Characteristic not implies quasiautomorphism-invariant

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 subgroup) need not satisfy the second subgroup property (i.e., quasiautomorphism-invariant subgroup)
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Statement

It is possible to have a group $G$ and a characteristic subgroup $H$ of $G$ such that $H$ is not a quasiautomorphism-invariant subgroup, i.e., there are quasiautomorphisms of $G$ that do not send $H$ to itself.

Related facts

• Commutator subgroup not is quasiautomorphism-invariant
• Center is quasiautomorphism-invariant
• Center not is 1-automorphism-invariant
• Quasiautomorphism-invariant not implies 1-automorphism-invariant

Proof

Further information: inner automorphism group of wreath product of groups of order p

Suppose $p$ is a prime number greater than $3$. Let $G$ be the group isomorphic to the inner automorphism group of the wreath product of two groups of order $p$. $G$ is a group of order $p^{p}$ with an elementary abelian normal subgroup $N$ of order $p^{p-1}$, an element of order $p$ acting on it from outside, and every non-identity element of $G$ has order $p$.

Consider the subgroup $H=[G,G]$. $H$ is a group of order $p^{p-2}$, contained inside the elementary abelian normal subgroup. It is a characteristic subgroup.

However, we can construct a quasiautomorphism $\sigma$ of $G$ that does not preserve $H$ as follows: the restriction of $\sigma$ to $N$ is an automorphism of $N$ that fixes the center $Z(G)$ (which is cyclic of order $p$) but does not send $H$ to itself, and $\sigma$ fixes every element of $G$ outside $H$. Note that we need $p>3$ to ensure that $H$ is strictly bigger than $Z(G)$, which is necessary to be able to construct a $\sigma$ with the desired specifications.