Is there a homomorphism $\psi : \text{Aut}(G) \to \mathcal G$ with $\ker \psi = \text{Inn}(G)$? (Besides mapping each automorphism to its corresponding coset in $\text{Aut}(G) / \text{Inn}(G)$.)

Any group $G$ acts on itself via conjugation: $g * h = ghg^{-1}$. So there is a corresponding homomorphism $\varphi : G \to \text{Sym}(G)$ defined by $\varphi(g) = (h \mapsto g*h)$. The kernel of this action is clearly $Z(G)$, so $Z(G) \trianglelefteq G$. The image of $\varphi$ is clearly $\text{Inn}(G)$, the set of all conjugation automorphisms of $G$, so $\text{Inn}(G) \leq \text{Sym}(G)$. By the first isomorphism theorem, $G / Z(G) \cong \text{Inn}(G)$.

It follows that $\text{Inn}(G) \leq \text{Aut}(G)$. What we DON'T get from this argument is that $\text{Inn}(G)$ is normal in $\text{Aut}(G)$. So far I've only seen proofs that analyze what happens when you conjugate an inner automorphism by an automorphism: Inner automorphisms form a normal subgroup of $\operatorname{Aut}(G)$, Set of all inner automorphisms is a normal subgroup. But is there a homomorphism $\psi : \text{Aut}(G) \to \mathcal G$ (for some other group $\mathcal G$) with $\ker \psi = \text{Inn}(G)$?

An obvious choice is the canonical map $\pi : \text{Aut}(G) \to \text{Aut}(G)/\text{Inn}(G) = \text{Out}(G)$ that maps each element to its corresponding coset. But its codomain will not be a group unless we first prove that $\text{Inn}(G) \trianglelefteq \text{Aut}(G)$.

EDIT: To be clear, I am not asking for any arbitrary proof that $\text{Inn}(G)$ is normal. I am looking for a homomorphism with kernel $\text{Inn}(G)$ besides the obvious one.

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    $\begingroup$ Of course the map $\psi$ you seek would have to have image naturally isomorphic to $\operatorname{Out}(G)$, so you would still have to end up more or less defining the outer automorphism group. And I don't know a way to do this aside from the typical quotient, which as you say necessitates showing that the inner automorphism group is normal. $\endgroup$ Commented Sep 24, 2020 at 4:43
  • $\begingroup$ Good point. I'm hopeful there is an alternative homomorphism. Finding a homomorphism with kernel $Z(G)$ could also be done with the canonical map from $G \to G / Z(G)$, but there is the alternative of the conjugation action of $G$ on itself, with image isomorphic to $G / Z(G)$. $\endgroup$
    – jskattt797
    Commented Sep 24, 2020 at 5:07
  • 4
    $\begingroup$ I'm not sure if this works, but you can try to consider the action of the automorphism on conjungacy classes of $G$. Every inner automorphism acts trivially on it, I don't know if it works in the other direction as well. $\endgroup$
    – sss89
    Commented Sep 24, 2020 at 8:14
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    $\begingroup$ To clarify the question: Instead of proving that $\operatorname{Inn}(G)$ is normal directly, you want to find a homomorphism $\phi:\operatorname{Aut}(G)\rightarrow \mathcal{G}$ which has kernel $\operatorname{Inn}(G)$, and hence obtain normality via the first isomorphism theorem? [If so, nice question! But I think it needs rewritten a bit to make this clear.] $\endgroup$
    – user1729
    Commented Sep 24, 2020 at 8:28
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    $\begingroup$ @sss89 I'm doubtful about whether it is true that $\text{Inn}(G)$ is equal to the kernel of the action on conjugacy classes for arbitrary groups $G$. The reason for my doubt is that in the proofs that I know for special cases --- $G =$ a free group or a surface group --- the special nature of the group $G$ plays a big role. But on the other hand, I don't know of any counterexamples. $\endgroup$
    – Lee Mosher
    Commented Sep 24, 2020 at 12:35

3 Answers 3


Edit, 9/25/20: The suggestion I made at the end works.

Proposition: Let $G$ be a group of order $n$ (which may be infinite). Then $\text{Inn}(G)$ is precisely the kernel of the action of $\text{Aut}(G)$ acting on the set $\text{Hom}_{\text{HGrp}}(F_n, G)$ of (simultaneous) conjugacy classes of $n$-tuples of elements of $G$.

