Determine Aut(G)/Inn(G), given G = D4

[NasuSama]

Homework Statement

Determine $Aut(G)/Inn(G)$, given $G = D4$.

2. The attempt at a solution

Let $D_4 = \{e, x, y, y^2, y^3, xy, xy^2, xy^3\}$

Found that $Inn(G) = \{\phi_e, \phi_x, \phi_y, \phi_{xy}\}$ and $\mathrm{Aut}(G)$ consists of 8 bijective functions. But I can't find $Aut(G)/Inn(G)$.

 

[micromass]

What is the order of the quotient?

 

[NasuSama]

What is the order of the quotient?

Then, it must be 2. This means there are 2 functions, but I found 4, which is quite strange.

 

[NasuSama]

I have a problem. Why would this link says that there are 4 outer automorphisms? I don't get it...

http://books.google.com/books?id=sb...#v=onepage&q=outer automorphism of d4&f=false

 

[micromass]

They're saying that there are 4 elements in $Aut(G)$ that are also in $Inn(G)$. And that there are 4 elements in $Aut(G)$ that are not in $Inn(G)$.

 

[NasuSama]

So this means that I need to find two cosets. This is the implication of what the writer is trying to show.

 

[NasuSama]

Let's give a good try to determine the set of outer automorphism.

Well, there are four automorphisms that map $y$ to $y$ while there are other four automorphisms that map $y$ to $y^3$. So the first four belong to one set while the remaining four belong to another set. Thus, we obtain two sets in $\mathrm{Aut}(G)/\mathrm{Inn}(G)$.

Is this even right? I am assuming that is the answer.

 

[NasuSama]

I wonder if the one I have is true for $\mathrm{Out}(D_4)$