4. Given the sets $E=\left\{e \left\lvert\, e=m+\frac{1}{6}\right., m \in \mathbf{Z}\right\}, F=\left\{f \left\lvert\, f=\frac{n}{2}-\frac{1}{3}\right., n \in \mathbf{Z}\right\}$, $G=\left\{g \left\lvert\, g=\frac{p}{2}+\frac{1}{6}\right., p \in \mathbf{Z}\right\}$, then the sets $E, F, G$ satisfy the relationship ( ).
A. $E=F=G$
B. $E \subset F=G$
C. $E \subset F \subset G$
D. $F \subset G \subset E$
4. B.
$$
e=m+\frac{1}{6}=\frac{6 m+1}{6}, f=\frac{n}{2}-\frac{1}{3}=\frac{3 n-2}{6}=\frac{3(n-1)+1}{6}, g=\frac{p}{2}+\frac{1}{6}=\frac{3 p+1}{6}
$$

Since $6 m+1$ represents an integer that leaves a remainder of 1 when divided by 6, and $3(n-1)+1$ and $3 p+1$ represent integers that leave a remainder of 1 when divided by 3, then $E \subset F=G$