Is permutation in a subgroup of $S_6$?

Question

Consider the permutation $\sigma = (123)(46)$ Suppose that the vertices of a regular hexagon are labeled in counterclockwise order with the numbers $1$ through $6$. Then the dihedral group $D_6$ of symmetries of the hexagon can be identified with a subgroup of $S_6$. Is $\sigma \in D_6$?

My attempt: Note that $D_6$ is generated by rotations and reflections, so $D_6 = \{e, s, r, ..., r^5, rs, r^2s, ..., r^5s\}.$ Intuitively, I think, if we change the label of 1 vertex, then clearly the labels of other vertices would change too, but $\sigma$ moves all elements except $5$. Therefore $\sigma \notin D_6$. Is there a more formal way to show this?

Answer 1

I am tempted to use a possibly less formal but a shorter argument:

The vertices labeled $1$ and $6$ are adjacent, but their images under $\sigma$, that is, the vertices labeled $2$ and $4$ are not. Therefore $\sigma$ is not a symmetry of the hexagon.

The idea being that adjacent is equivalent to closest neighbor. And a symmetry preserves distances, hence the closest neighbor -relation.

Plenty of alternatives:

• A symmetry will map pairs of diametrically opposite points to other such pairs. So if vertex number $5$ is a fixed point, its opposite vertex $2$ should also be fixed. If pressed to add something to your answer, I might use this!
• If $\sigma$ were a symmetry, so would $\sigma^3$. But $\sigma^3=(46)$ only switches two vertices, which is absurd given that we are looking at a hexagon (to see the absurdity we may want to go back to looking at adjacency, possibly meaning that this is no different from the other approaches).