Criteria for groups to be abelian

This is a review of some elementary criteria for a group to be abelian.

Lemma. Let G be a group. Then the following are equivalent:

  1. G is abelian,
  2. The map G \to G given by g \mapsto g^2 is a group homomorphism;
  3. The map G \to G given by g \mapsto g^{-1} is a group homomorphism;
  4. The diagonal G \subseteq G \times G is normal.

Proof. We prove that each criterion is equivalent to (1).

For (2), note that (gh)^2 = ghgh, which equals gghh if and only if gh = hg.

For (3), note that (gh)^{-1} = h^{-1}g^{-1}, which equals g^{-1}h^{-1} if and only if gh = hg.

For (4), clearly \Delta_G \colon G \hookrightarrow G \times G is normal if G is abelian. Conversely, note that (e,h)(g,g)(e,h^{-1}) = (g,hgh^{-1}), which is in the diagonal if and only if gh = hg. \qedsymbol

3 thoughts on “Criteria for groups to be abelian

    • Fascinating; I have never heard of this. Are you saying that this is true for any group G? Do you know where I can read more about this?

      • Hello, Remy 🙂
        Yes, I meant that any group G such that the three maps g\mapsto g^{28}, g\mapsto g^{50}, g\mapsto g^{81} are three group endomorphisms must be an abelian group.

        You can find conditions of this type among the exercises in Herstein’s algebra textbook (which is now a bit too old-fashioned, but contains a fantastic wealth of good exercises). For example you can prove that if g\mapsto g^{n} is an endomorphism for three consecutive values of n, then G must be abelian; or that if g\mapsto g^{3} and g\mapsto g^{5} are both endomorphisms, then G must be abelian.
        I liked these exercises a lot as a student, so I was very pleased to see the topic addressed fully in this nice elementary paper:
        Abelian Forcing Sets, Joseph A. Gallian and Michael Reid, The American Mathematical Monthly, Vol. 100, No. 6 (Jun. – Jul., 1993), pp. 580-582

Leave a Reply

Your e-mail address will not be published. Required fields are marked *