Characteristic not implies sub-isomorph-free in finite group

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

The center of a non-abelian group of odd prime cube order
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 G.
 * $$H$$ 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.