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) does not always satisfy a particular subgroup property (i.e., normality-preserving endomorphism-invariant subgroup)
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Statement

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

Related facts

Similar facts

• Center not is fully invariant
• Center not is fully invariant in class two p-group

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 $A$ be a nontrivial cyclic group and $C$ a centerless group containing a normal subgroup isomorphic to $A$, say $B$. Consider the direct product $G:=A\times C$.

Clearly, $H:=A\times \{e\}$ (as an embedded direct factor) is the center of $A\times C$.

Now consider the endomorphism $\alpha$ of $G=A\times C$ which composes the projection from $G$ onto $A$ with an isomorphism from $A$ to $\{e\}\times B$. Note that:

• $\alpha$ is normality-preserving: Its image, $\{e\}\times B$ is a cyclic normal subgroup of a direct factor $\{e\}\times C$. Since direct factor implies transitively normal and cyclic normal implies hereditarily normal, all subgroups of the image $\{e\}\times B$ are normal in $G$. In particular, the image of any normal subgroup in $G$ is normal.
• The endomorphism does not send $H=A\times \{e\}$ to itself: Rather, it gets mapped to $\{e\}\times B$.

Find a group $A$ with an abelian normal subgroup $B$.

Particular example

The above generic example can generate a particular example by setting $A$ as cyclic group:Z3 and $C$ as symmetric group:S3. The group $G$ 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 $p$-groups, but the following slight modification does: instead of requiring $C$ to be a centerless group, simply require that $B$ not be contained in the center of $C$. In this case, the center of $G=A\times C$ becomes $H=A\times Z(C)$ where $Z(C)$ is the center of $C$. We construct the endomorphism in the same way (project to $A$, compose with isomorphism to $\{e\}\times B$) and note that since $B$ is not contained in $Z(C)$, this endomorphism does not send $H$ to within itself.

Particular examples for finite p-groups

• $p=2$: We can take $A$ as cyclic group:Z4 and $C$ as dihedral group:D8 to get direct product of D8 and Z4. Alternatively, we can take $A$ as cyclic group:Z4 and $C$ as quaternion group to get direct product of Q8 and Z4.
• $p$ odd: Take $A$ as the cyclic group of prime-square order and $C$ as semidirect product of cyclic group of prime-square order and cyclic group of prime order.