In order to prove this, I first proved that the set of all automorphisms from a group $G$ to $G$ form a group under composition: The identity homorphism is an automorphism because sends $x$ from $G$ to $x$ from $G$:

$\phi_e(x) = x \in G$

The inverse of an homomorphism exists:


The associativity works because composition of functions is always associative.

Closure property:

$$\phi\circ\gamma(a+b) = \phi(\gamma(a+b)) = \phi(\gamma(a)+\gamma(b)) = \phi(\gamma(a)) + \phi(\gamma(a)) = \phi\circ\gamma(a)+\phi\circ\gamma(b)$$

So, an Inner automorphism is defined as a function $f$ such that

$f(x) = a^{-1}xa$

for a fixed element $a$ from $G$.

I'm supposed to prove that the set of all these automorphisms form a normal subgroup of $G$, that is:

$$gfg^{-1}\in N$$ for all $g$

where $N$ is the set of all inner automorphisms of $G$, and $g$ is an automorphism of $G$.


Ok, so what I learned from this is that

we have a group $Aut(G)$ made of all the automorphisms of $G$, and we want to show that the subgroup of $Aut(G)$ made of all the inner automorphisms, is normal. That is, given an inner automorphism $\phi_a(x) = a^{-1}xa$ for a fixed $a$ in $G$ and $x\in G$.

So, we need to show that, given $g$ as an automorphism from $Aut(G)$ and $\phi_a$ an inner automorphism from the subgroup $N$ of inner automorphisms, we must have:

$$g^{-1}\phi_ag \in N$$

for all $g$

Am I rigth?

  • 4
    $\begingroup$ For clarity, instead of $f$, let $\varphi_a$ denote the inner automorphism $\varphi_a(x)=a^{-1}xa$. Then show $g\varphi_a g^{-1}=\varphi_{g(a)}$, hence $g\varphi_a g^{-1}$ is inner. $\endgroup$
    – Ben West
    Aug 22, 2015 at 22:04
  • 4
    $\begingroup$ It's a normal subgroup of $\operatorname{Aut}G$, not of $G$. $\endgroup$
    – Bernard
    Aug 22, 2015 at 22:11
  • 2
    $\begingroup$ Your title is deceptive. The question is about inner automorphisms from a group to itself, not to another group. In fact, one only uses the terminology "automorphisms" when talking about isomorphisms from an object to itself. $\endgroup$
    – Lee Mosher
    Aug 22, 2015 at 22:33
  • $\begingroup$ $g(a^{-1}xa)g^{-1} = ga^{-1}xag^{-1} = (ag^{-1})^{-1}x(ga^{-1})^1$ but I don't think this help. Can I argue that $ag^{-1}$ is in $G$? $\endgroup$ Aug 22, 2015 at 23:43
  • 1
    $\begingroup$ @GuerlandoOCs No, $g$ is an automorphism of $G$, not an element of $G$, so $$g\varphi_ag^{-1}(x)=g\varphi_a(g^{-1}(x))=g(a^{-1}g^{-1}(x)a)=g(a^{-1})g(g^{-1}(x))g(a)=g(a)^{-1}xg(a)=\varphi_{g(a)}(x).$$ $\endgroup$
    – Ben West
    Aug 23, 2015 at 0:03

3 Answers 3


For any given $a \in G$, conjugacy by $a$, namely the map $\varphi_a \colon G \to G$ defined by $\varphi_a(g):=aga^{-1}$, is an automorphism of $G$. In fact, $\varphi_a(g)=\varphi_a(g') \Rightarrow aga^{-1}=ag'a^{-1} \Rightarrow g=g'$ (injectivity); then, for evey $g\in G, \varphi_a(a^{-1}ga)=g$ (surjectivity); finally, $\varphi_a(gg')=agg'a^{-1}=aga^{-1}ag'a^{-1}=(aga^{-1})(ag'a^{-1})=\varphi_a(g)\varphi_a(g')$ (homomorphism). Therefore, $\Phi:=\{\varphi_a, a \in G\} \subseteq \operatorname{Aut}(G)$.

Now, $(\varphi_a\varphi_b)(g)=\varphi_a(\varphi_b(g))=a(bgb^{-1})a^{-1}=(ab)g(ab)^{-1}=\varphi_{ab}(g)$, whence $\varphi_a\varphi_b=\varphi_{ab} \in \Phi$ (closure); then, $(\varphi_a\varphi_{a^{-1}})(g)=\varphi_a(\varphi_{a^{-1}}(g))=a(a^{-1}ga)a^{-1}=g$, whence $\varphi_a^{-1}=\varphi_{a^{-1}}\in \Phi$ (closure by inverses). Therefore, $\Phi \le \operatorname{Aut}(G)$.

