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

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David C. Ullrich

Posts: 21,553
Registered: 12/6/04
Re: name for definition in group theory
Posted: Mar 24, 2013 12:44 PM
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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
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 non-trivial automorphism.

>
>>
>>Paul Epstein





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