Paul
Posts:
411
Registered:
7/12/10


Re: name for definition in group theory
Posted:
Mar 24, 2013 12:49 PM


On Sunday, March 24, 2013 4:44:44 PM UTC, David C. Ullrich wrote: > 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
Thanks again. I'm just posting to confirm that I totally understand your second proof. Unfortunately, I don't have the background (or have forgotten the background) to follow your first proof, but your first proof is good motivation for me (as I said earlier).
Paul Epstein

