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I have a question about abelian groups (or rather $X$-groups):

Suppose $A$ is an abelian group (or rather an $X$-group) which is artinian and noetherian. $f$ is an endomorphism (or rather an $X$-endomorphism) of $A$, i.e. $f\in \mathrm{End}(A)$. Also know that $f(f(A))=f(A)$. Prove that $A=\ker(f)\dot{+}f(A)$, i.e. $A$ is the direct sum of those two subgroups.

My idea is firstly show $A=\ker(f)+f(A)$ which I have proved. So next is to prove that $\ker(f)\bigcap f(A)$ is trivial, i.e. $= 0$. This part I got astray. I thought I should use the noetherian condition here. Since, A is noetherian, then there exists an subgp C of A and $C\subset f(A)$, s.t. C is maximal in the sense of $\ker(f)+C=0$, then find an subgp T, s.t. $C satisfy $T\bigcap f(A)=0$ then contradict with maximality of C to reach a contradiction. But I get astray here to find this T.

Maybe I use the wrong method. Can anyone help me?

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    @Green Iden: You are welcome. After the deadline for your homework assignment, you could post your solution as answer.2010-12-20

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