Can the group of rotation be considered as a cyclic group

Question

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.

Answer 1

In case of a rotation group, this is called (in mathematics) an infinitesimal generator and is not a generator in the sense you mention.

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 $t \in \mathbb R$.

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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.