# Very very quick quiestion about D_4?

1. Nov 22, 2008

### sutupidmath

So, i was wondering if i wanted to find the total number of automorphisms of D_4(octic group) then what would that number be?

My answer is 4. Here, is how i reasoned about it. SInce D_4 is generated by only two elements (1234) and (24) i assume it is sufficient to fix their images, once we have done this, we have well defined one automorphism. NOw, we know that the isomorphism preserves the order of each element. THis means that the images of (1234) can be only (1234) itself and (1432). So, all isomorphisms (automorphisms) would start like this:

(1234)-->(1234) and (24)-->(24)
(1234)-->(1234) and (24)-->(13)
(1234)-->(1432) and (24)-->(24)
(1234)-->(1432) and (24)-->(13)

So, we would end up having only 4 automorphisms. Is this correct?

Edit: I just realized that there should be 8 automorphisms, but how do i go about finding the other 4?

Last edited: Nov 22, 2008
2. Nov 22, 2008

### ptr

Consider a square, and label the vertices 1, and 2, and 3, and 4. Then, proceed to record the automorphisms as you rotate it and reflect it along its axes of symmetry. Then, you'll be able to find all eight automorphisms if you do not forget, as some might, though I trust you wouldn't, the identity automorphism.

3. Nov 22, 2008

### sutupidmath

It looks to me that what you are indeed sayin' is not find it's automorphisms, but rather the permutations that go along with it: Like

$$D_4=\{(1),(1234),(13)(24),(1432),(24),(14)(23),(13),(12)(34)\}$$ but this is the octic group itself

If this is what you are saying, then these are not it's automorphisms. But we rather need to find all the mappings $$\theta:D_4->D_4$$ such that it forms an automorphism. I know how to find the 4 inner automorphisms, but i am faling to find the 4 others. Because since D_4 is generated only by two elements $$\rho, \phi$$ then to find its inner automorphisms we only look at the conjugates of these two generators. THen since the classes of conjugation for these two elements are $$C(\rho)=\{\rho, \rho^2\}$$ and $$C(\phi)=\{\phi,\rho^2\phi\}$$ then there are only 4 such automorphisms.

Can anyone else enlighten me?

Last edited: Nov 22, 2008
4. Nov 22, 2008

### ptr

Forgive me, I am probably wrong, but I thought that the mapping from 1234 to the permutations that go along with the reflections and rotations described were the automorphisms, and the automorphisms formed the automorphism group of the dihedral group that we are talking about. In which case, the mapping from 1234 to each of the elements of the dihedral group discussed is the eight elements of the automorphism group of the dihedral group. To find it out in terms of rotations and reflections would simply be to work it out with a square. Sorry again.

5. Nov 22, 2008

### sutupidmath

Ok, now i am confused. YOu are saying that the 'mappings' defined as below

$$\theta: (1234)->(1234),(1234)->(12), (1234)-->(12)(34)....$$ form a group automorphism?

Well, first of all this is not even well defined, since how are we gonna find where the other generator (24) is being mapped by this isomorphism.??? I believe that we need to fix the image of the other generator as well, secondly, say for example that (24)->(12) as a counterexample to this one, would be: let's suppos that (1234)-->(13) and (24)-->(13)

Without doing any further reasoning, this is not possible since under an isomorphism the orde of the elements is preserved. and obviously this is not the case here.

6. Nov 22, 2008

I think you were on the right track in your first post, but you missed two other possible reflections: (24) -> (12)(34) and (14)(23).

7. Nov 22, 2008

### sutupidmath

we, yeah, i was thinking something simmilar, but i took as a trial the mapping (12)-->(13)(24), and this didn't work, so for some stupid reasons i thought it won't work for the others as well,and was too lazy to check it as well..lol...

Thnx.

Last edited: Nov 22, 2008
8. Nov 22, 2008