Automorphisms of groups

I set a question to my group theory class. It is quite easy until suddenly you find yourself falling off a cliff into deep water. Indeed, I don’t know the answer to the second part of this question.

Show that a finite group of order greater than 2 has a non-trivial automorphism. What can you say about infinite groups?

For the first part, we divide groups into three classes:

  • If the group is non-abelian, then conjugation by an element not in the centre is a non-identity (inner) automorphism.
  • If the group is abelian with exponent greater than 2, then the map taking every element to its inverse is a non-identity automorphism.
  • Suppose that G is abelian with exponent 2, written additively. Define scalar multiplication by elements of the two-element field F={0,1} by 0.x=0 and 1.x=x. It is readily checked that this makes G into a vector space over F. (The only non-trivial instance of an axiom is (1+1).x=1.x+1.x, which holds because the group has exponent 2.) Now choose a basis for G over F. Since |G|>2, there are at least two basis vectors; the map which interchanges the first two and fixes the others extends to a non-trivial automorphism of G.

Almost the entire argument goes through for infinite groups, all the way down to where it says “Now choose a basis …”. What happens next depends on our set-theoretic universe. If we work in ZFC (that is, if we assume the Axiom of Choice), then every vector space has a basis, and the argument concludes. But there are models of ZF in which there exist vector spaces with no basis. Can one get by with less?

If G*, the vector space dual of G (the set of linear maps from G to F) is non-zero, then we can choose a non-zero element f of G*, and construct a transvection as follows: pick a non-zero vector a in the kernel of f, and map x to x+f(x)a; this is an automorphism. The existence of a non-zero linear functional is weaker than the existence of a basis, since if there is a basis then we can map its elements to F arbitrarily and extend to a linear map.

So we can divide the question as follows. For each of the following statements, does there exist a model of ZF set theory in which it is true?

  • Every non-zero vector space has non-zero dual space, but there is a vector space which doesn’t have a basis.
  • Every vector space containing two linearly independent vectors has a non-scalar automorphism, but there is a non-zero vector space with zero dual space.
  • There exists a vector space containing two linearly independent vectors, all of whose automorphisms are scalars.

And in particular, do these statements hold for vector spaces over the two-element field?

I will post this on MathOverflow and report responses here.

About Peter Cameron

I count all the things that need to be counted.
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4 Responses to Automorphisms of groups

  1. Ralph says:

    How can an entity call itself a vector space if it doesn’t have a basis? It would be like one of those unnameable real numbers about which one can only say that its digits cannot be calculated — like Chaitin’s omega, but no, much worse. His number can be named and discussed, even though you can never know how close you are to the value. (Sorry, sir, there is no delta currently available to go with that epsilon. No, I don’t know when we will be getting them in.)

    Much worse than that. I tell you, something that cannot be discussed does not exist. Did a tree fall in the forest? Ask one of the angels standing on this pin. Maybe he will know.

  2. Asaf Karagila says:

    An answer is given here:

    • Thanks. That indeed answers the main question. But I think it does leave open which of my three subsidiary questions have affirmative answers (at least one does). Unless I missed something in the replies.

  3. Matt Daws says:

    The Mathoverflow link is:

    Peter: I think your MO question only asks the final of your 3 questions here (and it seems that Asaf completely answered that). Any chance you could edit your MO question to ask these additional questions? (Of course, this assumes I have read correctly; often an unwarranted assumption…)

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