# Center not is normality-preserving endomorphism-invariant

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., center) doesnotalways satisfy a particular subgroup property (i.e., normality-preserving endomorphism-invariant subgroup)

View subgroup property satisfactions for subgroup-defining functions View subgroup property dissatisfactions for subgroup-defining functions

## Contents

## Statement

It is possible to have a group such that the center is not normality-preserving endomorphism-invariant, i.e., there exists a normality-preserving endomorphism of (i.e., sends normal subgroups to normal subgroups) but is not contained in .

## Related facts

### Similar facts

### Opposite facts

Below are some properties *weaker* than being normality-preserving endomorphism-invariant, that the center in fact satisfies:

Property | Meaning | Proof that center satisfies the property |
---|---|---|

strictly characteristic subgroup | invariant under all surjective endomorphisms | center is strictly characteristic |

finite direct power-closed characteristic subgroup | finite direct power of subgroup closed in corresponding power of whole group | center is finite direct power-closed characteristic |

direct projection-invariant subgroup | invariant under projections to direct factors | center is direct projection-invariant |

characteristic subgroup | invariant under all automorphisms | center is characteristic |

## Proof

### Generic example

Let be a nontrivial cyclic group and a centerless group containing a normal subgroup isomorphic to , say . Consider the direct product .

Clearly, (as an embedded direct factor) is the center of .

Now consider the endomorphism of which composes the projection from onto with an isomorphism from to . Note that:

- is normality-preserving: Its image, is a cyclic normal subgroup of a direct factor . Since direct factor implies transitively normal and cyclic normal implies hereditarily normal,
*all*subgroups of the image are normal in . In particular, the image of any normal subgroup in is normal. - The endomorphism does not send to itself: Rather, it gets mapped to .

Find a group with an abelian normal subgroup .

### Particular example

The above generic example can generate a particular example by setting as cyclic group:Z3 and as symmetric group:S3. The group is thus direct product of S3 and Z3.

### Modification of generic example for finite p-groups

The generic example outlined above does not work for finite -groups, but the following slight modification does: instead of requiring to be a centerless group, simply require that *not* be contained in the center of . In this case, the center of becomes where is the center of . We construct the endomorphism in the same way (project to , compose with isomorphism to ) and note that since is not contained in , this endomorphism does not send to within itself.

### Particular examples for finite p-groups

- : We can take as cyclic group:Z4 and as dihedral group:D8 to get direct product of D8 and Z4. Alternatively, we can take as cyclic group:Z4 and as quaternion group to get direct product of Q8 and Z4.
- odd: Take as the cyclic group of prime-square order and as semidirect product of cyclic group of prime-square order and cyclic group of prime order.