If a 2D object has vertical reflection symmetry and horizontal reflection symmetry does that imply it has 2-fold symmetry?

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From the group symmetries of a square, or $D_4$, we know that $H \circ V = R_2$, where

  • $H$ indicates a flip/reflection about a horizontal axis
  • $V$ indicates a flip/reflection about a vertical axis
  • $R_{180}$ indicates a $180^\circ$ rotation about the point where said horizontal and vertical axes intersect.

I know this means that performing $V$ followed by performing $H$ on a $2D$ object is equivalent to just performing $R_{180}$ on said object (independent of whether said object has any symmetries$^1$). Can I also conclude from this that any object with vertical and horizontal reflection symmetry must also have $2$-fold (i.e. $R_{180}$) symmetry?

In general, can I conclude that if an object has two symmetries $f, g \in D_4$, then it must also have the compositions of those symmetries, namely $f \circ g$ symmetry and $g \circ f$ symmetry? Even if has no other symmetries in $D_4$? (Not sure if relevant, but $\{H, V, R_{180}\}$ is closed under composition.)

Intuitively, it feels like this should be true for any $2D$ object, not just squares. If an object has $f$ symmetry and $g$ symmetry and $f \circ g = h$, then I can:

  1. perform $g$ on it, get back the same object,
  2. then perform $f$ on it, get back the same object.

The net effect is that I performed just $h$ on the object and got back the same object, which means it has $h$ symmetry, no?

Reference: http://mathonline.wikidot.com/the-group-of-symmetries-of-the-square

(Note: the reference uses $\mu_1$ instead of $V$, $\mu_2$ instead of $H$, and $\rho_2$ instead of $R_{180}$.)

$^1$For example, the image of lowercase letter p has neither $V$ nor $H$ symmetry, but I can still perform $V$ on it to get q, then perform $H$ to get d. This would have been equivalent to performing $R_{180}$ on p.