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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.

amWhy
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Steve
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    $PGL_2(\mathbb{F}_3)$ naturally acts on $\mathbb{P}^1(\mathbb{F}_3)$, which has $4$ elements. This gives a homomorphism and now you can try to show that it's injective and surjective. – Qiaochu Yuan May 10 '16 at 19:02
  • @QiaochuYuan Thanks, that means somehow I need to show that the kernel of the homomorphism is trivial to show injectivity? And is there such a thing to show that it must be surjective too? – Steve May 10 '16 at 19:13
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    For a proof see here, at $(9)$. – Dietrich Burde May 10 '16 at 19:18

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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.

Kenta S
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