# Can the group of rotation be considered as a cyclic group

I read from some physics books that the 2D rotation can be expressed in exponential form:

$$R(\theta) = e^{X(\theta)}, \theta \in R$$

In physics, X relates to the momentum of a particle, and is defined to be the generator of the rotation. Some authors also state that X is the generator of the rotation group .

Is that statement correct, and if it is, can we consider the rotation group ($$N\ge2$$) as a cyclic group. As if I understand correctly, only cyclic groups have a generator.

• A group is called cyclic if all elements are powers of one single element, i.e., if it can be generated by a single element. This answers your question. Nov 3, 2020 at 12:36
• As @DietrichBurde says, cyclic groups are groups generated by a single element. But be careful, in some languages, "cyclic" means "cyclic and finite". And there is another word for "cyclic infinite". (For example in French, "cyclique" means that the group is also finite, and we call "monogène" for an infinite cyclic group). So maybe the answer can depend on which language you consider :) Nov 3, 2020 at 12:43
• @TheSilverDoe: Do you mean that there is more than 1 definition of cyclic group? Nov 3, 2020 at 13:41
• @Rekkhan In English no. In English, "cyclic group" means that the group is generated by a single element. I just wanted to underline the fact that if you are not English, maybe "cyclic" in your language means another thing ! Nov 3, 2020 at 13:42

For a cyclic group, there is a generator $$G$$ such that the elements of the group are $$G^k$$ with $$k \in \mathbb Z$$. For a rotation group, there is an "infinitesimal generator" $$X$$ such that the elements of the group are $$e^{tX}$$ with $$t \in \mathbb R$$.
If $$R(t) = e^{tX}$$ is the rotation group, then the derivative is $$R'(t) = \lim_{t \to 0}\frac{R(t) - I}{t} = X$$ (the identity matrix $$I = R(0)$$) and that is the sense it is an "infinitesimal generator". We have $$R(t) = I + tX + O(t^2)$$ so if $$t$$ is very small, then rotation is approximately $$I+tX$$. So $$X$$ is the "tangent direction" to the rotation.
• No, in general $X$ does not belong to the Lie group, but to the tangent space -- the corresponding Lie algebra. Nov 3, 2020 at 16:16