Proof of isomorphism between $\text{PGL}_2(\mathbb{F}_5)$ and $S_5$

Question

This question has been asked here before but I don't think any of the previous answers are clear to someone like me who only has an elementary background in abstract algebra. So can I take the time to ask once again: Why do we have $PGL_2(\mathbb{F}_5) \cong S_5$?

So far I have tried to find an action of $GL_2(\mathbb{F}_5)$ on a set with 5 elements but have had no luck. However if you let $GL_2(\mathbb{F}_5)$ act on the projective line $P^1(\mathbb{F}_5)$ then we get a homomorphism to $S_6$ whose kernel is the set of scalar matrices which is exactly $Z(GL_2(\mathbb{F}_5))$. So we get an isomorphism from $PGL_2(\mathbb{F}_5)$ to a subgroup of $S_6$ . I then tried to consider the action of $S_6$ on $S_6:PGL_2(\mathbb{F}_5)$ and tried to use that to show the isomorphism but it didn't help.

Does anyone know how it might be possible to proceed from here or am I going down completely the wrong track? Any hints are much appreciated!

EDIT: Here is the full question as requested:

Show that the groups $SL_2(\mathbb{F}_4)$ and $PSL_2(\mathbb{F}_5)$ both have order 60. Use this and some results from previous questions to show that they are both isomorphic to the alternating group $A_5$. Show that $SL_2(\mathbb{F}_5)$ and $PGL_2(\mathbb{F}_5)$ both have order 120, that $SL_2(\mathbb{F}_5)$ is not isomorphic to $S_5$, but $PGL_2(\mathbb{F}_5)$ is.

The previous questions which the question refers to (and I was able to do) were:

1. Let $G$ be a group of order 60 which has more than one Sylow 5-subgroup. Show that $G$ is simple.

2. Let $G$ be a simple group of order 60. Deduce that $G \cong A_5$, as follows. Show that $G$ has six Sylow 5-subgroups. By considering the conjugation action of the set of Sylow 5-subgroups, show that $G$ is isomorphic to a subgroup $G \leq A_6$ of index 6. By considering the action of $A_6$ on $A_6:G$, show that that there is an automorphism of $A_6$ taking $G$ to $A_5$.

Answer 1

Here's a proof of what I said in the comments (which solves your problem, but doesn't provide an explicit isomorphism).

Let $H\le S_n$ be a subgroup of index $n$, with $n\ge5$. Consider the action of $S_n$ on the cosets of $H$. The kernel of this action is trivial (since the only possible kernel is $A_n$, but $A_n\not\subset H$). Thus, $H$ acting on its own cosets is also a faithful action. Since $H$ is a subgroup, it fixes the trivial coset ($H$ itself). The action on the non-trivial cosets gives an injection of $H$ into $S_{n-1}$. Order considerations show this is an isomorphism.

Answer 2

If we consider the conjugacy classes of $S_6$ of the subgroups isomorphic to $S_5$ we discover the astonishing fact that there are in fact two different classes. One is the classic known class of naturally embedded copies of $S_5$ in $S_6$ where one of the six moved points of $S_6$ is left invariant; but there is another class that acts on six points without leaving one of them invariant. A representative of this class (of 6 conjugate subgroups) is the group generated by $(1,2,3)(4,5,6)$ and $(1,3,4,6,2,5)$. This group act transitively on the points $\{1 \ldots 6\}$ and so can be related to the action of $\operatorname{PGL(2,\Bbb{F}_5)}$ on the six lines through the origin in $\Bbb{F}_5^2$.