A friend pointed out to me << In place of "the automorphisms that fix all but a finite number of elements" you want "the automorphisms that fix a subspace of finite codimension."
The corrected questions, with slightly better notation: ------------------------------------------------------------------------------------------------ By analogy with the finite groups PSL(n,p) (n >=2), which are simple except for PSL(2,2) = S_3 and PSL(2,3) = A_4, (and, PSL(n,p) is uniquely determined by (n,p) except for the isomorphic pair PSL(3,2) = PSL(2,7)): For any prime p, let F(p) be the finite field of size p. Let F(p)^oo be the vector space of countably infinite dimension over F(p). Let G(oo,p) be the group of automorphisms Aut(F(p)^oo). This has a normal subgroup: those automorphisms that fix a subspace of finite codimension. Call this subgroup GL_f(oo,p). The quotient group Q(p) = GL(oo,p)/GL_f(oo,p) still has a normal subgroup (of size p-1), namely the nonzero scalars Sc(oo,p) = {kI in Q(p) : k in (F(p)*}, where I is the identity on Q(p). So, let PQ(p) denote 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