Characteristic not implies isomorph-normal in finite group

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a finite group, every subgroup satisfying the first subgroup property (i.e., characteristic subgroup) need not satisfy the second subgroup property (i.e., isomorph-normal subgroup)
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Statement

Statement with symbols

It is possible to have a group $G$ and a subgroup $H$ of $G$ such that $H$ is a characteristic subgroup of $G$ but is not isomorph-normal in $G$: there exists a subgroup $K$ of $G$ isomorphic to $H$ that is not normal in $G$.

Proof

Example of the dihedral group

Further information: dihedral group:D8

Let $G$ be the dihedral group of order eight, given by:

$G=\langle a,x\mid a^{4}=x^{2}=1,xax=a^{-1}\rangle$.

Let $H$ be the center of $G$. $H$ is a subgroup of order two generated by $a^{2}$.

• $H$ is characteristic.
• $H$ is not isomorph-normal: The subgroup $\langle x\rangle$ of $G$ is isomorphic to $H$, but is not normal in $G$, because conjugation by $a$ sends it to the subgroup $\langle a^{2}x\rangle$.