Proof. Suppose $\varphi \in \text{Aut}(G)$ acts trivially. Consider its action on the $n$-tuple given by every element of $G$. Fixing this $n$-tuple means fixing it up to conjugacy, which means there is some $g \in G$ such that $\varphi(h) = ghg^{-1}$ for all $h \in G$, which says precisely that $\varphi \in \text{Inn}(G)$. On the other hand, every element of $\text{Inn}(G)$ clearly acts trivially. $\Box$

Of course we can do much better than considering every element of $G$; it suffices to consider a generating set. But this construction is at least "canonical."

Here's an approach that maybe will seem like it doesn't tell you anything new but I'll extract something slightly more concrete out of it, which generalizes the suggestion to look at conjugacy classes. $\text{Out}(G)$ occurs naturally as the automorphism group of $G$ in a category we might call the homotopy category of groups $\text{HGrp}$. This category can be defined concretely as follows:

  • objects are groups $G$, and
  • morphisms $f : G \to H$ are conjugacy classes of homomorphisms, where two homomorphisms $f_1, f_2 : G \to H$ are identified (homotopic) iff there exists $h \in H$ such that $h f_1 = f_2 h$.

For example:

  • $\text{Hom}_{\text{HGrp}}(\mathbb{Z}, G)$ is the set of conjugacy classes of $G$
  • $\text{Hom}_{\text{HGrp}}(G, S_n)$ is the set of isomorphism classes of actions of $G$ on a set of size $n$
  • $\text{Hom}_{\text{HGrp}}(G, GL_n(\mathbb{F}_q))$ is the set of isomorphism classes of actions of $G$ on $\mathbb{F}_q^n$

and so forth.

Now we can prove the more general fact that composition in this category is well-defined (that is, that the homotopy class of a composition of morphisms only depends on the homotopy class of each morphism), which implies in particular that the automorphism group $\text{Aut}_{\text{HGrp}}(G)$ of $G$ in this category is really a group, and of course this group is $\text{Out}(G)$.

So far this is just a slight extension and repackaging of the proof via conjugating by an inner automorphism, but the point is that this construction tells you what conjugating by an inner automorphism means. The homotopy category of groups has a second description, as follows:

  • objects are Eilenberg-MacLane spaces $K(G, 1) \cong BG$, and
  • morphisms $f : BG \to BH$ are homotopy classes of continuous maps.

We get the ordinary category of groups if we instead insist that Eilenberg-MacLane spaces have basepoints and our morphisms and homotopies preserve basepoints. So the passing to conjugacy classes has to do with the extra freedom we get from throwing out basepoints. Here the incarnation of conjugacy classes $\text{Hom}(\mathbb{Z}, G)$ is the set of free homotopy classes of loops $S^1 \to BG$.

Anyway, all this suggests the following generalization of looking at conjugacy classes: we can look at the entire representable functor

$$\text{Hom}_{\text{HGrp}}(-, G) : \text{HGrp}^{op} \to \text{Set}.$$

By the Yoneda lemma, the automorphism group of this functor is precisely $\text{Aut}_{\text{HGrp}}(G) \cong \text{Out}(G)$. What this says is that an outer automorphism of $G$ is the same thing as a choice, for each group $H$, of an automorphism (of sets) of $\text{Hom}_{\text{HGrp}}(H, G)$, which is natural in $H$. We can furthermore hope that it's possible to restrict attention to a smaller collection of groups $H$; for example (and I haven't thought about this at all) maybe it's possible to restrict to the free groups $H = F_n$, which means looking at $\text{Hom}_{\text{HGrp}}(F_n, G)$, the set of conjugacy classes of $n$ elements of $G$ (under simultaneous conjugacy).

  • 4
    $\begingroup$ Could make explicit in a non-categorial way the definition of the stated action in the opening Prop, please? $\endgroup$
    – user810157
    Commented Sep 26, 2020 at 12:55
  • 1
    $\begingroup$ @user750041: the action is pointwise on tuples. $\endgroup$ Commented Sep 26, 2020 at 16:54

Following @sss89's hint in the comments.

