Date: Mar 24, 2013 12:25 PM
Author: David C. Ullrich
Subject: Re: name for definition in group theory
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...

>

>Paul Epstein