Given the field F_p of prime order p, we can form the group
PSL(n,p) = SL(n,p)/{+-I}, where SL(n,p) is the group of vector
space automorphisms of (F_p)^n having determinant = 1.
Ignoring n=1, since PSL(1,p) is the trivial group:
It's known that PSL(n,p) is simple for all n >= 2 and p prime,
except for the two cases (n,p)=(2,2) and (n,p)=(2,3).
Also, PSL(n,p) is isomorphic to PSL(n',p') implies (n,p)=(n',p'),
except for one case: PSL(2,3) is isomorphic to PSL(7,2).
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QUESTIONS: What about n = oo ? Can we find an analogous simple
group?
I don't know how to define the determinant of an arbitrary
automorphism of the vector space over F_p of countable dimension.
Let's call this (F_p)^oo.
Its group of automorphisms GL(oo,p) is well-defined.
This has a normal subgroup: those automorphisms that fix all but a
finite number of elements. Let's call this GL_f(oo,p).
The quotient group Q_p = GL(oo,p)/GL_f(oo,p) still has a normal
subgroup, namely the nonero scalars Sc(oo,p) = {[kI] in Q_p : k in (F_p)*},
so let PQ(p) = Q_p/Sc(oo,p)
Q1. Is there any chance that PQ(p) is simple for some primes p, or
maybe all of them?
Q2. Is there some reason to believe that PQ(p) is isomorphic to
PQ(p') always implies p=p' ?
--Dan
