Show that $5\mathbb{Z}-5\mathbb{Z}=5\mathbb{Z}$. My proof.
Lemma. $\mathbb{Z}-\mathbb{Z}=\mathbb{Z}$.
Proof. ($\Rightarrow$) Let $z\in \mathbb{Z}-\mathbb{Z}$. Then, there is $z_{1},z_{2} \in\mathbb{Z}$ such that $z=z_{1}-z_{2}$. So, $z\in\mathbb{Z}$.Thus,$\mathbb{Z}-\mathbb{Z} \subset\mathbb {Z}$. 
($\Leftarrow$) Let $z'\in\mathbb{Z}$. Then, $z'-0\in\mathbb{Z}-\mathbb{Z}$. Thus, $\mathbb {Z}\subset\mathbb{Z}-\mathbb{Z}$. 
     Therefore, $\mathbb{Z}-\mathbb{Z}=\mathbb{Z}$. 
Now, since we have $\mathbb{Z}-\mathbb{Z}=\mathbb{Z}$, $5\mathbb{Z}-5\mathbb{Z}=5(\mathbb{Z}-\mathbb{Z})=5\mathbb{Z}$.
Can you check my proof?
 A: Rather than trying to use a lemma, it may be better just to imitate it in writing the direct proof.
Using your notation, we first prove $5\mathbb{Z} - 5\mathbb{Z} \subset 5\mathbb{Z}$.
Proof: An arbitrary element in $5\mathbb{Z} - 5\mathbb{Z}$ is of the form $5a - 5b$ for $a, b \in \mathbb{Z}$. By the distributive law of multiplication over subtraction, we have $5a-5b = 5(a-b) \in 5\mathbb{Z}$, where the final inclusion follows from the closure of $\mathbb{Z}$ under subtraction. Thus, the desired containment holds.
Next, we show that $5\mathbb{Z} \subset 5\mathbb{Z} - 5\mathbb{Z}$.
Proof: An arbitrary element in $5\mathbb{Z}$ is of the form $5a$ for some $a \in \mathbb{Z}$. Next, observe that:
$$5a = 5a - 0 = 5a - 5(0) \in 5\mathbb{Z} - 5\mathbb{Z}$$
where the final inclusion follows because $a, 0 \in \mathbb{Z}$. Thus, the desired containment holds.
So: Combining the two proofs above, we have the desired equality of sets. QED.
A: Here is a proof based on your own comment.
First, let $z\in 5{\Bbb Z}-5{\Bbb Z}$.  Then $z=5z_1-5z_2$ for some integers $z_1$ and $z_2$.  But then $z=5(z_1-z_2)$, and $z_1-z_2$ is an integer, so $z\in5{\Bbb Z}$.
Conversely, let $z\in5{\Bbb Z}$.  Then $z=5z_1$ for some integer $z_1$, and so
$$z=5(z_1-0)=5z_1-5\times0\ .$$
But $z_1\in{\Bbb Z}$ so $5z_1\in5{\Bbb Z}$; and $0\in{\Bbb Z}$ so $5\times0\in5{\Bbb Z}$; so $z\in5{\Bbb Z}-5{\Bbb Z}$.  This completes the proof.
Simpler version of the second part: let $z\in5{\Bbb Z}$; then $z=z-5\times0$ which is in $5{\Bbb Z}-5{\Bbb Z}$.
