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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).

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  • $\begingroup$ 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 $G_z$. $\endgroup$ – Bungo Dec 15 '15 at 23:53
  • $\begingroup$ @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 $\endgroup$ – abstractGone Dec 16 '15 at 0:35
  • $\begingroup$ Oh, is the answer the subgroup of $D_6$ whose elements have order $3$? $\endgroup$ – abstractGone Dec 16 '15 at 0:59

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