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:
well…
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 $G$.
UPDATE:
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?
Best Answer
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 $\operatorname{Inn}(G)$.