1
$\begingroup$

Let $Q$ be a (plane) quadrilateral. Show that the order of its group of symmetries $S(Q)$ is less than or equal to $8$.

I tried saying that a quadrilateral cannot have more than 4 rotations, because it only has 4 vertices. As for reflections, well there are only two diagonals, for the vertices and edges, so that's 4. Is that a rigorous proof? (obviously not)

Also does the $s r s^{-1} = r^{-1}$ property hold for irregular polygons? What are the names of the groups of symmetries of irregular polygons? Are they all isomorphic to some dihedral group?

$\endgroup$

1 Answer 1

2
$\begingroup$

Take two points on the polygon with an edge between them. These two points must map to the endpoints of an edge, of which there are four. This can be done in two ways for each edge, so there are eight possibilities. Once this edge is mapped, the other edges are completely determined because a point is adjacent to exactly two edges.

In general this symmetry group will be a subgroup of the dihedral group because the square is the "most symmetric" quadrilateral. If we are using the orthogonal group to construct our symmetries, the symmetries must preserve both length and angles, and the only quadrilaterals with all lengths and angles equal are the squares.

$\endgroup$
2
  • $\begingroup$ Ah thanks, that's a much better way of thinking about it. $\endgroup$
    – user85798
    Jan 17, 2015 at 15:16
  • $\begingroup$ @LTS you're welcome. $\endgroup$ Jan 17, 2015 at 15:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .