Example 2 The set $M=\{u \mid u=12 m+8 n+4 l, m, n, l \in \mathbf{Z}\}$ and
$$
N=\{u \mid u=20 p+16 q+12 r, p, q, r \in \mathbf{Z}\}
$$

The relationship between them is ( ).
A. $M=N$
B. $M \not \subset N$ and $N \not \subset M$
C. $M \subset N$
D. $M \supset N$
Subsection 13 Through simplification, recognize the characteristics of the elements in these two sets, and then make a judgment.
Solution 1 $12 m+8 n+4 l=4(3 m+2 n+l)$, and $3 m+2 n+l$ can take any integer, so set $M$ represents the set of multiples of 4, i.e., $M=\{u \mid u=4 k, k \in \mathbf{Z}\}$. $20 p+16 q+12 r=$ $4(5 p+4 q+3 r)$, set $p=-q=k, r=0$, we get $N=\{u \mid u=4 k, k \in \mathbf{Z}\}$. Therefore, $M=N$, option A should be selected.

Solution 2 In the conclusions provided for selection in this problem, all are inclusion relationships between two sets. The general method to prove the inclusion relationship between sets is “if $a \in A \Rightarrow a \in B$, then $A \subseteq B$”; the general method to prove the equality relationship between sets is “if $\left\{\begin{array}{l}A \subseteq B \\ B \subseteq A,\end{array}\right.$ then $A=B$”.
Solution 2 If $u \in M \Rightarrow u=12 m+8 n+4 l$. Set $m=r, n=2 q, l=5 p$, then $u=$ $20 p+16 q+12 r \in N \Rightarrow M \subseteq N$. If $u \in N \Rightarrow u=20 p+16 q+12 r$. Set $p=$ $-q=2 n+l, r=m$, then $u=12 m+8 n+4 l \in M \Rightarrow N \subseteq M$. By $\left\{\begin{array}{l}M \subseteq N \\ N \subseteq M\end{array} \Rightarrow M=\right.$ $N$, so option A should be selected.