Exercise 6 from Section 1.4 of Beachy, Blair: Abstract Algebra, 2006.
Let $m$ and $n$ be positive integers such that $m\mid n$. Show that for any integer $a$, the congruence class $[a]_m$ is the union of the congruence classes $[a]_n, [a+m]_n, [a+2m]_n, \dots, [a+n-m]_n$.
This is an exercise from Abstract Algebra by Beachy and Blair involving congruence classes. I understand that the congruence classes are sets, and I'm assuming the best approach would be to explore the meaning of that union of congruence classes and hopefully realize it equals $[a]_m$. other than that I can't find my bearings at all. I'm not sure what this really means. any hints/advice appreciated, I am not asking for the answer.
Hint: $[a]_m= \{a+lm:l\in \mathbb Z\}$, whereas $[a+km]_n=\{a+km+ldm: l\in \mathbb Z\}$, where $d=\frac nm$ ...
In the latter case you can see that you sort of have an $m$ equivalence class (with some missing elements...) The missing elements get filled in when $k$ ranges from $0$ to $d-1$...
You may want to look at some concrete examples... For instance $m=2$ and $n=8$... Then, say $a=1$.
$[1]_2=\{1+2l: l\in \mathbb Z\}$, whereas $[1]_8=\{1+8l: l\in \mathbb Z\}$, $[1+2]_8=[3]_8=\{3+8l:l\in \mathbb Z\}$, $[1+4]_8=[5]_8=\{5+8l:l\in \mathbb Z\}$ and $[1+6]_8=[7]_8=\{7+8l: l\in \mathbb Z\}$...
The union of the last four sets equals the first (namely the odd integers)...
The key is that modding out by $n$ is less precise (than by $m$): we need to account for the equivalence classes at increments of $m$ to make up for that...