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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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Why do you keep posting questions when you obviously feel the answers given to you are unacceptable? – Uticensis Mar 14 '11 at 9:07
@Billare: I am not forcing you to answer it!! Leave the question which you don't want to answer!! I can not post these questions for MO!! These things are written in the paper, without proof; which is correct, because the author knows that such facts have been proved. But if I didn't clarified with that, I would like to get some direction for that from someone. – user8186 Mar 14 '11 at 9:19
I think Billare means that it is polite to accept answers to questions you posted. There is nothing wrong with this question IMO. – Thomas Rot Mar 14 '11 at 12:40
Could someone tell me the subgroups lattice of $PSL(n,p^k)$ or could someone tell me sources I should read to know all about $Aut(PSL(n,p^k))$ and its subgroups lattice. Thanks in advance. – Kakalotte Apr 23 '11 at 15:27
@user9974: you should post that as a new question, not as an answer to an existing question. – Willie Wong Apr 23 '11 at 16:13

1 Answer 1

up vote 7 down vote accepted

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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