Characteristic not implies fully invariant in odd-order class two p-group

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

For any (odd) prime $p$, there exists a $p$-group $G$ of class two and a characteristic subgroup of this group that is not fully invariant.

The construction also works for $p = 2$, but for $p = 2$, there are already examples of abelian groups with characteristic subgroups that are not fully invariant.

Related facts

• Characteristic not implies fully invariant in finite abelian group
• Characteristic equals fully invariant in odd-order abelian group
• Characteristic not implies fully invariant in class three maximal class p-group
• Center not is fully invariant in class two p-group
• Socle not is fully invariant in class two p-group

Proof

Let $p$ be an odd prime. Let $P$ be any non-abelian group of order $p^3$ with center $Z$. There are two possibilities for $P$: a group of prime-square exponent, and a group of prime exponent. In both such groups, there is an element $x$ of order $p$ outside $Z$.

Define $G = P \times C$ where $C$ is the cyclic group of order $p$ with generator $y$. The center of $G$ is the subgroup $H = Z \times C$. Then:

• $H$ is characteristic in $G$, because center is characteristic.
• $H$ is not fully invariant in $G$: Consider the retraction with kernel $P \times \{ e \}$ and with image generated by the element $(x,y)$. This is an endomorphism of $G$, but it does not send $H$ to itself, since the element $(e,y)$ gets sent to $(x,y)$, which is outside $H$.