Orbits in graph

Question

Can anyone help me visualize orbits in graph.

Let $G=K_4$ with vertex set $V=\{1,2,3,4\}$ and edge set $E\left(K_4\right)=\{12,13,14,23,24,34\}$

Now if i take permutation $\sigma=(1~2~3~4)\in Sym(V)$ ($Sym(V)$ is permutation group on $V$.

https://math.stackexchange.com/a/423948/520264 in this answer he says that if there is $k$-cycle in permutation then there is $\Big\lfloor\dfrac{k}{2}\Big\rfloor$ orbits of edges between vertices of $k$- cycle.

Here we have $4$-cycle in $\sigma=(1~2~3~4)$ by above formula there is $\Big\lfloor\dfrac{4}{2}\Big\rfloor=2$ orbits of edges between vertices of $4$-cycle.

Which are these two orbits?

Answer 1

$k=4$ is a bit too small to visualize, so here is a picture for $k=8$:

Imagine that we are looking at a permutation $\sigma$ that permutes these $8$ vertices in a cycle, and we want a graph fixed by this permutation. (Maybe there are also other vertices we're not looking at, because we don't care what they do for the moment.) Then:

• The $8$ red edges are an orbit of the action of $\sigma$ on $E(K_n)$. (They are either all present or all absent, if we want $\sigma$ to leave the graph unchanged.)
• The $8$ blue edges are another orbit.
• The $8$ purple edges are a third orbit.
• The $4$ orange edges are a fourth orbit.

Essentially, if you rotate the picture above by $45^\circ$, then each edge is moved to another edge of the same color.

In general, there are $\lfloor \frac k2\rfloor$ orbits characterized by the length of an edge, when we draw the vertices in the cycle as a regular $k$-gon.