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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 ?

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    $\begingroup$ There's the trivial automorphism, and then the one that switches the two non-identity elements. My point here is that there is an automorphism which sends an element to another element in a different conjugacy class. That is, it's not an inner automorphism. $\endgroup$ – Josh B. Sep 13 '18 at 23:17
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    $\begingroup$ But the inner automorphisms form a normal subgroup of the group of all automorphisms; and I think I’ve seen the quotient called the “group of outer automorphisms”. But I’m not a group theorist, and what do I know? $\endgroup$ – Lubin Sep 14 '18 at 0:04
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    $\begingroup$ Indeed, the terminology surrounding this is a bit strange: $\mathrm{Aut}(G)/\mathrm{Inn}(G)$ is usually called the "outer automorphism group of $G$", but if someone speaks of "an outer automorphism of $G$", they usually mean an automorphism that is not inner. Note that, using this terminology, outer automorphisms are not actually element of the outer automorphism group. For this reason, the phrase "group of outer automorphisms" is usually completely avoided as, in the usage above, outer automorphisms do not form a group. $\endgroup$ – verret Sep 14 '18 at 1:27
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    $\begingroup$ @Berci I've never seen your usage before. The outer automorphism group is almost universally defined as $\mathrm{Aut}(G)/\mathrm{Inn}(G)$, see for example en.wikipedia.org/wiki/Outer_automorphism_group. In particular, it does not "contain" the group of inner automorphisms, since it is a quotient of $\mathrm{Aut}(G)$, not a subgroup. $\endgroup$ – verret Sep 14 '18 at 1:29
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    $\begingroup$ Ok, I had a misconception, sorry.. With the correct terminology then, the outer automorphisms don't form a group, as e.g. the identity doesn't belong there. But the 'outer automorphism group' is defined as the quotient group of the automorphism group by the (normal sub)group of inner automorphisms. A generic element of the outer automorphism group is thus represented by an automorphism. If that automorphism is a conjugation, then it represents the identity element of the outer automorphism group. $\endgroup$ – Berci Sep 14 '18 at 8:34
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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:

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

  2. 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 $\operatorname{Out}(\pi_1(\Sigma_g))\cong\operatorname{MCG}^{\pm}(\Sigma_g)$.

  3. 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 $H\rtimes_{\alpha_1}K\cong H\rtimes_{\alpha_2}K$.

  4. 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 $\operatorname{Out}(G)\cong H$.

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  • $\begingroup$ And so does one have $ \operatorname{Out}(\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}))\cong\mathbb{Z}/2\mathbb{Z} $? $\endgroup$ – Sylvain Julien Sep 14 '18 at 9:40
  • $\begingroup$ I don't know much about Galois groups. Why would you think that this was true? $\endgroup$ – user1729 Sep 14 '18 at 11:28
  • $\begingroup$ You talk about "true symmetries" of the considered group, and I read here and there that in some sense, the identity and the complex conjugation are the only elements of the absolute Galois group of the rationals you can describe explucitly without the axiom of chouce. $\endgroup$ – Sylvain Julien Sep 14 '18 at 12:03
  • $\begingroup$ Apparently it was an open problem in the 70s whether $\operatorname{Out}(\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ is trivial or not (so I guess your question is hard). See Masatoshi Ikeda, On the group automorphisms of the absolute Galois group of the rational number field, Arch. Math (1975) 26: 250 link $\endgroup$ – user1729 Sep 14 '18 at 13:16
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    $\begingroup$ No idea. These questions are better addressed via a new question. $\endgroup$ – user1729 Sep 17 '18 at 8:06

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