I have a question about something I believe the naming convention for group automorphisms suggests. From my understanding the inner automorphisms defined on a group $G$ are those automorphisms $\varphi: G \to G$ where there is some $g \in G$ where $\varphi$ is equivalent to the action of $g$ on $G$ by conjugation.

Those automorphisms which are not inner automorphisms ($\text{Inn }G$), are outer automorphisms. To me, this suggests as if (informally) there is a larger group $A$ where $G \trianglelefteq A$, and those "outer" automorphisms are simply equivalent to the action of some element $a \in A/G$ on $G$ by conjugation (It's called outer because $a$ is outside of $G$). Note that I'm stating $G$ is a normal subgroup of $A$ since we want $G$ to be normalized by all elements of $A$ so that the action of $a$ by conjugation limited to $G$ becomes an automorphism.

This idea is not discussed in the textbook I am studying. So my question is if the following theorem is valid, and how one goes around proving it.

Proposition: For every group $G$ there exists groups $G'$ and $A'$ such that $G \cong G'$ and $G' \trianglelefteq A'$ where the following condition holds: $$ \text{Aut }G \cong \text{Inn}\ A' $$

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    $\begingroup$ Techically, outer automorphisms are equivalence classes of automorphisms modulo $\mathrm{Inn}(G)$, since $\mathrm{Out}(G) = \mathrm{Aut}(G)/\mathrm{Inn}(G)$. $\endgroup$ – Arturo Magidin Apr 27 at 21:01
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    $\begingroup$ @ArturoMagidin - I think it depends on the author. Rotman in "Introduction to the Theory of Groups" defines an outer automorphism as any non-inner automorphism. $\endgroup$ – HallaSurvivor Apr 27 at 21:05
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    $\begingroup$ @HallaSurvivor Just checked and you are correct. Even so, people usually talk about "the" outer automorphism of $S_6$, even though under that definition there are actually $6!$ of them... $\endgroup$ – Arturo Magidin Apr 27 at 21:08
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    $\begingroup$ @Soroushkhoubyarian -- As Arturo has mentioned, your proposition is false. That said, your idea is not entirely without merit. You might want to look into the holomorph of a group. This is a larger group which has elements for $G$ and for every automorphism of $G$ (and $G$ is normal, conjugating by an automorphism does what you expect, etc.). The reason I had Rotman open was because I know he discusses the holomorph. Indeed you can read more in his "Introduction to the Theory of Groups", chapter 7.2 $\endgroup$ – HallaSurvivor Apr 27 at 21:11
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    $\begingroup$ Regarding the "outer automorphism" terminology, in certain studies the terminology is used as @ArturoMagidin suggested, namely elements of the quotient group $\mathrm{Out}(G) = \mathrm{Aut}(G)/\mathrm{Inn}(G)$. This is particularly true in the study of $\mathrm{Out}(F_n)$, the outer automorphism group of the rank $n$ free group. $\endgroup$ – Lee Mosher Apr 27 at 21:17

My original answer misread the question and so the answer is orthogonal to what you are actually asking. As it had some up-votes, I will keep it deleted.

Your proposed statement is false. Here's a counterexample:

Let $G=\mathbb{Z}$. Then there is no group $A$ such that $\mathrm{Inn}(A)\cong\mathrm{Aut}(G)\cong C_2$. In particular, there can be no group $A'$ satisfying the conditions you give.

That no such $A$ exists follows from the following well-known proposition:

Proposition. Let $G$ a group, and let $N\leq Z(G)$. If $G/N$ is cyclic, then $G$ is abelian.

This has been proven many times in this site.

In particular, $\mathrm{Inn}(A)$ cannot be cyclic and nontrivial, because $\mathrm{Inn}(A)\cong G/Z(G)$; this is the case in the example above, since $\mathrm{Aut}(\mathbb{Z})$ is cyclic of order $2$. Thus, there is no group $A$ such that $\mathrm{Inn}(A)\cong\mathrm{Aut}(\mathbb{Z})$. In particular, there can be no group $A$ that contains a normal copy of $\mathbb{Z}$ and satisfies $\mathrm{Inn}(A)\cong\mathrm{Aut}(\mathbb{Z})$.

Likewise, using primitive roots, we have that $\mathrm{Aut}(\mathbb{Z}_n)$ is cyclic for any $n$ that is an odd prime power, twice an odd prime power, $n=2$, and $n=4$; for all those values of $n$ except for $n=2$, there can be no $A$ with $\mathrm{Inn}(A)\cong \mathrm{Aut}(\mathbb{Z}_n)$. In particular, there is no group $A$ that contains a copy of $\mathbb{Z}_n$ as a normal subgroup and has inner automorphism group equal to the automorphism group of $\mathbb{Z}_n$. So the proposition cannot be rescued even if we restrict to finite groups.

On the other hand, Takao Matsumoto proved that every group is isomorphic to the outer automorphism group of some group (Any group is represented by an outerautomorphism group. Hiroshima Math. J. 19 (1989), no. 1, 209–219, MR 1009671 (90g:20051)), and more recently it was proven that every group is the outer automorphism group of a simple group.

As to your initial paragraph: if $G$ is centerless, then you can certainly take $A=\mathrm{Aut}(G)$; then $G\cong \mathrm{Inn}(G)\triangleleft \mathrm{Aut}(G)$, and the elements not in $G$ act by conjugation on $\mathrm{Inn}(G)$ inducing an automorphism of $G$ that is not inner.

  • $\begingroup$ Thanks for the complete and detailed answer. $\endgroup$ – Soroush khoubyarian Apr 27 at 21:36
  • $\begingroup$ More citations on realising groups as outer automorphism groups can be found in this MathOverflow question, while in it's accepted answer YCor proves that every finite group is the outer automorphism group of a finite group, which is rather nice. $\endgroup$ – user1729 Apr 28 at 9:23

Here are some further finite counterexamples. One such is an extra-special group $G$ of order $p^3$ and exponent $p$ for an odd prime $p$.

Then ${\rm Aut}(G) = p^2\!:\!{\rm GL}(2,p) = {\rm AGL}(2,p)$ is a split extension of an elementary abelian group of order $p^2$ by ${\rm GL}(2,p)$ with the natural induced action.

Here is sketch proof that no group $A$ exists with $G \unlhd A$ and ${\rm Aut}(G) \cong {\rm Inn}(A) = A/Z(A)$. I can give more details if requested.

If $Z(G) \not\le Z(A)$, then $G \cap Z(A) = 1$, so $A/Z(A)$ has a normal subgroup isomorphic to $G$. But ${\rm Aut}(G)$ has no such subgroup.

So $Z(G) \le Z(A)$ and in fact $Z(A) \cap G = Z(G)$, and $G/Z(A)$ is the unique elementary abelian normal subgroup of $A/Z(A)$ of order $p^2$. But elements of $A$ that map onto elements of ${\rm GL}(2,p)$ with determinant not equal to $1$ induce nontrivial actions on $Z(G)$, which contradicts $Z(G) \le Z(A)$.


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