Full invariance does not satisfy image condition

This article gives the statement, and possibly proof, of a subgroup property (i.e., fully invariant subgroup) not satisfying a subgroup metaproperty (i.e., image condition).
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Statement

Suppose $G$ is a group, $K$ is a fully invariant subgroup of $G$, and $\phi :G\to H$ is a surjective homomorphism. Then, $\phi (K)$ need not be fully invariant in $H$.

Proof

Example of an Abelian group of prime-cube order

(This example uses additive notation).

Suppose $G$ is the direct product of a cyclic group $A$ of order $p$ and a cyclic group of order $B$ of order $p^2$. Define:

• $K={\mathrm{\Omega }}_1(G)$ (see omega subgroups of a group of prime power order), i.e., $K$ is the subgroup comprising all the elements:

$\{x\in G\mid px=0\}$.

• $\phi$ is the quotient map by the normal subgroup ${\mho }^1(G)$ (see agemo subgroups of a group of prime power order), i.e., $\phi$ is the quotient map by the subgroup:

$N:=\{y\in G\mid \mathrm{\exists }x,px=y\}$.

• Observe that $K$ is fully invariant in $G$ (more generally, all omega subgroups are fully invariant). However, $\phi (K)$ is a subgroup of order $p$ in $\phi (G)$ which is elementary abelian of order $p^2$ -- hence $\phi (K)$ is not fully invariant in $\phi (G)$.

Example of a non-abelian group of prime-cube order

Further information: Prime-cube order group:p2byp, Subgroup structure of prime-cube order group:p2byp

Let $p$ be an odd prime. Suppose $A$ is a cyclic group of order $p^2$ and $B$ is a cyclic group of order $p$, with $B$ acting on $A$ via multiplication by $p+1$. Then, the semidirect product of $A$ by $B$ is a non-Abelian group of order $p^3$. Call this group $P$. Define ${\mathrm{\Omega }}_1(P)$ (see omega subgroups of a group of prime power order) as the subgroup generated by all elements of order $p$ in $P$. By the fact that Omega-1 of odd-order class two p-group has prime exponent, ${\mathrm{\Omega }}_1(P)$ is a subgroup of prime exponent. This forces it to be a subgroup of order $p^2$ generated by the elements of $B$ and the multiples of $p$ in $A$. All the omega subgroups are fully characteristic, so ${\mathrm{\Omega }}_1(P)$ is fully characteristic.

The center of $P$, namely $Z(P)$, simply comprises the multiples of $p$ in $A$. Thus, in the quotient map $P\to P/Z(P)$, the image of ${\mathrm{\Omega }}_1(P)$ is cyclic of order $p$, while the whole group is elementary Abelian of order $p^2$. Thus:

• ${\mathrm{\Omega }}_1(P)$ is fully characteristic in $P$.
• The image of ${\mathrm{\Omega }}_1(P)$ in $P/Z(P)$ is not fully characteristic in $P/Z(P)$.