If we take $(3, 2),$ we get, for example, $\langle 123 \rangle \langle 45 \rangle \langle 12345 \rangle = \langle 1324 \rangle$, but we cannot use the cyclic type $(4,1)$ with $30$ elements.
This means we have only all even permutations in our subgroup, so it is $A_5$.