Normal-extensible not implies normal

This article gives the statement and possibly, proof, of a non-implication relation between two automorphism properties. That is, it states that every automorphism satisfying the first automorphism property (i.e., normal-extensible automorphism) need not satisfy the second automorphism property (i.e., normal automorphism)
View a complete list of automorphism property non-implications | View a complete list of automorphism property implications
Get more facts about normal-extensible automorphism|Get more facts about normal automorphism

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., normal subgroup) need not satisfy the second subgroup property (i.e., normal-extensible automorphism-invariant subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about normal subgroup|Get more facts about normal-extensible automorphism-invariant subgroup

EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property normal subgroup but not normal-extensible automorphism-invariant subgroup|View examples of subgroups satisfying property normal subgroup and normal-extensible automorphism-invariant subgroup

Statement

In terms of automorphism properties

A normal-extensible automorphism of a group (i.e., an automorphism that can always be extended for any embedding of the group as a normal subgroup of a bigger group) need not be a normal automorphism, i.e., it need not send every normal subgroup to itself.

In terms of subgroup properties

A normal subgroup of a group need not be a normal-extensible automorphism-invariant subgroup: i.e., there may be normal-extensible automorphisms of the group that do not leave the normal subgroup invariant.

Statement with symbols

We can have a group $G$ and a normal-extensible automorphism $\sigma$ of $G$ that is not a normal automorphism: in other words, there exists a normal subgroup $N$ of $G$ such that $\sigma (N)\ne N$.

Related facts

• Centerless and maximal in automorphism group implies every automorphism is normal-extensible
• Normal-extensible not implies inner
• Normal-extensible not implies extensible

Stronger facts

• Normal not implies normal-extensible automorphism-invariant in finite: This is the stronger version, and the example outlined here in fact shows the stronger version.

Applications

• Normal not implies semi-strongly potentially characteristic
• Normal not implies strongly potentially characteristic

Facts used

1. Centerless and maximal in automorphism group implies every automorphism is normal-extensible
2. Automorphism group of direct power of simple non-abelian group is wreath product of automorphism group and symmetric group: Let $S$ be a simple non-abelian group and $S^n$ be a direct product of $S$ with itself $n$ times. The automorphism group of $S^n$ is $\mathrm{A}\mathrm{u}\mathrm{t}(S)\wr \mathrm{S}\mathrm{y}\mathrm{m}(n)$, where the latter has the usual action on $n$ letters.

Proof

Further information: general linear group:GL(3,2)

Let $S=GL(3,2)$, i.e., $S$ is the general linear group of order three over the field of two elements. Consider $G=S\times S$.

We know that $S$ is simple and non-abelian (hence centerless) and every automorphism of $S$ is inner, so $S=\mathrm{A}\mathrm{u}\mathrm{t}(S)$.

By fact (2), the automorphism group of $G$ is $\mathrm{A}\mathrm{u}\mathrm{t}(S)\wr \mathbb{Z}/2\mathbb{Z}$. Because every automorphism of $S$ is inner, this is $G⋊\mathbb{Z}/2\mathbb{Z}$, with the action being the coordinate exchange automorphism. Thus, $G$ is centerless and maximal in its automorphism, so by fact (1), every automorphism of $G$ is normal-extensible.

Let $N$ be the normal subgroup of $G$ defined by the first direct factor, i.e., $N=S\times 1$. Let $\sigma$ be the coordinate exchange automorphism of $G$, i.e., $\sigma (a,b)=(b,a)$. Then, since every automorphism of $G$ is normal-extensible, $\sigma$ is normal-extensible. However, $\sigma (N)\ne N$, since $\sigma (N)=1\times S$ is the other direct factor.