6. Set
$M=\{u \mid u=12 m+8 n+4 l$, where $m, n, l \in Z\}$,
$N=\{u \mid u=20 p+16 q+12 r$, where $p, q, r \in Z\}$

The relationship between them is
(A) $M=N$;
(B) $M \nsubseteq N, N \nsubseteq M$;
(C) $M \subset N$;
(D) $M \supset N$.
6. (A)

For any element $u$ in $N$, we have
$$
u=20 p+16 q+12 r=12 r+8(2 q)+4(5 p) \in M,
$$

thus $N \subseteq M$.
On the other hand, for any element $u$ in $M$, we have
$$
u=12 m+8 n+4 l=20 n+16 l+12(m-n-l) \in N,
$$

thus $M \subseteq N$. Therefore, $M=N$.