What is the outer automorphism group? I try to understand what the notion of outer automorphism group is. I know an inner automorphism of a group G is a map  $ \iota_{g} $ attached to some element  $ g $ of  $ G $ that sends an element  $ h $  to  $ ghg^{-1} $. Two elements a  $ a $ and  $ b $ are said to be conjugate if there is some  $ g $ in G such that  $ b=gag^{-1} $.
So is an outer automorphism of  $ G $ an automorphism of  $ G $ that sends an element  $ a $ to an element  $ b $ that is not conjugate to  $ a $? Or is it something else ? Incidentally, is it true that the outer automorphism group of the absolute Galois group of the rationals consists of the identity and the complex conjugation ? 
 A: I think of the outer automorphism group $$\operatorname{Out}(G):=\frac{\operatorname{Aut}(G)}{\operatorname{Inn}(G)}$$ as being "all automorphisms, apart from the obvious ones we always get". So removing the "obvious" inner ones, which in many ways look like the group itself, allows us to study the "true" symmetries of the group.
Therefore, outer automorphisms are not automorphisms, but are equivalence classes of automorphisms. This is extremely similar to viewing elements of groups as words over the generators - elements are not words, but are equivalence classes of words.
For example, consider the free group on two generators, $F(a, b)$. Now, the derived subgroup of $F(a, b)$ is a characteristic subgroup and so automorphisms are preserved under the abelianization map $F(a, b)\rightarrow \mathbb{Z}\times\mathbb{Z}$. So there exists a map $\phi: \operatorname{Aut}(F(a, b))\rightarrow\operatorname{Aut}(\mathbb{Z}\times\mathbb{Z})$. Now, suppose $\alpha_1=\alpha_2\pmod{\operatorname{Inn}(F(a, b))}$. Then there exists an inner automorphism $\gamma$ (corresponding to conjugation by some $g\in F(a, b)$) such that $\alpha_1=\alpha_2\gamma$. That is, there exists some $g\in F(a, b)$ such that for all $h\in F(a, b)$ we have that $\alpha_1(h)=g^{-1}\alpha_2(h)g$. Therefore, under the abelinisation map $\alpha_1$ and $\alpha_2$ act identically. Hence, we can factor the above map $\phi$ as:
$$\operatorname{Aut}(F(a, b))\rightarrow\operatorname{Out}(F(a, b))\rightarrow\operatorname{Aut}(\mathbb{Z}\times\mathbb{Z}).$$
A similar "factorisation" always holds, no matter the group. However, the free group of rank two is special, as Nielsen in the 1920s proved that the map $\operatorname{Out}(F(a, b))\rightarrow\operatorname{Aut}(\mathbb{Z}\times\mathbb{Z})$ is an isomorphism. Which is really pretty and means that $\operatorname{Out}(F(a, b))\cong\operatorname{GL}_2(\mathbb{Z})$.
Outer automorphism groups are extremely natural groups to study. For example:


*

*[the example stated above.] $\operatorname{Out}(F_2)\cong\operatorname{GL}_2(\mathbb{Z})$, where $F_2$ is the free groups of rank two.

*Let $\Sigma_g$ be a closed, orientable surface of genus $g$, and write $\operatorname{MCG}^{\pm}(\Sigma_g)$ for its extended mapping class group (so the quotient group of the space of all homeomorphisms, rather than of just the orientation-preserving ones). Then $\operatorname{Out}(\pi_1(\Sigma_g))\cong\operatorname{MCG}^{\pm}(\Sigma_g)$.

*Let $H$ and $K$ be groups and suppose $\alpha_1, \alpha_2\in\operatorname{Aut}(H)$. If ${\alpha_1}={\alpha_2}\pmod{\operatorname{Inn}(G)}$ then $H\rtimes_{\alpha_1}K\cong H\rtimes_{\alpha_2}K$.

*There does not exist any group $G$ such that $\operatorname{Aut}(G)\cong\mathbb{Z}$. However, for every group $H$ there exists some group $G$ such that $\operatorname{Out}(G)\cong H$.