Denoted with $\operatorname{Cl}(a)$ the conjugacy class of $a\in G$, let's consider the natural action of $\operatorname{Aut}(G)$ on $X:=\{\operatorname{Cl}(a), a\in G\}$, namely: $\sigma\cdot \operatorname{Cl}(a):=\operatorname{Cl}(\sigma(a))$. This is indeed an action because:

  1. good definition: $a'\in \operatorname{Cl}(a)\Rightarrow \sigma\cdot\operatorname{Cl}(a')=\operatorname{Cl}(\sigma(a'))$; now, $\sigma$ is (in particular) a surjective homomorphism, and hence $\operatorname{Cl}(\sigma(a'))=\sigma(\operatorname{Cl}(a'))=\sigma(\operatorname{Cl}(a))=\operatorname{Cl}(\sigma(a))=\sigma\cdot \operatorname{Cl}(a)$, and the map is well-defined;
  2. by construction, $\operatorname{Cl}(\sigma(a))\in X, \forall\sigma\in\operatorname{Aut}(G),\forall a\in G$;
  3. $Id_G\cdot \operatorname{Cl}(a)=\operatorname{Cl}(Id_G(a))=\operatorname{Cl}(a), \forall a\in G$;
  4. $(\sigma\tau)\cdot\operatorname{Cl}(a)=\operatorname{Cl}((\sigma\tau)(a))=\operatorname{Cl}(\sigma(\tau(a))=\sigma\cdot(\operatorname{Cl}(\tau(a)))=\sigma\cdot(\tau\cdot\operatorname{Cl}(a)), \forall \sigma,\tau\in\operatorname{Aut}(G), \forall a\in G$

The point-wise stabilizer under this action is given by:

\begin{alignat}{1} \operatorname{Stab}(\operatorname{Cl}(a)) &= \{\sigma\in\operatorname{Aut}(G)\mid \operatorname{Cl}(\sigma(a))=\operatorname{Cl}(a)\} \\ &= \{\sigma\in\operatorname{Aut}(G)\mid \sigma(\operatorname{Cl}(a))=\operatorname{Cl}(a)\} \\ \end{alignat}

and the kernel of the equivalent homomorphism $\phi\colon \operatorname{Aut}(G)\to \operatorname{Sym}(X)$ by:

\begin{alignat}{1} \operatorname{ker}\phi &= \bigcap_{a\in G}\operatorname{Stab}(\operatorname{Cl}(a)) \\ &= \{\sigma\in\operatorname{Aut}(G)\mid \sigma(\operatorname{Cl}(a))=\operatorname{Cl}(a), \forall a\in G\} \\ \end{alignat}

Now, $\operatorname{Inn}(G)=\{\varphi_b,b\in G\}$, where $\varphi_b(g):=b^{-1}gb$, and hence:

\begin{alignat}{1} \varphi_b(\operatorname{Cl}(a)) &= \{\varphi_b(gag^{-1}), g\in G\} \\ &= \{b^{-1}gag^{-1}b, g\in G\} \\ &= \{(b^{-1}g)a(b^{-1}g)^{-1}, g\in G\} \\ &= \{g'ag'^{-1}, g'\in G\} \\ &= \operatorname{Cl}(a), \forall a\in G \\ \end{alignat}

whence $\varphi_b\in \operatorname{ker}\phi, \forall b\in G$, and finally $\operatorname{Inn}(G)\subseteq \operatorname{ker}\phi$. Conversely, let $\sigma\in \operatorname{ker}\phi$; then, $\sigma(\operatorname{Cl}(a))=\operatorname{Cl}(a), \forall a\in G$; in particular:

\begin{alignat}{1} \sigma(\operatorname{Cl}(a))\subseteq\operatorname{Cl}(a), \forall a\in G &\Rightarrow \forall g\in G,\exists g'\in G\mid \sigma(gag^{-1})=g'ag'^{-1}, \forall a\in G \\ &\Rightarrow \exists g''\in G\mid \sigma(a)=g''ag''^{-1}, \forall a\in G \\ &\Rightarrow \exists g''\in G\mid \sigma(a)=\varphi_{g''}(a), \forall a\in G \\ &\Rightarrow \exists g''\in G\mid \sigma=\varphi_{g''} \\ &\Rightarrow \sigma\in \operatorname{Inn}(G) \\ &\Rightarrow \operatorname{ker}\phi\subseteq \operatorname{Inn}(G) \\ \end{alignat}

Therefore, by the double inclusion, $\operatorname{Inn}(G)=\operatorname{ker}\phi$.

