# Group which has only one automorphism

35 views

### µ

Jun 4, 2007, 8:53:40 AM6/4/07
to
Hi everybody,

here's the theorem:
let G be a group whose only automorphism is identity. Then G has at most
two elements.

I know only the following proof which uses AC:

1) inner automorphisms (those of the form x->g x g^(-1)) are all
identity, so G is abelian;

2) thus x->x^(-1) is an automorphism of G, hence is identity; this shows
that every non-zero element of G has order 2, so G has a natural
structure of vector space over the field F_2 which has two elements;

3) suppose G has at least 3 elements. Let x and y two non-zero elements
of G. The family (x,y) is free over F_2. *Consider a basis B of G over
F_2 such that x and y belong to B*. Let f be the only vector space
endomorphism of G such that f(x)=y, f(y)=x and f(z)=z for every other
element z of the basis B. Then f is a group automorphism of G which is
not identity, a contradiction. Hence G has at most two elements.

Is it possible to avoid this strong use of AC?
I have no idea of the answer neither how two begin to study this problem...

--

### quasi

Jun 4, 2007, 11:04:13 AM6/4/07
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Yes, easy.

Now that you've proved that every nonidentity element has order 2,
suppose a,b are 2 distinct nonidentity elements.

Define a map f:G->G by:

f(a)=b, f(b)=a, f(x)=x for all x other than a,b.

Then f is an automorphism which not the identity.

Hence G has at most 2 elements.

quasi

### avital

Jun 4, 2007, 10:19:40 AM6/4/07
to
On Jun 4, 10:04 am, quasi <q...@null.set> wrote:

This is not an automorphism. f(ax) = ax (since ax is not a or b) but
f(a)f(x) = bx.

### quasi

Jun 4, 2007, 11:50:06 AM6/4/07
to
On Mon, 04 Jun 2007 14:19:40 -0000, avital <avital...@gmail.com>
wrote:

>On Jun 4, 10:04 am, quasi <q...@null.set> wrote:

Yep. I take it back -- it's not as simple as just transposing 2
elements. Lots of things will have to move -- it's a big job.

You need the axiom of choice.

quasi

### avital

Jun 4, 2007, 10:59:14 AM6/4/07
to
On Jun 4, 10:50 am, quasi <q...@null.set> wrote:
> On Mon, 04 Jun 2007 14:19:40 -0000, avital <avitaloli...@gmail.com>

> wrote:
>
>
>
> >On Jun 4, 10:04 am, quasi <q...@null.set> wrote:

Do you? This question is very interesting. If you do, can you show
that (The only group that has only one automorphism is of order 2) ==>
AC?

I'm not quite sure you need AC actually, but I'm still thinking about
it.

Avital.

### quasi

Jun 4, 2007, 1:03:19 PM6/4/07
to
On Mon, 04 Jun 2007 14:59:14 -0000, avital <avital...@gmail.com>
wrote:

It's not obvious.

But you need _something_.

To phrase the question more fully, consider the following two
statements:

(1) Every group with more than 2 elements has a nontrivial
automorphism.

(2) Every vector space has a basis.

Clearly (2) implies (1). The question is, does (1) imply (2)?

I'm not sure, but perhaps replace (2) with the following apparently
weaker version of (2)

(2') Every vector space over the field Z_2 has a basis.

Immediate question -- is (2') really weaker than (2)? It seems so, but
again, I'm not sure.

In any case, consider the revised question, does (1) imply (2')?

Assuming that the answer to the above question is "yes", a natural
line of attack to try to prove it would be as follows ...

Start with a vector space V over Z_2, and let f be a nontrivial
automorphism of the additive group of V. Using f, somehow "construct"
a basis of V. How to do that beats me. Maybe use the axiom of choice?
(Just kidding).

quasi

### Axel Vogt

Jun 4, 2007, 12:03:25 PM6/4/07
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Do you really need AC to see G ~= (F2*x + F2*y) + H
as _direct_ summands of vectorspaces over F2 and Hom
'commuting' with finite direct sums?

