Characteristic not implies sub-isomorph-free 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., sub-isomorph-free subgroup)
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Statement

We can have a finite group $G$ and a characteristic subgroup $H$ of $G$ such that $H$ is not a sub-isomorph-free subgroup of $G$. In other words, there is no ascending chain of subgroups starting at $H$ and ending at $G$ with each member an isomorph-free subgroup of its successor.

Related facts

• Characteristic not implies isomorph-free in finite
• Characteristic not implies sub-(isomorph-normal characteristic) in finite

Proof

The center of a non-abelian group of odd prime cube order

Further information: Prime-cube order group:U3p, Prime-cube order group:p2byp

Let $p$ be an odd prime. Let $G$ be the non-abelian group of order $p^{3}$ and exponent $p$. Let $H$ be the center of $G$. Then, we have:

• $H$ is characteristic in $G$.
• $H$ is not sub-isomorph-free in $G$: In fact, no proper subgroup of $G$ containing $H$ is isomorph-free. The subgroups of order $p^{2}$ are all elementary abelian, while $H$ itself is isomorphic to many other cyclic groups of prime order.

Another example is to take $G$ as the non-abelian group of order $p^{3}$ and exponent $p^{2}$, and $H$ to be the center of $G$. In this case:

• $H$ is characteristic in <mah>G</math>.
• $H$ is not sub-isomorph-free in $G$: $H$ is not isomorph-free, because there are other cyclic groups of order $p$. There is only one isomorph-free proper subgroup of $G$ properly containing $H$, namely, an elementary abelian subgroup of order $p^{2}$. But $H$ is not isomorph-free in this subgroup.