Order statistics-equivalent not implies 1-isomorphic
It is possible to have two finite groups and are Order statistics-equivalent finite groups (?) in the sense that have the same order statistics (i.e., for any natural number, and have the same number of elements of order equal to that natural number) but such that and are not 1-isomorphic groups, in the sense that there is no bijection between them that is a 1-homomorphism of groups (?) both ways.
Further information: groups of order 16#Element structure
The key thing to observe is that the order statistics capture only the number of elements of each order, whereas a 1-isomorphism preserves the structure of cyclic subgroups. In particular, if and are 1-isomorphic, they have the same number of elements of order that are squares. Thus, if we find two groups with the same order statistics but where the number of elements of order that are squares differs, then we have the required counterexample.
We have a few such counterexamples for groups of order 16:
- SmallGroup(16,3) and direct product of Z4 and V4 both have the same order statistics: element of order , elements of order , and elements of order . However, in SmallGroup(16,3), two of the elements of order are squares, whereas in direct product of Z4 and V4, exactly one of the elements of order is a square.
- Consider the three groups direct product of Z4 and Z4, nontrivial semidirect product of Z4 and Z4, and direct product of Q8 and Z2. All these groups have the same order statistics: element of order , elements of order , and elements of order . However, no two of them are 1-isomorphic. The first group has three elements of order that are squares, the second group has only two elements of order that are squares, and the third group has only one element of order that is a square.