# To complete the proof that $\operatorname{PSL}(2,\Bbb F_5)\cong A_5$

Elementary isomorphism between $\operatorname{PSL}(2,5)$ and $A_5$ I want to fill in the details of user David Speyer's algebraic proof that $\operatorname{PSL}(2,\Bbb F_5)\cong A_5$. I haven't actually been formally introduced to the concept of a projective linear group or a projective space (I'm reading Dummit & Foote's Abstract Algebra, I'm sure it won't talk anything deep about them), but I've learnt about linear fractional transformations and cross-ratios. I hope that helps:

Of course, I won't assume that $\operatorname{PSL}(2,\Bbb F_5)$ is simple. So we let $\operatorname{PGL}(2,\Bbb F_5)$ acts naturally on $\Bbb P^1(\Bbb F_5)$ (I don't exactly know how it actually works, but I've learnt about complex Möbius transformations, so I'm sure it is $\Bbb F_5\cup\{\infty\}$):$$\operatorname{PGL}(2,\Bbb F_5)\times\Bbb P^1(\Bbb F_5)\rightarrow \Bbb P^1(\Bbb F_5)$$$$(f,x)\mapsto f(x)$$where $f:x\mapsto\frac{ax+b}{cx+d}$ can be identified with the coset with representative $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ as in $\operatorname{PGL}(2,\Bbb F_5):=\operatorname{GL}(2,\Bbb F_5)/Z(\operatorname{GL}(2,\Bbb F_5))$

As a permutation on $6$ elements of $\Bbb P^1(\Bbb F_5)$, I checked for involutions in $S_6$. $15$ of them are products of three disjoint transpositions. They have no fixed point:$$\{(0\infty)(12)(34),(0\infty)(13)(24),(0\infty)(14)(23),(1\infty)(23)(04),(1\infty)(24)(03),(1\infty)(34)(02),(2\infty)(13)(04),(2\infty)(14)(03),(2\infty)(34)(01),(3\infty)(12)(04),(3\infty)(14)(02),(3\infty)(24)(01),(4\infty)(12)(03),(4\infty)(13)(02),(4\infty)(23)(01)\}$$ Now since a linear fractional transformation is completely decided by its action on $3$ points, I can write them in the form $x\mapsto\frac{ax+b}{cx+d}$ using cross-ratios because it is invariant under these transformations: for example $(0\infty)(12)(34):0\mapsto\infty,\infty\mapsto0,1\mapsto2$ so $[x,0;\infty,1]=[f(x),\infty;0,2]$, so the map is $x\mapsto\frac{2}{x}$. The $15$ involutions corresponds to (in consistent order with the permutations above):$$\{\begin{pmatrix}0&2\\1&0\end{pmatrix},\begin{pmatrix}0&3\\1&0\end{pmatrix},\dots,\begin{pmatrix}1&1\\1&4\end{pmatrix},\begin{pmatrix}1&2\\1&4\end{pmatrix},\dots,\dots,\begin{pmatrix}2&4\\1&3\end{pmatrix},\begin{pmatrix}2&3\\1&3\end{pmatrix},\begin{pmatrix}3&3\\1&2\end{pmatrix},\begin{pmatrix}3&4\\1&2\end{pmatrix},\dots,\dots,\begin{pmatrix}4&2\\1&1\end{pmatrix},\begin{pmatrix}4&1\\1&1\end{pmatrix}\}$$ where the $\dots$ are involutions $(0\infty)(14)(23),(1\infty)(34)(02),(2\infty)(13)(04),(3\infty)(24)(01),(4\infty)(12)(03)$ which I omitted since I tried to write them explicitly as linear fractional maps by investigating cross-ratios and then realised they induced different permutations: e.g. $(0\infty)(14)(23):0\mapsto\infty,\infty\mapsto0,1\mapsto4$, so I had $x\mapsto\frac{4}{x}$, but then it would in turn fixes $2$ and $3$, which is absurd.

