Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find all the symmetries of the $ℤ\subset ℝ$.

I'm not sure what is meant with this. My frist thought was that every bijection $ℤ→ℤ$ is a symmetry of $ℤ$.

My second thought was that if I look at $ℤ$ as point on the real line, then many bijections would screw up the distance between points. Then I would say that the set of symmetries contains all the translation: $x↦x+a$ and the reflections in a point $a∈ℤ$, which gives, $x↦a-(x-a)$.

share|cite|improve this question
up vote 4 down vote accepted

It probably means the latter, in this case. Note that we can also reflect over the midpoint of two (successive) points in $\Bbb Z$. Otherwise, you've got it covered.

Addendum: It might be fun for you to prove that every such symmetry of $\Bbb Z$ can be obtained as a composition of finitely-many instances of the two maps $x\mapsto x+1$ and $x\mapsto-x$.

share|cite|improve this answer
So in general, you've got the elements $t,r$ with $t:ℤ→ℤ:x↦x+1$ and $r:ℤ↦ℤ:x↦-x$. And then the group of symmetries is generated by $t$ (infinite order) and $s$ (order 2). Right ? – Kasper Mar 16 '13 at 20:13
@Kasper: That's hilarious! I literally just finished adding that fact to my answer when your comment showed up. Yes, that's true. – Cameron Buie Mar 16 '13 at 20:16

Both of you thoughts are correct, but in different contexts. If we regard $\mathbb{Z}$ as just a set with no extra structure then the symmetries of that set are just the bijections, (as in this case, we are thinking as $\mathbb{Z}$ as a bag of points). If on the other hand, we add the extra structure of the distances, then the symmetries are more restricted and the symmetries are those maps that preserve distance.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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