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How to prove that $Aut(PSL(2,7))=PGL(2,7)$? Is this result extendible to $PSL(n,q)$ where $q=p^n$?

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OK, here is a quick direct argument to show that Aut(PSL$(2,7)$) = PGL$(2,7)$. Let $\alpha \in {\rm Aut}({\rm PSL}(2,7)$. Now PSL$(2,7)$ acts 2-transitively by conjugation on its 8 Sylow $7-$subgroups. Since these must also be permuted under the action of $\alpha$, $\alpha$ is induced by conjugation by an element $a \in S_8$. By multiplying $a$ by an inner automorphism, we may assume that it fixes a specific Sylow $7-$subgroup $S$ of PSL$(2,7)$. Since the full automorphism group of $S$ (i.e. the cyclic group of order 6) is induced on $S$ within PGL$(2,7)$, by

multiplying $a$ be an element of PGL$(2,7)$, we may assume that $a$ centralizes $S$. Then, by multiplying $a$ by an element of $S$, we can assume that $a$ fixes some other point as an element of $S_8$, but then the fact that it centralizes $S$ forces $a=1$.

The same argument works for ${\rm Aut}({\rm PSL}(2,p)$ for any odd prime $p$. For ${\rm PSL}(n,p^k)$ with $k>1$ there are also field automorphisms, and for $n \ge 3$, there is also the graph automorphism (induced by inverse-transpose on ${\rm SL}(n,p^k)$), but you would probably need to learn some more general theory of classical groups or groups of Lie type to understand that.

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