Suppose I have an abelian group $C$, with a group homomorphism $d\colon C\to C$ such that $d^2=0$.
I need to show that the image of $d$ is contained in the kernel of $d$.
My original attempt was to say that
$$\operatorname{Im}(d)= \{ d(x) | x \in C \} = 0$$
since $d(x)=0$ for all $x \in C$, and
$$\ker(d) = \{ x ∈ C | d(x) = 0 \} = C$$
since $d(x)=0$ for all $x \in C$, then $\operatorname{Im}(d) = 0 \in C = \ker(d)$.
But I'm convinced this is wrong. To be honest I don't fully understand what the question is asking.