Do there exist cyclic groups that are continuous/uncountable over an interval?

Question

For my specific line of research, I am applying the Taylor series for sec(x), that is $$ 1 + \frac{x^2}{2} + \frac{5x^4}{24} \space ... $$ Even though $x$ is considered the same geometrically for, say, $2\pi + \frac{\pi}{3}$ and $\frac{\pi}{3}$, the Taylor series seems to "want" $x$ to be below $\frac{\pi}{2}$. I have a basic understanding of group theory, and am aware of cyclic groups that are finite, such as, say, $\mathbb{Z}_6$, which cycles over: $$0 \space \space \space 1 \space \space \space 2 \space \space \space 3 \space \space \space 4 \space \space \space 5 $$ What I am not aware of are continuous groups which are also cyclical. I am not aware of the notation convention, so forgive me, but is there a group that could be written as something of the form $\mathbb{R}_{2\pi} $ which cycles (and is continuous) over the interval: $$ 0 \space \space \space ... \space \space \space 2\pi $$ I am aware that this group may not satisfy the angle requirement that I need, but I am looking for something of this form. Any references to reading material would of course also be appreciated. Thank you in advance.

Answer 1

A cyclic group is a group that can be generated by one element. There are only finite or countable cyclic groups, and they are exactly $\Bbb{Z}_n$ for a positive integer $n$, or $\Bbb{Z}$ in the infinite case. None of these are "continuous" in the sense you mean.

You can take the half-open unit interval, $[0,1)$, and it will admit a group structure with the operation of addition modulo 1 (e.g., $\frac{3}{4}+\frac{3}{4} = \frac{1}{2}$). One mathematical way of formulating it is to look at it as the following quotient group: take the additive group of the reals, $\Bbb{R}$, modulo the (normal) subgroup $\Bbb{Z}$.

Answer 2

Perhaps you might be thinking about the circle group $S^1$, which is the set of unit norm complex numbers, where the group operation is complex multiplication. This group is often parameterized as follows, using polar notation for complex numbers: $$S^1 = \{\cos(\theta) + i \, \cos(\theta) \mid 0 \le \theta < 2\pi\} $$ Using this parameterization one can easily work out, using the rules of complex multiplication and trigonometric identities, that the multiplication operation is given by $$\bigl(\cos(\theta) + i \, \sin(\theta)\bigr) \bigl(\cos(\phi) + i \, \sin(\phi)\bigr) = \cos(\theta+\phi) + i \, \sin(\theta+\phi) $$ and in the case that $\theta+\phi \ge 2 \pi$, the right hand side can be replaced by $$\cos(\theta+\phi-2\pi) + i \, \sin(\theta+\phi-2\pi) $$ However, this is not an example of what group theorists call a cyclic group.