Questions:

-The answer given by David Speyer suggested an action of $\operatorname{PGL}(2,\Bbb F_5)$ by conjugation on the $5$ involutions with fixed points, and then it gives us $\operatorname{PGL}(2,\Bbb F_5)\cong S_5$. I don't get it. I've checked above that there are indeed $5$ such involutions. Why is this action faithful?

-Let's assume I already know $\operatorname{PGL}(2,\Bbb F_5)\cong S_5$. Can I view $\operatorname{PSL}(2, \Bbb F_5)$ as a subgroup of $\operatorname{PGL}(2,\Bbb F_5)$ by saying that they are sets of linear fractional transformations? By doing so, I immediately concludes that $\operatorname{PSL}(2, \Bbb F_5)\cong A_5$ since $A_5$ is the unique subgroup of index $2$ in $S_5$.

• @JyrkiLahtonen Hmm... that comment repeats what David Speyer explained and reflects what I've just done in my attempt. It does not help at all. – user441558 Sep 24 '17 at 7:38
• Ok. Sorry. ${}{}$ – Jyrki Lahtonen Sep 24 '17 at 8:03
• I have a new observation: suppose $\tau\in PGL(2,\Bbb F_5)$ stabilizes, i.e. centralizes, the $5$ involutions outside $PGL(2,\Bbb F_5)$. Suppose $\tau$ fixes an element of $\Bbb P^1(\Bbb F_5)$, then it fixes the whole projective line. $\tau=id$. What if $\tau$ does not fix anything? I'm trying to arrive at a contradiction... – user441558 Sep 24 '17 at 8:10

You need to show that any element of $PGL(2,\Bbb F_5)$ commuting with each of the five involutions you list (that are outside $PGL(2,\Bbb F_5)$) is the identity. Then the conjugation action on these five permutations is faithful.
Then $PSL(2,\Bbb F_5)$ is a subgroup of $PGL(2,\Bbb F_5)$, and as $S_5$ has the unique index $2$ subgroup $A_5$, then it must be isomorphic to $A_5$.
• So... 'any element of $PGL(2,\Bbb F_5)$ commuting with each of the $5$ involutions... is the identity' is what I don't understand. I can only see that $\operatorname{Stab}_{PGL(2,\Bbb F_5)}\sigma$ where $\sigma$ is one of the $5$ involutions there, has order at most $48$. Why do these stabilizers intersect trivially? – user441558 Sep 24 '17 at 7:46
I think I have it now. Let $\operatorname{PGL}(2,\Bbb F_5)$ acts on $$\{(0\infty)(14)(23),(1\infty)(34)(02),(2\infty)(13)(04),(3\infty)(24)(01),(4\infty)(12)(03)\}$$ by conjugation:$$(\tau,\sigma)\mapsto\tau\sigma\tau^{-1}$$ Suppose $\tau$ stabilizes all of the $5$ involutions. Then$$(\tau(0)\tau(\infty))(\tau(1)\tau(4))(\tau(2)\tau(3))=(0\infty)(14)(23)$$$$(\tau(1)\tau(\infty))(\tau(3)\tau(4))(\tau(0)\tau(2))=(1\infty)(34)(02)$$$$(\tau(2)\tau(\infty))(\tau(1)\tau(3))(\tau(0)\tau(4))=(2\infty)(13)(04)$$ Suppose $\tau$ fixes $0$. By the first equality, $\tau$ fixes $\infty$. By the second equality, $\tau$ fixes both $1$ and $2$. By the third equality, $\tau$ fixes both $3$ and $4$. i.e. $\tau$ fixes everything. $\tau=id$. By symmetry, we claim that if $\tau$ fixes a point on $\Bbb P^1(\Bbb F_5)$, it fixes everything.
So we suppose that $\tau$ does not fix anything. By symmetry, we may assume that $\tau(0)=\infty$. Then $\tau(\infty)=0$ by the first equality. Then $\tau(2)=1$ by the secdond equality and $\tau(2)=4$ by the third equality. Contradiction.
Thus we must have $\tau=id$. i.e. the action is faithful.