Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel 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.
 A: Let $x\in C$ be in the image of $d$, so there exists a $y\in C$ with the property that $d(y)=x$. Consider the element $d^2(y)$ which we know much be $0$ because $d^2=0$. It follows that $d(x)=d(d(y))=d^2(y)=0$ and so $x\in\ker d$. Hence $\mbox{Im }d\subset\ker d$.
A: You already have a good answer, but I want to emphasize one particular misunderstanding here. For group homomorphisms, $d^2=0$ does not imply that $d=0$ in general! So you shouldn't immediately conclude that $d(x)=0$ for all $x\in C$. This happens in other situations too - for example the matrix
$$\begin{pmatrix}0&1\\0&0\end{pmatrix}$$
is non-zero but squares to zero. You could interpret this matrix as a linear map  $\mathbb{R}^2\to\mathbb{R}^2$ using the standard basis (note that a linear map is an example of a homomorphism of abelian groups!), and then you would have a map that squares to zero, but isn't zero.
What you know is that if $x\in C$, then $d(d(x))=d^2(x)=0$; then the proof works by noticing that elements of the image are precisely those of the form $d(x)$ for $x\in C$.
