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Why is $PGL_2(5)\cong S_5$? And is there a set of 5 elements on which $PGL_2(5)$ acts?

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  • $\begingroup$ Related: math.stackexchange.com/questions/57986/… $\endgroup$ – vadim123 Apr 29 '13 at 18:24
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    $\begingroup$ If it helps, $PGL_2(\mathbb{F}_5)\cong \operatorname{Aut}(PSL_2(\mathbb{F}_5))$ and $S_5\cong \operatorname{Aut}(A_5)$ and $PSL_2(\mathbb{F}_5)\cong A_5$. (source) $\endgroup$ – Alexander Gruber Apr 29 '13 at 18:32
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    $\begingroup$ I think this might be explained in some of those exceptional automorphisms of S6 papers. S5 acts on its 6 Sylow 5-subgroups, embedding it weirdly in S6. That embedding is the same as PGL(2,5) on its projective plane. So it shows S5 is isomorphic to PGL(2,5) instead of vice versa. $\endgroup$ – Jack Schmidt Apr 29 '13 at 20:50
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As David Speyer explains,

there are 15 involutions of $P^1(\mathbb F_5)$ without fixed points (one might call them «synthemes»). Of these 15 involutions 10 («skew crosses») lie in $PGL_2(\mathbb F_5)$ and 5 («true crosses») don't. The action of $PGL_2(\mathbb F_5)$ on the latter ones gives the isomorphism $PGL_2(\mathbb F_5)\to S_5$.

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