Let's now prove that $\Phi \unlhd \operatorname{Aut}(G)$. Well, $\forall a,b \in G, \forall \sigma \in \operatorname{Aut}(G)$, we get: $(\sigma^{-1}\varphi_a\sigma)(b)=\sigma^{-1}(\varphi_a(\sigma(b)))=\sigma^{-1}(a\sigma(b)a^{-1})=$ $\sigma^{-1}(a)b\sigma^{-1}(a^{-1})$; call $\tau:=\sigma^{-1} \in \operatorname{Aut}(G)$, then $(\sigma^{-1}\varphi_a\sigma)(b)=\tau(a)b\tau(a^{-1})=\tau(a)b\tau(a)^{-1}=\varphi_{\tau(a)}(b)=\varphi_{\sigma^{-1}(a)}(b)$, so that $\sigma^{-1}\varphi_a\sigma=\varphi_{\sigma^{-1}(a)} \in \Phi$. $\Box$

$\Phi$ is precisely $\operatorname{Inn}(G)$.


To prove "$\mathrm{Inn}(G)$ is a normal subgroup of $\mathrm{Aut}(G)$", you need to prove "for each $\phi_a\in \mathrm{Inn}(G)$ and $\theta \in \mathrm{Aut}(G)$, we have $\theta \circ \phi_a \circ \theta^{-1} \in \mathrm{Inn}(G)$". To verify this, simply evaluate $(\theta \circ \phi_a \circ \theta^{-1})(x)$ and use some properties that $\theta$ and $\phi_a$ are known to have.


First, you used additive notation for the group operation, but we needn't suppose that the group $G$ is abelian, the result holds in the general case.

Second, group automorphisms $Aut(G)$ is a different group from $G$ itself (so, inner automorphisms do not form a subgroup in $G$...).

The set/class of all inner automorphisms $Inn(G)$ form a normal subgroup in $Aut(G)$, not in $G$ itself!

Conjugation operation, you called it $f(x)$, more precisely $f_a(x)$ or $x^a$, is a group homomorphism $G\rightarrow Aut(G)$ (not necessarily isomorphism or even mono- or epi-). And the whole image of a group homomorphism is always a normal subgroup in the codomain (the proof is a bit verbose).

In this case the whole image of the conjugation operation is exactly the set of all inner automorphisms (forming a subgroup in $Aut(G)$).

$$ f: G\rightarrow Aut(G) $$

$$ f_a(x) := a^{-1}xa $$

$$ f_G = Inn(G) $$

$$ Inn(G) \triangleleft Aut(G) $$

So we can indeed always speak about outer automorphisms $Out(G)$ of a group $G$, which is possibly trivial, because $Inn(G)$ exhaust the whole $Aut(G)$ (all automorphisms are inner), or, conversely, equal (isomorphic) to $Aut(G)$, because $Inn(G)$ is trivial (in case of abelian groups).

  • $\begingroup$ PS: Actually, we should use pairing (carrier, oepration(s)) to denote groups without ambiguity, but we assume, as always, that the carrier and the algebraic structure are denoted by the same symbol. $\endgroup$
    – Andrew
    Aug 23, 2015 at 0:03
  • $\begingroup$ Also, $Inn(G)$ is isomorphic to $G/Z(G)$, quotient group to the kernel of the conjugation homomorphism. Maybe you meant this. $\endgroup$
    – Andrew
    Aug 23, 2015 at 0:08
  • $\begingroup$ Also, it is considerably a bad idea to denote $Inn(G)$ by $N$, because just N suggests that it is a subgroup of G, but it isn't. $\endgroup$
    – Andrew
    Aug 23, 2015 at 0:24
  • $\begingroup$ And also, to prove this result you don't have to do it directly by definition of normality (that it contains also all the conjugate elements). Because you must have strict distinction which elements are from $G$ and which are from $Aut(G)$. For example, your $g$, performing conjugation in $Aut(G)$ is also an automorphism (not necessarily inner), but not and element of $G$. $\endgroup$
    – Andrew
    Aug 23, 2015 at 0:26
  • $\begingroup$ You write "And the whole image of a group homomorphism is always a normal subgroup in the codomain". This is false. Take, as a counterexample, any non-normal subgroup $H \le G$ and consider the inclusion map $H \to G$ given simply by $i(h) = h$. $\endgroup$ Jun 27, 2018 at 13:59

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