4. For the imaginary number $\mathrm{i}\left(\mathrm{i}^{2}=-1\right)$, consider the set $S=$ $\left\{\mathrm{i}, \mathrm{i}^{2}, \mathrm{i}^{3}, \mathrm{i}^{4}\right\}$. It is easy to see that the product of any two elements in $S$ is still in $S$. Now, we define the multiplicative identity $\theta$ in $S$: for any $a \in S$, we have $a \theta=\theta a=a$. Then $\theta$ is ( ).
(A) $\mathrm{i}$
(B) $i^{2}$
(C) $\mathrm{i}^{3}$
(D) $i^{4}$
4.D.

From $\mathrm{i}^{4}=1$, we know that for $k=1,2,3,4$, $\mathrm{i}^{4} \cdot \mathrm{i}^{k}=\mathrm{i}^{k} \cdot \mathrm{i}^{4}=\mathrm{i}^{k}$.
By definition, $\mathrm{i}^{4}$ is the identity element.