I'm having some conceptual issues with the infinite cyclic group $C_\infty$. Finite groups $C_n$ have a clear representation as integers $0,1,\cdots,n-1$ under addition $(\operatorname{mod} n)$, or as the rotation group of the $n$-gon for $n\geq 3$. The rotation group of a circle, which is what I interpreted $C_\infty$ to be, has uncountable order since any real angle $[0,2\pi)$ is valid. This would make it isomorphic to $[0,2\pi)$ under addition $(\operatorname{mod} 2\pi)$. But online it says $(\mathbb{Z},+)$ is also isomorphic, which doesn't make sense to me because it has order $\aleph_0$. Also, the first group has two inverses $0$ and $\pi$, while this group only has $0$.

I'm guessing my interpretation is wrong. The textbook never defines what $C_\infty$. What exactly is it?

  • $\begingroup$ "The rotation group of a circle, which is what I interpreted $C_\infty$ to be". Well, there we are. $\endgroup$ – Lord Shark the Unknown May 25 at 15:36
  • $\begingroup$ Is there a good geometric interpretation then? The best one I can think of is rotating a circle by an integer number of radians. $\endgroup$ – Ovinus Real May 25 at 15:39
  • $\begingroup$ The order of the rotation group of a circle is $2^{\aleph_0}$. Whether or not this is $\aleph_1$ is not decidable from the axioms of set theory. $\endgroup$ – DanielWainfleet May 25 at 15:56
  • $\begingroup$ Should have just said uncountable, sorry $\endgroup$ – Ovinus Real May 25 at 16:05

The (up to isomorphism) infinite cyclic group is just $\mathbb{Z}$ under addition.

You can visualize it as the group of integer shifts of the integers.

You can also visualize it as the rotations of the circle through integer numbers of radians, but that's not pretty geometrically since the orbit of any point on the circle is dense.

The group of all rotations of the circle is not cyclic.


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.