Why is $PGL_2(5)\cong S_5$?

group-theory finite-groups permutations exceptional-isomorphisms groups-of-lie-type

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


Solution 1:

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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