From: mareg@primrose.csv.warwick.ac.uk () Subject: Re: Questions about PGL(n,q) Date: Wed, 25 Apr 2001 13:46:13 +0000 (UTC) Newsgroups: sci.math Summary: Simplicity of PLG_n(K) In article , "Jim Heckman" writes: >Does PGL(n,q) always have trivial center? If so, can this be >proved easily? Is PGL(n,q) always a subgoup of Aut(PSL(n,q)) ? Yes, fairly easily, and yes. If n=1, PGL(n,q) is trivial. If n=2, q=2, PGL(n,q) = S_3, PSL(n,q) = A_3. If n=2, q=3, PGL(n,q) = S_4, PSL(n,q) = A_4. So the answer is yes in those cases. In all other cases, PSL(n,q) is simple. Let Z be the nonzero scalars, so PGL(n,q) = GL(n,q)/Z. Let X/Z be the centralizer of PSL(n,q) in PGL(n,q). Since PGL(n,q)/PSL(n,q) is cyclic, so is X/Z. So PSL(n,q) acts on the group X, in such a way that its actions on the cyclic groups Z and X/Z are both trivial. It is easy to check that the group of automorphisms of X acting trivially on Z and X/Z is abelian, so the simplicity of PSL(n,q) implies that its action on X is trivial. Hence X is centralized by SL(n,q), so X=Z. It follows now that the answers to both of your questions is yes. This proof uses the fact that Z is the centralizer of SL(n,q) in GL(n,q), which is not hard to prove, and that PSL(n,q) is (usually) simple. But if you are going to study the groups GL(n,q) at all, then you probably want to learn about the simplicity of PSL(n,K) fairly early on! Derek Holt.