# How do I find all of the orbits and stabilisers of X?

Consider $D_{10}$ The group of symmetries of the regular pentagon. Let $\sigma= (12345)$ and $\tau =(13)(45)$ being rotation by $72^{\circ}$ and reflection (with 2 being the fixed point) respectively. Therefore $D_{10} = ({1, \sigma, \sigma^2, \sigma^3, \sigma^4, \tau, \sigma\tau, \sigma^2\tau, \sigma^3\tau, \sigma^4\tau})$.

Let $D_{10}$ act on the set $X =\left \{ 1,2,3,4,5 \right \}$ of vertices by permutation.

How do I find all the orbits of $X$?

How do I find all the stabilisers of $X$?

I'd like to thank you in advance for any help/guidance you can give me.

• Consider the subgroup of $D_{10}$ consisting of the rotations. Is there any pair of vertices $v,w\in X$ such that $v$ cannot be rotated to $w$? – Bungo Dec 5 '15 at 2:07
• P.S. $360/5 = 72$, not $108$ :-) – Bungo Dec 5 '15 at 2:11
• @Bungo thanks man, I was being silly. I'm still not too sure. Surely $v$ can be rotated to any $w$? – Geometry Dec 5 '15 at 2:16
• Yes, exactly. We can move any vertex to any other vertex by some rotation. So there is just one orbit, namely all of $X$. Now, how does the size of the orbit relate to the size of the stabilizer of a particular vertex in the orbit? – Bungo Dec 5 '15 at 2:18
• @Bungo I'm really not sure :( – Geometry Dec 5 '15 at 2:25

Summarizing the comments and discussion in chat:

Given any vertices $v,w \in X$, there is a rotation that takes $v$ to $w$, so there is just one orbit, namely all of $X$.

Let's denote the group by $G = D_{10}$.

Now choose any vertex $v \in X$. The orbit containing $v$ is all of $X$, as noted above. This orbit has size $|X| = 5$. Let us denote the stabilizer of $v$ by $G_v$. By the orbit-stabilizer theorem, the index of $G_v$ equals the size of the orbit containing $v$. In other words, $|G:G_v| = |X| = 5$. Since $|G:G_v| = |G| / |G_v|$, this means that $|G| / |G_V| = 5$. Since $|G| = |D_{10}| = 10$, it follows that $|G_v| = 2$.

Now $G_v$ is a group, so it contains the identity. (This makes sense since clearly the identity stabilizes every vertex.) Since $|G_v| = 2$, there is one additional element of $G$ which stabilizes $v$. None of the nontrivial rotations stabilize any vertex, so it must be one of the "flips". In particular, if we draw a line from the vertex $v$ to the midpoint of the opposite side, and flip the pentagon along that axis, then $v$ is fixed (stabilized) by that flip.

To summarize, each vertex of the pentagon is fixed by two elements of $D_{10}$: one is the identity, and the other is the flip along the axis which connects the vertex to the opposite side.

I'll leave it to you to work out which element of $D_{10}$ (in terms of $\tau$ and $\sigma$) corresponds to the flip that fixes each vertex.