The aim of this short note is to prove the following result:

Theorem. Let G be a group of finite Morley rank with Aut G acting transitively on G/{1}. Then G is either abelian or a bad group.

Bad groups were first defined by Cherlin [Ch]: these are groups of finite Morley rank without solvable and nonnilpotent connected subgroups. They have been investigated by the author [Ne 1], Borovik [Bo], Corredor [Co], and Poizat and Borovik [Bo-Po]. They are not supposed to exist, but we are far from proving their nonexistence. This is one of the major obstacles to proving Cherlin's conjecture: infinite simple groups of finite Morley rank are algebraic groups.

If the group G of the theorem is finite, then it is well known that G ≈ ⊕Zp for some prime p: clearly all elements of G have the same order, say p, a prime. Thus G is a finite p-group, so has a nontrivial center. But Aut G acts transitively; thus G is abelian. Since it has exponent p, G ≈ ⊕Zp.

The same proof for infinite G does not work even if it has finite Morley rank, for the following reasons:

1) G may not contain an element of finite order.

2) Even if G does contain an element of finite order, i.e. if G has exponent p, we do not know if G must have a nontrivial center.