Class-preserving not implies subgroup-conjugating

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., class-preserving automorphism) need not satisfy the second automorphism property (i.e., subgroup-conjugating automorphism)
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Statement

A class-preserving automorphism of a group (i.e., an automorphism that preserves conjugacy classes of elements) need not be a subgroup-conjugating automorphism (i.e., it need not send every subgroup to a conjugate subgroup).

Related facts

Weaker facts

• Class-preserving not implies inner

Proof

An infinite group example

Further information: Finitary symmetric group is conjugacy-closed in symmetric group

One example of this is the finitary symmetric group on an infinite set: the group of all permutations that move only finitely many elements. Consider the automorphism on this group induced via conjugation by an infinitary permutation (a permutation that moves infinitely many elements). This automorphism sends every element to an element in the same conjugacy class, but is not an inner automorphism.

In fact, it does not even send every subgroup to a conjugate subgroup. To see this, divide the infinite set into two infinite subsets of equal cardinality. Consider the subgroups of finitary permutations on these two subsets. There is an infinitary permutation that conjugates one subgroup to the other; however, in the group of finitary permutations, these subgroups are clearly not conjugate.

A finite group example

Constructing an example involving a finite group is somewhat more tricky. One construction is as follows. Consider the ring $\mathbb {Z} /8\mathbb {Z}$. Let $A$ be the additive group of this ring, and $G$ the multiplicative group of units. Let $E$ be the semidirect product $A\rtimes G$.

Now, we use three facts:

• The stability group of $1\triangleleft A\triangleleft E$ corresponds to the elements that are 1-cocycles for the action of $G$ on $A$
• The elements in this stability group that come from inner automorphisms, are those that correspond to 1-coboundaries
• The elements in this stability group that send every element to an element in the same conjugacy class, correspond to 1-cocycles such that for each element, it looks like a 1-coboundary. In other words, it is a 1-cocycle $\varphi$ for the action of $G$ on $A$, such that for every $g\in G$, there exists $a\in A$ (depending on $g$) such that we have:

$\varphi (g)=g.a-a$

A class automorphism that is not inner corresponds to a 1-cocycle that is not a 1-coboundary. The construction, specifically, is:

$\varphi (1)=\varphi (7)=0,\varphi (3)=\varphi (5)=4$

We now argue that the class automorphism arising thus is not subgroup-conjugating. For this, consider $G$ as a subgroup of order four in $E$. The class automorphism $\sigma$ arising this way is a stability automorphism that sends $G$ to another complement, say $G_{1}$, of $A$. If these two subgroups are conjugate, the conjugation must preserve cosets of $A$ (because $E/A$ is Abelian) and hence, the inner automorphism sending $G$ to $G_{1}$ has the same effect on every element as $\sigma$, so $\sigma$ is an inner automorphism, contradicting the assumption.