Full invariance does not satisfy image condition

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

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

Proof

Example of a non-Abelian group of prime-cubed order

Further information: Prime-cube order group:p2byp

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 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 $P/Z(P)$.