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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

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

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    What is an (abelian) X-group?2010-12-19
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    @Pete L. Clark, Let X be an arbitrary (possible empty) set and let G be a group. We say that G is an X-group (or group with operator set X) provided that for each $x\in X$ and $g\in G$, there is defined an element $g^x\in G$ such that if g,h $\in G$, then $(gh^x)=g^xh^x$. When X is empty set, G is as normally defined group. For here X is relevant, I will remove "X".2010-12-19
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    Did you already look at the subgroups $\mathrm{Ker}(f^n)$?2010-12-19
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    @Green Iden: okay, I think you mean **group with operators**, as in http://en.wikipedia.org/wiki/Group_with_operators. (You seem to have a typo in your definition.) But, as I think you may have been trying to say above, for the problem at hand the most general case would be if $X = \varnothing$, since then the homomorphism $f$ is unrestricted.2010-12-19
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    Further comments: when you say Noetherian and Artinian, I guess you mean as a $\mathbb{Z}$-module? Do you know which abelian groups are both Noetherian and Artinian (hint: Noetherian modules are finitely generated).2010-12-19
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    @Pete: $f$ does not fulfill the condition $f\circ f = f$, but only $f\circ f(A) = f(A)$.2010-12-19
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    @jug, do you mean the poset of $ker(f^n)$? $ker(f^{n+i})\ge ker(f^n)$ so use noetherian?2010-12-19
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    @Green Iden: Yes.2010-12-19
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    @jug: thanks for mentioning that, I deleted my comment. But I still think that thinking about projection-like operators would be helpful here.2010-12-19
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    @Pete: Yes, the proof is quite similar.2010-12-19
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    @jug, Thanks a lot for your hint. I got it later with the hint. I proved it by firstly show $ker(f^n)\bigcap f(A)=0$, then by noetherian, the result follows trivially.2010-12-20
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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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