Stablizer of $D_{6}$

I am hoping someone can help me with this exercise as I do not really understand as there is so much information that I am confused. Firstly, what is a $$G$$ space? It seems to be related to group actions, but I can't actually find what it means. I've heard of $$G$$-set's, but G space no.

So each $$r^k$$ rotates the plane counterclockwise by $$\frac{2pi k}{3}$$ radians, and $$s$$ conjugates it. Given any complex $$z$$, I need to find the elements $$g$$ of $$D_6$$ such that $$g$$'s action on $$z$$ takes it to $$z$$ itself. That's what $$G_z$$ means.

BUT, I don't see any rotation except for $$k=3$$ allowing $$r^k \cdot z = z$$ to be satisfied. Since $$r^{3}$$ rotates by $$\frac{2\pi}{3}$$ which is a full rotation (which is what we are looking for).

• I've never heard of a $G$-space either. A google search brings up this book, which seems to indicate that a $G$-space is a $G$-set which is also a topological space. Presumably $G_z$ is the stabilizer of the point $z \in \mathbb C$. Then, according to the notation in your first paragraph, $(G_z)$ would be the set of subgroups of $G$ which are conjugate to $G_z$. – Bungo Dec 15 '15 at 23:53
• @Bungo So only k=3 work for rotations. So this eliminates r,r^2,r^4,r^5. Is there a nice way to look at rs, r^2s,r^3s,r^4s, and r^5s? So we look at rs, then we need to make sure that rs.z = z right? If so, how do we see if r.s = z? or r^k . s = z I mean – abstractGone Dec 16 '15 at 0:35
• Oh, is the answer the subgroup of $D_6$ whose elements have order $3$? – abstractGone Dec 16 '15 at 0:59