Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

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

 Messages: [ Previous | Next ]
 David C. Ullrich Posts: 21,553 Registered: 12/6/04
Re: name for definition in group theory
Posted: Mar 24, 2013 12:25 PM

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

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