Prove that $S$ is a subring of $\mathbb{Z}_{28}$

Question

Question: $S=\{0,4,8,12,16,20,24\}.$ Prove that $S$ is a subring of $\mathbb{Z}_{28}$

Confusion 1: This might be a dumb question, but when we refer to $[4]$ in $S$, for example, is that the congruent class of $4$ modulo $28$ in our case? Because I assume $[4]=\{x\in\mathbb{Z}:x=4+nk\}$, and $n$ doesn't have to be $28$.

Confusion 2: In order to prove that set $S$ is closed under addition and multiplication, is there any other way than going case by case? I noticed that all elements in $S$ are multiples of $4$, but I don't really see how that would help me.

Thank you.

Answer 1

Since you are in $\mathbb{Z}_{28}$, everything is modulo $28$ i.e. $a=[a]=\{a+28k:k\in\mathbb{Z}\}$ for all $a\in\{0,2,\dots,27\}$.

Now you have noticed that all elements in $S$ are multiples of $4$, then it follows that if you add any two elements in $S$, the answer must still be a multiple of $4$, which must be an element of $S$. This shows that $S$ is closed under addition. Multiplication works the same way.

Edit: If the sum of two elements is a multiple of $4$, since $28$ is also a multiple of $4$, the sum modulo $28$ is again a multiple of $4$, which means it is in $S$.