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My professor said: "If $A$ is an abelian group and End(A) is a ring under addition and composition of group homomorphisms, then $A$ can be regarded as an $End(A)$-module under the multiplication defined by: $$ f(a) = fa, \forall f \in End(A),\forall a \in A $$ "

Because A is an additive abelian group, the only conditions remaining to prove that A is an $End(A)$-module, is $$ End(A) \times A \rightarrow A \\ (f,a) \rightarrow fa$$

And

  1. $f(a+b) = fa + fb$

  2. $(f + g)a = fa + ga$

  3. $f(ga)=(fg)a$

  4. $I_{End(A)} a = a$

$\forall f,g \in End(A), \forall a \in A$

I am not sure if the above remaining conditions are written in a right form or not , I am confused between when to apply composition of functions and when to apply the given multiplication in that proof of $A$ can be regarded as a unitary $End(A)$-module(I have already proved that $(End(A), + , \circ)$ is a ring).could anyone please check the above conditions for me?

Also, I do not know if the above conditions are written in a correct way, how can I prove them, could anyone help me please?

Thanks!!

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    $\begingroup$ All of the properties follow instantly from the definion of the module action. Which one are you having trouble with? Seems like you're overthinking things. $\endgroup$ – rschwieb Oct 4 '17 at 17:39
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There is one mistake (apparently): the first one should be $f(a+b)=fa+fb$

I am not sure if the above remaining conditions are written in a right form or not

Modulo the typo above, they are in the right form. You could write the module action using $\cdot$ if it helps you keep things straight. E.g.: 1. $f\cdot (a+b) = f\cdot a + f\cdot b$

  1. $(f + g)\cdot a = f\cdot a + g\cdot a$

  2. $f\cdot (g\cdot a)=(f\circ g)\cdot a$

  3. $I_{End(A)}\cdot a = a$

I am confused between when to apply composition of functions and when to apply the given multiplication

Composition of functions only arises in the condition testing $(f\circ g)\cdot a=f\cdot(g\cdot a)$, which is the first place where two functions appear composed. For example, the left hand side is, by definition $(f\circ g)(a)=f(g(a))$, and the right side is $f\cdot(g(a))=f(g(a))$. Obviously these are identical.

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  • $\begingroup$ I do not understand why can not we write $f . (g.a) = (f.g).a$? $\endgroup$ – Intuition Oct 5 '17 at 0:57
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    $\begingroup$ @Intuition If you are using $\cdot$ for the new module action, then $f\cdot g$ is a meaningless expression. $\endgroup$ – rschwieb Oct 5 '17 at 1:08
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    $\begingroup$ @Intuition Yes, $R\times A\to A$. $\endgroup$ – rschwieb Oct 5 '17 at 1:10
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    $\begingroup$ @Intuition For example, $I\cdot a:=I(a)=a$ by definition of $I$. $f\cdot(a+b):=f(a+b)=f(a)+f(b)$ by definition of an endomorphism, and $f(a)+f(b)=f\cdot a +f\cdot b$ by definition of the module action. Earlier with composition: $(f\circ g)\cdot a:=(f\circ g)(a)=f(g(a))=f(g\cdot a)=f\cdot (g\cdot a)$. $\endgroup$ – rschwieb Oct 5 '17 at 18:32
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    $\begingroup$ @Intuition Juxtaposition means writing an operation without a symbol, like $xy$ for the product of $xy$. By jumbling that with $\cdot$ you seemed to be showing me that you didn't fully understand where you were going through these verifications. $\endgroup$ – rschwieb Oct 5 '17 at 18:35

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