Let $X \subseteq R^2$. A symmetry of $X$ is isometry $f: R^2 \to R^2$ such that $f(X) = X$.

For example, square has $8$ symmetries one of which is $R_{90}(a, b) = (-b, a)$.

Is an element of $X$ is both an image and pre-image of a symmetry? So, then shouldn't $R_{90}(a, b) = (a, b)$? I understand this isn't correct if we look at it geometrically, but I am trying to understand the algebraic gist of it.


In this case, $X$ is the whole square, so for $R_{90}$ to be a symmetry, we only need that the image of $X$ equals $X$, not that every point in $X$ is fixed by $R_{90}$. So while you're right in that $R_{90}(a,b) \neq (a,b)$, it is true that $$ X = \{R_{90}(x) : x \in X\}, $$ and that's all we need.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.