Let $G=(\mathbb{Z}/n\mathbb{Z}, +)$ be an abelian group where $n$ is nonzero composite integer
$$\mathbb{Z}/n\mathbb{Z}=\{\bar{0}, \bar{1}, .... , \overline{n-1} \}$$
Now the Lagrange theorem tells us if we want to construct a subgroup $H$ of $G$ then the order of $H$ must divide $n$. Since $n$ is composite, there is always a non-trivial subgroup since $G$ is abelian. As we know every subgroup contains the identity $\bar{0}$ and every element has an inverse in $H$ and also $H$ should be closed under $+$ . For example $H=\{\bar{0}, \bar{6}, \overline{12} , \overline{18} \}$ is subgroup of $\mathbb{Z}/24\mathbb{Z}$ all the element of $H$ are divisible by $6$ so the congruence class have a common property.
My question is what is the condition for the congruence class to form a subgroup of $\mathbb{Z}/n\mathbb{Z}$ ?does the congruence class have a common property when they form a subgroup ?