Why is SL(2,Z) the Outer-Automorphism Group of Z^2?
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#1
electroweak
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I know that the outer-automorphism group of Z^2 is SL(2,Z). Can someone please show me why this is the case? I think Aut(Z^2)=GL(2,Z), but what about Inn(Z^2)? Thanks.
Note that Z^2 is abelian, so Inn(Z^2) = ? and consequently Out(Z^2) = ?.
#3
electroweak
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Aut(Z^2)=GL(2,Z), and Inn(Z^2)=Z^2/center(Z^2)=1, so that Out(Z^2)=Aut/Inn=GL(2,Z), right? OK, I figured out what was confusing me; I was applying the Dehn-Neilson theorem (which only holds on hyperbolic surfaces) to the torus (a parabolic surface). This would have equated Out(Z^2) and SL(2,Z). Thanks for confirming my suspicions!
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).
Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?