Resent-Date: Thu, 11 May 95 16:02:34 BST Comments on problem 2: A group such as 2^3:7:3 has the required properties. The elements of order 3 are of two types, those that square the 7-elements and those that 4th power them. Therefore no automorphism can map one type to the other (its inverse). Rob Wilson and Richard Parker. ________________________ Replied: isaacs@math.wisc.edu Subject: Solution to prob 2 Resent-Date: Mon, 3 Jul 95 11:23:31 BST The nonabelian group H of order 21 almost meets the requirements of the problem. It has a cyclic derived factor group and an element of order 3 is not inverted by any automorphism. To build a derived length 3 example, just let H act appropriately on something abelian and coprime, for example an elementary abelian group E of order 8. If G is the semidirect product EH of order 168, then G still has cyclic derived factor and any automorphism of G induces an automorphism of the factor group H = G/E since E is characteristic. Thus no automorphism of G inverts an element of order 3 of H. To see why no automorphism of H inverts an element x of order 3, note that H is a normal subgroup of its automorphism group A and so we need to show that x is not conjugate to its inverse in A. But if x were conjugate to its inverse, its image in the factor group A/N would also be conjugate to its inverse (in A/N). We can let N be the centralizer of C in A, where C is the subgroup of order 7 in H. Then x maps to an element of order 3 in A/N and since A/N is abelian, this element is not conjugate to its inverse. Marty Isaacs