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I would like to ask - why does $\operatorname{Out}(G)$ := ${\operatorname{Aut}(G) / \operatorname{Inn}(G)}$ make sense? As $\operatorname{Out}(G)$ is outer automorphism group, I would expect it to include outer automorphisms, but for nontrivial $\operatorname{Inn}(G)$ the quotient is a group of cosets which are sets of functions, not functions. Isomorphism would be totally intuitive for me but equality I do not understand.

Thank you in advance!

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    $\begingroup$ Sure they are cosets but all the representatives are automorphisms which are not inner (if the coset is not the zero coset). I have only seen outer automorphisms to be automorphisms that arent inner automorphisms, so perhaps you are using a different definition? $\endgroup$ – user722227 Mar 2 at 0:22
  • $\begingroup$ That's my point, definition gives us sets, which are not functions, so it would mean that outer autom. group does not contain automorphisms. $\endgroup$ – Duck Mar 2 at 0:35
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    $\begingroup$ the key point is that $Inn(G)$ is a normal subgroup of $Aut(G)$. $\endgroup$ – janmarqz Mar 2 at 1:51
  • $\begingroup$ A similar question was asked here. You might find the answer/discussion useful. (Although to be honest I'm struggling to work out what your question actually is! The definition makes sense because, as janmarqz says, $\operatorname{Inn}(G)$ is normal, but surely you know this?) $\endgroup$ – user1729 Mar 2 at 7:38
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    $\begingroup$ This is indeed confusing, and results from inconsistent notation, but unfortunately the notation is standard, so we just have to live with it. The point is that the elements of the outer automorphism group are not themselves outer automorphisms (which are automorphisms that are not inner) but cosets. $\endgroup$ – Derek Holt Mar 2 at 8:41
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The set of outer automorphisms (taking an "outer automorphism" of a group to be an automorphism that is not inner) does not give a group. The easiest way to see this is to consider that the identity automorphism is an inner automorphism. But there can also be other cases where the composition of two outer automorphisms gives a nontrivial inner automorphism.

Thus it makes sense to consider equivalence classes of outer automorphisms. Looking at $\mathrm{Aut}(G)/\mathrm{Inn}(G)$, you have 1: an actual group, and 2: every nonzero equivalence class consists entirely of outer automorphisms.

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I want to just make a quick addition to the other answers:

Some people will conflate “outer automorphism” and “automorphism which is not inner” in casual math conversations. This may be a source of confusion — these are distinct ideas.

Although outer automorphisms are not actual functions on $G$, they do induce functions on the conjugacy classes of $G$. (Why?) You might hear talk of outer automorphisms acting like functions — this is one context in which such language occurs.

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So indeed one defines $Out (G) := Aut(G)/Int(G)$. But an element of $Out(G)$, an outer automorphism, is not an automorphism of $G$ (as you notice if I understood correctly) - the naming is confusing. It is just what it is, an automorphism up to composing with an inner automorphism.

By the way, there is no definition of when an automorphism of $G$ is an outer automorphism.

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  • $\begingroup$ So outer automorphisms group just simply do not contain functions and a problem is in naming? $\endgroup$ – Duck Mar 2 at 0:37
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    $\begingroup$ Yes, you can say like that. But this is very common in math, that sometimes you have a quotient set rather than a subset. For example, when working with integers modulo $3$, you can call the elements of the quotient set $\mathbb{Z} / 3 \mathbb{Z}$ "modular integers", say. And then you have the same naming "issue" - a modular integer is not an integer, the set of modular integers is a quotient set of the set of integers, not a subset of the set of integers. And in fact I saw that a lot of students, when they see something like $\mathbb{Z}/3 \mathbb{Z}$ for the first time, sometimes think that $\endgroup$ – Sasha Mar 2 at 0:41
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    $\begingroup$ that $\mathbb{Z} / 3 \mathbb{Z}$ is a subset of $\mathbb{Z}$, usually they think it is the subset consisting of $0,1,2$. But this is of course not good to think like this. So it is a bit abstract, I guess. $\endgroup$ – Sasha Mar 2 at 0:41
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Since the definition is everything that survives modding out by Inn, the definition is precisely as those equivalence classes of automorphisms which are not inner.

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  • $\begingroup$ Exactly, but what you mean are functions, which are not inner, so why do we produce cosets? Did I understood you correctly? $\endgroup$ – Duck Mar 2 at 0:32
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    $\begingroup$ "the definition is precisely as those which are not inner." That's wrong - the elements of $Out(G)$ aren't automorphisms but sets of automorphisms, namely cosets of $Inn(G)$. (There is of course a map of sets $Aut(G)\setminus Inn(G)\rightarrow Out(G)$, but it's neither injective nor surjective.) $\endgroup$ – Noah Schweber Mar 2 at 0:43
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    $\begingroup$ It may technically be wrong, but it is typical to use a coset representative in place of an entire coset. As in, $\Bbb Z_n=\{0,1,2,\dots,n-1\}$. @NoahSchweber $\endgroup$ – Chris Custer Mar 2 at 1:24

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