But what is the dual space of $Z_n$?

That would consist of all linear mappings from $Z_n$ into $Z$.

But there is only _one_ such mapping: the function that sends everything in $Z_n$ to $0$.

Suppose that we had linear $f$ such that $f(1) = k$.

Then $f(n * 1) = f(0) = 0$, and also $f(n * 1) = n * f(1) = n * k$.

So, in $Z$, we have $n * k = 0$, which means that $k = 0$ (given that $n >= 1$).

That implies that $f$ is the zero homomorphism.

That would consist of all linear mappings from $Z_n$ into $Z$.

But there is only _one_ such mapping: the function that sends everything in $Z_n$ to $0$.

Suppose that we had linear $f$ such that $f(1) = k$.

Then $f(n * 1) = f(0) = 0$, and also $f(n * 1) = n * f(1) = n * k$.

So, in $Z$, we have $n * k = 0$, which means that $k = 0$ (given that $n >= 1$).

That implies that $f$ is the zero homomorphism.