Derived subgroup not is quasiautomorphism-invariant

This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., commutator subgroup) does not always satisfy a particular subgroup property (i.e., {{{property}}}Property "Proves property satisfaction of" (as page type) with input value "{{{property}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.)
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Statement

It is possible to have a group $G$ such that the commutator subgroup (or derived subgroup) $[G,G]$ is not a quasiautomorphism-invariant subgroup, i.e., there exists a Quasiautomorphism (?) $\sigma$ of $G$ such that $\sigma (H)\neq H$ .

Related facts

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

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.