Let $G=U(1)=\{z\in \mathbf{C}:|z|=1\}$ be the set of unit complex numbers. This is a group: if you multiply two elements of $U(1)$ together you get another unit complex number. Geometrically, it's a circle in the complex plane.

Any $z\in U(1)$ can be written as $z={e}^{i\theta}$ for some $\theta $ . The map $i\theta \mapsto {e}^{i\theta}$ is going to be our exponential map in this example and it goes from the imaginary numbers $i\mathbf{R}$ to $U(1)$ . The commutator bracket on $i\mathbf{R}$ vanishes: $[i{\theta}_{1},i{\theta}_{2}]=-{\theta}_{1}{\theta}_{2}+{\theta}_{2}{\theta}_{1}=0$ , and the Baker-Campbell-Hausdorff formula reduces to the usual law of logarithms ${e}^{i{\theta}_{1}}{e}^{i{\theta}_{2}}={e}^{i({\theta}_{1}+{\theta}_{2})}$ .

If we translate the line of imaginary numbers so that it passes through the identity element $1\in U(1)$ then we get a tangent line to the circle $U(1)$ . This will be true in general: $\U0001d524$ will be (parallel to) the tangent space to $G$ at the identity.