$[a]_m \subseteq [a]_n$ if and only if $n\mid m$. Theorem. Let $a \in \mathbb{Z}$, then $[a]_m \subseteq [a]_n$ if and only if $n\mid
Can anyone help me perfect my proof, please?
Proof. We will start by assuming $n\mid m$ and show that $[a]_m \subseteq [a]_n$. Since $n\mid m$ we know $nk=m$ for some integer $k$ by the definition of divides. We will add some integer $a$ to both sides to get $a+nk=a+m$. By the definitions of modulo and congruence class, we know $a+nk$ is some element in $[a]_n$ is equal to some number $a+m$ which is some element in $[a]_m$, we know some element $x \in a$ is also some element $x \in b$. Thus, we know $[a]_m \subseteq [a]_n$ by the definition of subset. 
Conversely, we will now assume $[a]_m \subseteq [a]_n$ and show $n|m$. By the definition of subset, we know every element $x \in [a]_m$ is also in $[a]_n$. We also know that $[a]_m = a \bmod{m}$ by the definition of congruence class. Similarly, we know $a \pmod{m} \equiv a + mk$ for some integer $k$. Similarly, $a \pmod{m} \equiv \{..., a-m, a, a+m, a+2m, ... \}$ by the definition od modulo. Also, by the definition of congruence class, we know $[a]_n \equiv a \pmod{n}$, and by the definition of modulo, we know $a \pmod{n} \equiv a + nj$ for some integer $j$. We will now choose some element $a+m \in [a]_m$, and since we are assuming, $[a]_m \subseteq [a]_n$, we know $a+m \in [a]_n$. We will choose some element $a + nj \in [a]_n$ for some integer $j$, thus $a+m \in a+nj$ for some integer $j$. Thus there is some integer $j$ such that $a+m=a+nj$. Subtracting $a$ from both sides, we get $m=nj$. Thus, we know by the definition of divides that $n$ divides $m$ which proves if $a \in \mathbb{Z}$, then $[a]_m \subseteq [a]_n$ if and only if $n|m$ and concludes the proof.
Notes from my Professor:
For the first part: 
"You did not set up this argument correctly. You need to assume that some element $x$ is in $[a]_m$ and then show $x$ is in $[a]_n$."
For the second part: 
"A congruence class cannot be congruent to an integer."
"Your argument does not make sense, it is unclear what you are showing and how you are justifying it."
Please help!
Thank you!
 A: I would do the proofs at a global level. Looking at the elements level is, in my opinion, unnecessary here:


*

*First part: if $m=kn$, then $[a]_m=a+m\mathbf Z\subseteq a+n\mathbf Z$ since $m\in n\mathbf Z$, so $m\mathbf Z\subset n\mathbf Z$.

*Second part: if $[a]_m\subset [a]_n$ for all $a\in \mathbf Z$, in particular, $\;[0]_m=m \mathbf Z\subseteq [0]_n=n\mathbf Z$, so that $m\in n\mathbf Z$,which means that $n$ divides m.

A: For the first part, you want to show that: $n |m$ implies that: $[a]_m \subseteq [a]_n$. This means that if $x\in[a]_{m}$ implies that $x\in [a]_{n}$ (that you can assume is the definition of being a subset).
Suppose that $x \in [a]_{m}$, then $x=a+mk$ for some $k\in \mathbb{Z}$.
If $n$ divides $m$, that means that $m=qn$ for some $q \in \mathbb{Z}$. That is $x=a+(qk)n$. So $x =a$ $\bmod n$, in other words $x \in [a]_{n}$, as required.
Your argument for first bit is a bit unclear, you claim you are adding some integer $a$ to both sides, (I take it you mean any integer a), but then you are treating $a$ as being the representative of the equivalence class $[a]_{m}$. Moreover, since a is just an integer, $x \in a$ doesn't really make sense, as $x$ cannot be an element of the set $a$, as $a$ is an integer and not a set. ($x \in [a]_m$ makes sense though, as $[a]_m$ is a set.)