### Arturo Magidin

Jun 4, 2007, 12:06:16 PM6/4/07
to
In article <5cirerF...@mid.individual.net>,
Axel Vogt <&nor...@axelvogt.de> wrote:
>avital wrote:

[. G is an abelian group of exponent 2 with only one automorphism .]

>Do you really need AC to see G ~=3D (F2*x + F2*y) + H

>as _direct_ summands of vectorspaces over F2 and Hom
>'commuting' with finite direct sums?

I don't know. How would you prove the existence of such an H if G is
infinite?

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

### Randall Dougherty

Jun 4, 2007, 8:09:46 PM6/4/07
to

This has come up before ...

>From: r...@math.ohio-state.edu (Randall Dougherty)
>Newsgroups: sci.math
>Subject: Re: Rigid groups w.o. Axiom of Choice
>Date: 1 Apr 1997 23:46:26 GMT
>Message-ID: <5hs6oi\$7d4\$1...@mathserv.mps.ohio-state.edu>
>
>In article <5hjnci\$k...@vixen.cso.uiuc.edu>,
>>The recent thread on rigid (i.e. having no automorphism
>>aside from the identity) rings and fields got me wonder-
>>ing: can it be proven, without the Axiom of Choice, that
>>the only rigid group is Z/2Z, or even that there do not
>>exist arbitrarily large rigid groups? It's easy, without
>>the Axiom of Choice, to prove that a rigid group must be
>>an Abelian group of exponent 2. If we could prove, with-
>>out the Axiom of Choice, that any group of exponent 2 has
>>a group isomorphic to Z/2Z as a direct summand, it would
>>follow that Z/2Z is the only rigid group. But I don't
>>see, offhand, how to do this.
>
>Some form of AC is required for this, because one can use
>symmetric forcing methods (see chapter 21 of Jech's Set Theory)
>to produce a model of ZF where it fails. Starting in a ground
>model M of ZFC, let A be an infinite Abelian group of exponent 2.
>Let P be the forcing poset for adding independent Cohen-generic
>subsets s_a of omega for each a in A. The automorphisms of A
>(in the ground model) induce a corresponding group of automorphisms
>of P. For each finite subset E of A, one can consider the
>subgroup fix(E) of this automorphism group coming from the
>set of automorphisms of A fixing the elements of E (and therefore
>fixing all elements of the subgroup of A generated by E; this
>subgroup is still finite). The groups fix(E) generate a normal
>filter. Now the hereditarily symmetric names give a
>submodel N of the generic extension M[G] (G M-generic over P).
>N is a model of ZF, and in N one has the set B = {s_a : a in A}
>and the group operation defined by s_a +_B s_{a'} = s_{a +_A a'};
>this makes B into an infinite Abelian group of exponent 2.
>In M[G] one can see that B is isomorphic to A; however, in N
>no such isomorphism exists. In fact, one can show that, in N,
>B does not have an infinite proper subgroup, so B cannot have
>Z/2Z as a direct summand. Furthermore, in N, B is rigid.
>
>Producing a model in which there are arbitrarily large rigid groups
>looks more difficult, partly because "arbitrarily large" is
>a tricky concept in non-choice models. Perhaps some form
>of class-length iterated forcing will do the job.

Randall Dougherty rdo...@ccrwest.org
"I have yet to see any problem, however complicated, which when looked at in
the right way didn't become still more complicated." -- Poul Anderson

### Herman Rubin

Jun 4, 2007, 10:02:09 PM6/4/07
to
In article <46640b55\$0\$27202\$426a...@news.free.fr>,

The theorem is equivalent to every vector space G over
F_2 having the property that given any element w of G
there is a representation of G as a direct sum of {0, w}
and another vector space. This follows from the Boolean
prime ideal theorem, and I conjecture it is equivalent
to it.

--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558

### µ

Jun 5, 2007, 1:09:53 AM6/5/07
to
Randall Dougherty a écrit :

> This has come up before ...

[...]

Thanks. I hadn't found this old message.

--