Exploring the G-Set Action of C on G=(R,+)

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In summary, the homework statement states that there is a mapping from the vector space C to the set C. This mapping makes C into a G-set. The action of G on C is a G-set, which is a set of geometric transformations that are satisfied by 0. The orbits and stabilizers of a point in C are determined by the value of z and the angle a.
  • #1

Homework Statement

Let G = (R,+) and define az = eiaz for all z in C and a in G. Show that this definition makes C into a G-set, describe the action geometrically, and find the orbits and the stabilizers.

The Attempt at a Solution

A mapping G x C -> C, denoted (a,z) -> az = eiaz for z in C.
Let eiaz = (cos(a) + isin(a))(x+iy)
Thus az = a(x+iy) = (cos(a) + isin(a))(x+iy)
We can be called this an action of G as satisfied the follwing:
Since 0 is identity of G, 0z = (cos(0) + isin(0))(x+iy) = (x+iy) = z
Also this satisfies such as a(bz) = (ab)z for all z in C and for all a,b in G
Thus G acts on C, called a G-set.
I have no idea how to show and find orbits and stabilizers.
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  • #2
Looks ok so far. You might want to elaborate a little on why a(bz)=e^(ia)(e^(ib)z)=(ab)z. If you think of the complex plane C as R^2, with coordinates (x,y), what kind of geometric operation does e^(ia)z correspond to?
  • #3
Thanks very much,
I think I can modify to more correct..
But I still have no idea about orbits and stabilizers.
Could you give me any hints?
Are the orbits just circle?
like, r is distance of z and denoted |z| and
e^ia is the circle which is 1 of radius??
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  • #4
Yes, the orbits are circles. e^(ia)z is z rotated around the origin by an angle a. Write z in polar form z=r*e^(it) to see this.
  • #5
and the stabilizers are when a = 0?
  • #6
Depends on what value of z you are talking about. And the stabilizer of a given z is a subgroup. If you are saying it's the trivial subgroup {0} I don't agree with that.
  • #7
I do not know what you mean.
Yeah, I know the stabilizers are subgroup of G but
I don't know how I can apply to this problem..
I have to find in case of az = z, right?
  • #8
Right. 1) suppose z=0, what's the stabilizer group? 2) suppose z is not zero. What about a=2pi?
  • #9

I found something in my textbook,
If X is a G-set and x in X, the orbit of x is Gx = {ax | a in G}. combining this with "|Gx| = |G:S(x)| for each x in X" gives equivalent conditions that the orbit is a singleton:
Gx = {x} <=> ax=x for all a in G <=> S(x)=G.

And, when z = 0 and z is not zero, both cases seem just G = (R,+)..
  • #10
If z=0 then a0=0 for all a. So, yes, the stabilizer is G. If z is not equal to zero that's definitely not true. Which rotation angles take a point to itself? It's not just zero. What does a rotation by 2pi do?!
  • #11
Thank you very much!

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