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.

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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