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Topic: name for definition in group theory
Replies: 15   Last Post: Mar 26, 2013 11:35 AM

 Messages: [ Previous | Next ]
 Paul Posts: 780 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 non-trivial 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 non-trivial 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 non-trivial
>
> >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 real-valued.
>
> >
>
> >So every chi in K^* takes only the values 1 and -1.
>
> >Hence every non-trivial element of K^* has order 2.
>
> >
>
> >So. G is an abelian group and every non-trivial 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 non-trivial 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 non-trivial automorphisms.
>
> As above, since all the inner automorphisms of G
>
> are trivial, G must be abelian. So we will write G
>
>
>
>
> 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 non-trivial 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

Date Subject Author
3/24/13 Paul
3/24/13 David C. Ullrich
3/24/13 Paul
3/24/13 David C. Ullrich
3/24/13 Paul
3/24/13 Jose Carlos Santos
3/24/13 magidin@math.berkeley.edu
3/24/13 Jose Carlos Santos
3/24/13 Butch Malahide
3/24/13 Ken.Pledger@vuw.ac.nz
3/24/13 Paul
3/25/13 G. A. Edgar
3/25/13 G. A. Edgar
3/25/13 Paul
3/26/13 David C. Ullrich
3/25/13 David C. Ullrich