EDIT. As per the comments hereafter, I made a mistake in the final part of this answer, from "Conversely..." onwards. Therefore, so far the only inclusion $\operatorname{Inn}(G)\subseteq\operatorname{ker}\phi$ is actually proven.

EDIT (Dec 11, 2020)

I think that the inverse inclusion, and hence the claim, holds for the particular class $G=S_n$, as follows.

Each conjugacy class is a certain cycle structure, and then each stabilizer comprises all and only the automorphisms of $S_n$ which preserve a certain cycle structure, whence $\operatorname{Stab}(\operatorname{Cl}(\sigma))\le\operatorname{Inn}(S_n)$, for every $\sigma\in S_n$. But then, $\operatorname{ker}\phi=\bigcap_{\sigma\in S_n}\operatorname{Stab}(\operatorname{Cl}(\sigma))\le\operatorname{Inn}(S_n)$.

  • 1
    $\begingroup$ How is the existence of $g''$ proved where you first assert it towards the end of your proof? Existence of $g''$ is the key to proving that $\sigma \in \text{Inn}(G)$, but I don't see a justification for it. Presumably the intention is that $g'' = \sigma(g^{-1}) g'$, but $g'$ depends on $g$, so $g''$ looks like it is not well-defined independent of $g$. $\endgroup$
    – Lee Mosher
    Commented Sep 24, 2020 at 12:17
  • $\begingroup$ It's the $g'$ corresponding to $g=e$. $\endgroup$
    – user810157
    Commented Sep 24, 2020 at 12:20
  • $\begingroup$ Okay, but in that case one should also point out that $g'$ also depends on $a$. Your statement $$\forall g\in G,\exists g'\in G\mid \sigma(gag^{-1})=g'ag'^{-1}, \forall a\in G$$ properly written in prenex normal form, should be $$\forall a \in G, \forall g\in G,\exists g'\in G\mid \sigma(gag^{-1})=g'ag'^{-1}$$ $\endgroup$
    – Lee Mosher
    Commented Sep 24, 2020 at 12:24
  • $\begingroup$ @Lee Mosher, I see, so my $g''$ depends on $a$ and $ker\phi\subseteq Inn(G)$ doesn't follow, right? $\endgroup$
    – user810157
    Commented Sep 24, 2020 at 12:45
  • 7
    $\begingroup$ For a finite group $G$, the automorphisms sending every element to a conjugate are called class - preserving and form a group, denoted by $Aut_c(G)$. The question whether $Aut_cG=Inn(G)$ was posed in 1911 by Burnside. He also gave the answer constructing a group $G$ of order $p^6$ with $Aut_c(G)/Inn(G)=C_p^4$. See W. Burnside, ‘‘The Theory of Groups of Finite Order,’’ $\endgroup$
    – 1123581321
    Commented Sep 24, 2020 at 16:21

Well, you just need to check the criteria of normality. Let $F \in \text{Aut}(G)$ and $H \in \text{Inn}(G),\hspace{2mm} H(x) =hxh^{-1} $. Then for $x \in G$, you have

$$ F\circ H \circ F^{-1}(x) = F(hF^{-1}(x)h^{-1})=F(h)xF^{-1}(h) .$$

Which clearly means $\text{Inn}(G)$ is normal in $\text{Aut}(G)$

  • 1
    $\begingroup$ one needs to conjugate the inner one $H$ with a generic $F$ in ${\rm Aut}\ G$ $\endgroup$
    – janmarqz
    Commented Sep 24, 2020 at 4:52
  • 3
    $\begingroup$ I'm looking for a homomorphism with kernel $\text{Inn}(G)$ besides the obvious one. $\endgroup$
    – jskattt797
    Commented Sep 24, 2020 at 4:59
  • $\begingroup$ Regardless, I'm having trouble verifying that this argument is correct. $H(x) = hxh^{-1}$. So $F(H(x)) = F(hxh^{-1})$. So $H^{-1}(F(H(x))) = h^{-1}F(hxh^{-1})h$. How do you know $F(h) = h$? $\endgroup$
    – jskattt797
    Commented Sep 24, 2020 at 17:53
  • $\begingroup$ @jskattt797 I think the definition gives the conclusion $\endgroup$
    – Mikasa
    Commented Sep 24, 2020 at 18:00
  • $\begingroup$ @mrs can you please elaborate? $\endgroup$
    – jskattt797
    Commented Sep 25, 2020 at 4:28

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