Characteristic-isomorph-free not implies normal-isomorph-free in finite

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

We may have a group with a characteristic subgroup such that there are no other characteristic subgroups isomorphic to it, but there are other normal subgroups isomorphic to it.

Proof

Example of an Abelian group

Further information: prime-cube order group:p2timesp, agemo subgroups of a group of prime power order, omega subgroups of a group of prime power order

Let $p$ be any prime, and let $G:=\mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p^2\mathbb{Z}$. In other words, $G$ is a direct product of cyclic groups of order $p$ and $p^2$ respectively.

Consider the subgroup $H:=\stackrel{\mathrm{g}}{\mathrm{A}}(G)$: the set of all elements of $G$ that can be written as $p^{th}$ powers. This set is $0\times p\mathbb{Z}/p^2\mathbb{Z}$.

• $H$ is a characteristic subgroup: It is defined as the set of $p^{th}$ powers, so it is clearly invariant under conjugation.
• There is no other characteristic subgroup of $G$ of order $p$: All the subgroups of order $p$ in $G$ lie inside ${\mathrm{\Omega }}_1(G)=\mathbb{Z}/p\mathbb{Z}\times p\mathbb{Z}/p^2\mathbb{Z}$. Further, automorphisms of the form $(a,b)\mapsto (a+b,b)$ permute all the other subgroups.
• There are other normal subgroups of $G$ of order $p$: In fact, there are $p$ of them: the other $p$ subgroups of ${\mathrm{\Omega }}_1(G)$.

Thus, $H$ is characteristic-isomorph-free in $G$ but is not normal-isomorph-free in $G$.