Epicenter not is fully invariant

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This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., epicenter) does not always satisfy a particular subgroup property (i.e., fully invariant subgroup)
View subgroup property satisfactions for subgroup-defining functions | View subgroup property dissatisfactions for subgroup-defining functions

Statement

It is possible to have a group G such that the epicenter Z^*(G) of G is not a fully invariant subgroup of G.

Proof

Further information: nontrivial semidirect product of Z4 and Z4, subgroup generated by a non-commutator square in nontrivial semidirect product of Z4 and Z4, subgroup structure of nontrivial semidirect product of Z4 and Z4

Suppose G is the nontrivial semidirect product of Z4 and Z4, given explicitly as follows, where e denotes the identity element:

G := \langle x,y \mid x^4 = y^4 = e, yxy^{-1} = x^3 \rangle

As we can see from the subgroup structure of G, the biggest quotient of G that is a capable group is dihedral group:D8, and this arises as the quotient of G by the subgroup H = \langle y^2 \rangle, which is a subgroup generated by a non-commutator square in nontrivial semidirect product of Z4 and Z4. Thus, H is the epicenter of G.

H is not fully invariant in G. For instance, it is not invariant under the endomorphism with kernel \langle x\rangle that sends y to x.