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

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 $A_5$.

• Presumably the outer automorphism of ${\rm Sym}(6)$ permtues this copy of ${\rm PGL}(2,5)$ inside ${\rm Sym}(6)$ with the usual ${\rm Sym}(5)$, but that doesn't seem more elementary than the answers in the other thread. – anon Aug 15 '17 at 4:49
• groupprops.subwiki.org/wiki/PGL(2,5)_is_isomorphic_to_S5 may be of interest. Also, math.stackexchange.com/questions/376464/why-is-pgl-25-cong-s-5. Meanwhile, en.wikipedia.org/wiki/Projective_linear_group#Finite_fields says "there is no particularly natural set of 5 elements on which PGL(2, 5) acts." – Gerry Myerson Aug 15 '17 at 6:44
• Also, jaapsch.net/puzzles/pgl25.htm might be helpful. – Gerry Myerson Aug 15 '17 at 6:50
• Thanks everyone for all your help but we don't cover automorphism groups to much depth in the course I am doing so I have no idea how to show results for the automorphism group of alternating and projective special linear groups. The first part of this question asked me to show that $PSL_2(\mathbb{F}_5) \cong A_5$ which I managed to do but that bit of the question was structured and in steps. As an aside would it be possible to show that there is an order preserving map between some elements of order 2 in the projective group and all transpositions in $S_5$ and show the isomorphism that way? – Abdul Hadi Khan Aug 15 '17 at 8:15
• Can you post the entire question? I'm curious if you've shown an explicit isomorphism between $PSL(2,5)$ and $A_5$. – Steve D Aug 17 '17 at 16:16

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.
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 $\Bbb{F}_5^2$.