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1-closed transversal not implies permutably complemented


Property-theoretic statement

A subgroup having a 1-closed transversal need not be a permutably complemented subgroup.

Verbal statement

It is possible to have a group G and a subgroup H of G such that H has a left transversal that is a 1-closed subset of G but H does not have any permutable complement.


Example of a non-abelian group of prime exponent

Further information: prime-cube order group:U(3,p)

Let p \ne 2 be a prime number. Let G be a non-abelian group of order p^3 and exponent p (hence G is isomorphic to prime-cube order group:U(3,p)). Let H be the center of G. H is a normal subgroup of G and it has no permutable complement because nilpotent and non-abelian implies center is not complemented (which in turn follows from the fact that nilpotent implies center is normality-large). On the other hand, H does have a 1-closed transversal.

More generally, any subgroup of a group of prime exponent has a 1-closed transversal but is not necessarily permutably complemented.