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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.

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  • $\begingroup$ You want to prove that $\text{Im}(d) \subset \ker(d)$ which is equivalent to proving that for $x \in \text{Im}(d)$ (so $x = d(y)$ for some $y \in C$), $d(x) = 0$. Try to apply $d$ to $x = d(y)$. $\endgroup$
    – Krijn
    Oct 20, 2014 at 10:17

2 Answers 2

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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 $\mbox{Im }d\subset\ker d$.

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  • $\begingroup$ I get this up until saying that $x\in\ker d$. Could you just explain how this implies that $\mbox{Im }d\subset\ker d$? Thanks. $\endgroup$
    – user184543
    Oct 20, 2014 at 10:28
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    $\begingroup$ $d(x)=0$. That's the definition of $x$ being in the kernel of $d$. So because $x$ was chosen to be in the image of $d$, it means that $x\in\mbox{Im }d\implies x\in\ker d$ which is the definition of $\mbox{Im }d$ being a subset of $\ker d$. $\endgroup$
    – Dan Rust
    Oct 20, 2014 at 10:32
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    $\begingroup$ Small typo - in your first line, you mean $d(y)=x$. $\endgroup$
    – mdp
    Oct 20, 2014 at 10:33
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    $\begingroup$ @MattPressland thanks :) $\endgroup$
    – Dan Rust
    Oct 20, 2014 at 10:36
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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 $x\in C$.

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  • $\begingroup$ There are some cases where we can show that if $d^2=0$ then $d=0$, without $d$ being invertible (in fact $d$ is zero and invertible if and only if $C$ is the trivial group). $\endgroup$
    – Dan Rust
    Oct 20, 2014 at 10:40
  • $\begingroup$ @DanielRust Ah, of course - not quite sure what I was thinking when I wrote that! $\endgroup$
    – mdp
    Oct 20, 2014 at 10:50

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