Find an isomorphism $PGL_2(F_3) \cong S_4$

Question

I am struggling to find an explanation why this true, I know and I'm sorry that kind of question is commonly asked, although I couldn't find anything about this particular question.

Answer 1

As mentioned by Qiaochu, the isomorphism follows from the following observations:

(general observations)

• there is a homomorphism $\varphi_p\colon\mathrm{PGL}_2(p)\to S_{p+1}$ given by the action of $\mathrm{PGL}_2(p)$ on the $(p+1)$-element set $\mathbb{PF}_p^1$.
• $\varphi_p$ is injective, i.e., the action above is faithful. This is shown as follows: for any $g\in\mathrm{GL}_2(p)$, note that $g(1,0)=(a,0)$ and $g(0,1)=(0,b)$ for some constants $a,b\in\mathbb F_p^\times$. But now $g(1,1)=(a,b)$ so $a=b$.
• $\mathrm{PGL}_2(p)$ has order $(p^2-1)(p^2-p)/(p-1)=p(p^2-1)$, see, e.g. Order of general- and special linear groups over finite fields. Moreover, $S_{p+1}$ has order $(p+1)!$.

Now when $p=3$, the order of $\mathrm{PGL}_2(3)$ is $3\cdot(3^2-1)=24$ and $S_4$ has order $24$, so together with the injectivity of $\varphi_3$, we deduce $\varphi_3$ must be an isomorphism.