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I am taking an introductory course of group theory, and outer automorphism group is briefly mentioned in class, here are "two" definitions:

An automorphism of a group which is not inner is called an outer automorphism.
Outer automorphism group of a group G is the quotient Aut(G) / Inn(G).

Are they contradicting each other? because it seems to me that the 1st definition is suggesting that Outer automorphism is Aut(G) - Inn(G)....and since identity is in Inn(G)...I don't know how can we "group" all the outer automorphisms (I am trying to identify the similarity with Inn(G), which is a subgroup of Aut(G))?

the 2nd definition looks very weird to me...I mean it make some sense when G is abelian, we have Inn(G) = {e}, so that elements in the outer automorphism group are indeed outer automorphisms...but when Inn(G) = {$\phi_a,\phi_b...$} where $\phi_a(g) = aga^{-1}$, isn't outer automorphism just a coset of Inn(G)?...why are we calling {$\psi\phi_a,\psi\phi_b...$} an element of outer automorphism group?

also we proved that Inn(G) is a normal subgroup of Aut(G)...why does outer automorphism group has to be a quotient group?...can't it just be a left/right cosets?

I am try to do a bonus question in my assignment...but since there are very little information given in class about outer automorphism, I figure I should get a better understand of it first.

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    $\begingroup$ Both definitions can make sense depending on context. If we declare that an automorphism $\phi$ is outer if and only if it is not inner, then the set of outer automorphisms do not form a group (what would be the identity?) But since Inn(G) is normal in Aut(G), the cosets of Inn(G) do form a group. Any automorphism in Inn(G) is obviously inner. Any automorphism in any of the other cosets is outer. The cosets collect the automorphisms (inner and outer) in such a way that if two automorphisms $\phi$ and $\psi$ are in the same coset, then $\phi \psi^{-1}$ is an inner automorphism. $\endgroup$
    – user169852
    Commented Nov 12, 2015 at 3:38
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    $\begingroup$ The first usage is a slight abuse of notation. These are two slightly different concepts and they really shouldn't have the same name. $\endgroup$
    – Qiaochu Yuan
    Commented Nov 12, 2015 at 5:24

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Well, it depends of the definition you are using. I assume you know well inner automorphisms and automorphisms. From the definition it is easy to see that $Inn(G)\triangleleft Aut(G)$. This allows us to construct a quotient group from this $Aut(G)/Inn(G)$ which is noted $Out(G)$ and called the outer automorphism group.

To me, an outer automorphism is not an automorphism,it is an element of $Out(G)$ so this is a class in $Aut(G)$ modulo $Inn(G)$. To avoid confusion I would suggest to simply call an element in $Aut(G)$ which is not in $Inn(G)$ "non-inner automorphism", leaving the denomination "outer automorphism" to the elements of $Out(G)=Aut(G)/Inn(G)$.

With this convention an element of the outer automorphism group is an outer automorphism (which is reassuring). With your definition a non-trivial element of the outer automorphism group is a set of outer automorphisms. I would definitely suggest that you do not use the first definition you gave us for an outer automorphism essentially because your remark is correct with this definition the set of outer automorphisms is the set of automorphisms that are not inner and is not a group (which is very problematic because we have a group called outer automorphisms group).

Of course there is a semantic problem with my definition because there is no natural action (at least when $G$ is non-abelian) of $Out(G)$ on $G$ (a group of * automorphisms of $G$ does not act on $G$). In particular when $f$ is an outer automorphism (with my definition) one cannot take $g\in G$ and evaluate $f(g)$ (it just makes no sense).

However, one can show that $Out(G)$ acts on the set conjugacy classes of $G$, on the set of classes of irreducible representations of $G$, on the various group cohomology of $G$... So it has many interesting (and meaningful) actions. Those actions make the outer automorphism group a very interesting group to understand the group $G$. Let me remark (as an answer to your last question) that $Aut(G)$ also acts on the sets I have talked about, the thing is that in each case $Inn(G)$ acts trivially on each, in other words studying $Out(G)$ instead of $Aut(G)$ is really like focusing on the interesting data contained in $Aut(G)$ and that is what is relevant in studying $Aut(G)/Inn(G)$.

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We should note the difference between what we mean by: an outer automorphism and the group of outer automorphisms.

  • Automorphism $\phi$ of $G$ is outer means $\phi\notin Inn(G)$.

  • The set of outer automorphisms is not a group (identity is not an outer automorphism), hence we can not call it outer automorphism group.

  • But the set of outer automorphism, modulo inner automorphisms, i.e. $Aut(G)/Inn(G)$ is a group, which is called outer automorphism group.

(At what place you are confusing?)

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  • $\begingroup$ hmm I guess it's because this group does not follow the convention: element of an Automorphism group is an automorphism, element of an inner automorphism is an inner automorphism, element of a permutation group is a permutation etc, but what we call an element of outer automorphosim group? $\endgroup$
    – watashiSHUN
    Commented Nov 12, 2015 at 16:57
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    $\begingroup$ @watashiSHUN: An element of an outer automorphism group is a coset $\phi$Inn(G) of Inn(G). If this coset is not Inn(G) itself then it is a collection of non-inner automorphisms which are "similar" in the sense that they differ only by an inner automorphism. In other words, if $\phi$ and $\psi$ are both in the same coset of Inn(G), then $\phi = \psi \theta$, where $\theta$ is an inner automorphism. So the elements of a coset are different from each other but they are closely related. On the other hand, two automorphisms from different cosets do NOT simply differ by an inner automorphism... $\endgroup$
    – user169852
    Commented Nov 13, 2015 at 0:47
  • $\begingroup$ [continued] so in a sense, they are "more" different than if they were in the same coset. $\endgroup$
    – user169852
    Commented Nov 13, 2015 at 0:48

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