We know that the subgroups of $\frac{\mathbb{Z}}{n\mathbb{Z}}$ are defined as;
Let $n$ be a positive integer.
Let $d$ be a positive divisor of $n$, say $n = kd$, $k \in \mathbb{Z}$.
Then $ \\{ \overline{0}, \overline{d}, \overline{2d} , ..., \overline{(k-1)d} \\} $ is a subgroup of
$\frac{\mathbb{Z}}{n\mathbb{Z}}$.
In my exercise I get that $12$ is divisible by $1,2,3,4, 6$ and $12$, this would result in $$\{\overline{0}, \overline{1}, \overline{2}, ..., \overline{11}\}$$ but by the property of divisibility does this not mean that for every $n$ we choose for $\frac{\mathbb{Z}}{n\mathbb{Z}}$, that the subgroup is always $\\{\overline{0}, \overline{1}, \overline{2}, ..., \overline{n-1}\\}$, since every n is divisible by $1$?