
Re: name for definition in group theory
Posted:
Mar 24, 2013 12:44 PM


On Sun, 24 Mar 2013 10:25:18 0600, David C. Ullrich <ullrich@math.okstate.edu> wrote:
>On Sun, 24 Mar 2013 08:15:15 0700 (PDT), Paul <pepstein5@gmail.com> >wrote: > >>Does anyone know the name for the following property of a group G: G has no nontrivial automorphisms. ? >>Thank you > >These groups are referred to as "groups of order 1 or 2". > >There must be a very elementary proof of this. I know >no group theory; here's a not quite elementary proof >using a big result from harmonic analysis: > >A topological group is a group together with a topology >such that the group operations are continuous. >A (continuous) character of a topological group G >is a continuous homomorphism of G into the unit >circle in the complex plane. If G is a topological >group then the set of continuous characters is >denoted G^*; note that G^* is itself a group, >with multiplication defined pointwise. > >Now, if G is a locally compact abelian (LCA) group >then the Pontryagin Duality Theorem states that >G is isomorphic to its second dual (G^*)^*. >That's the nontrivial part. > >(Oops, there's a missing definition there. If G is >a LCA group then there is a natural topology on >the group G^*; it turns out that G^* is also LCA.) > >Ok. Assume G is a group with no nontrivial >automorphisms. Since all the inner automophisms >of G are trivial, G must be abelian. > >Give G the discrete topology. Now G is an LCA >group. Let K = G^*. (K is compact, not that >we need that here.) Then G is isomorphic to >K^*. > >Now, if chi is a character of K then chi^*, the >complex conjugate, is also a character of K. >The map chi > chi^* is an automorphism of >K^*. This automorphism must be trivial, so >every chi in K^* must be realvalued. > >So every chi in K^* takes only the values 1 and 1. >Hence every nontrivial element of K^* has order 2. > >So. G is an abelian group and every nontrivial element >of G has order 2. This means that G is a vector space >over the field Z_2 = {0,1}. > >If dim(G) = 0 or 1 then G = 1 or 2. If dim(G) > 1 then >G has a nontrivial automorphism. > >The elementary proof would start by noting that G must >be abelian, as above, and then invoke some structure >theorem or something to deduce that every element >has order 2, since any cyclic group of order greater >than 2 has an automorphism...
No wait! There's a proof even more elementary than that, using no group theory, nothing but the definitions:
Say G is a group with no nontrivial automorphisms. As above, since all the inner automorphisms of G are trivial, G must be abelian. So we will write G additively, as traditional.
Since G is abelian, the map x > x is an automorphism. Since this must be trivial, we have x + x = 0 for all x. Hence G is a vector space over Z_2. And now as above, if dim(G) = 0 or 1 then G = 1 or 2, while if dim(G) > 1 then G has a nontrivial automorphism.
> >> >>Paul Epstein

