# Thread: Is that an abelian group?

1. ## Is that an abelian group?

I read the following in a book. If A be an abelian group and End A be defined as set of all endomorphism f :A---->A , End A along sum of functions produce an abelian group. My problem is as following.
a,b belong to A , (f+g)(ab) = f(ab)+g(ab) = f(a)f(b)+g(a)g(b)

(f+g)(a) (f+g)(b)=(f(a)+g(a))(f(b)+g(b))
Does (f+g)(ab) equal to (f+g)(a) (f+g)(b)? In other words does f+g belong to End A?

2. You are confusing the binary operations in the two sets A and End(A). The definition of + in End(A) is: $\forall\;\mathrm{f},\mathrm{g}\in\mathrm{End}(A)$, (f+g):AA where (f+g)(a) = f(a)g(a) for all aA.

This may look strange, but that’s because you are writing product of elements in A multiplicatively.

However, since A is an Abelian group, why not denote the binary operation in A additively? To avoid confusion, let’s write $\oplus$ for the operation in End(A) and + for the operation in A. So we have

$\forall\;f,g\in\mathrm{End}(A),\ (\mathrm{f}\oplus\mathrm{g})(a) = \mathrm{f}(a) + \mathrm{g}(a)$ for all aA.

Now that looks more intuitive. Hence, for any endomorphisms f and g and any $a,b\in A$:

$(\mathrm{f}\oplus\mathrm{g})(a+b)=\mathrm{f}(a+b)+ \mathrm{g}(a+b)=\mathrm{f}(a)+\mathrm{f}(b)+\mathr m{g}(a)+\mathrm{g}(b)$

$(\mathrm{f}\oplus\mathrm{g})(a)+(\mathrm{f}\oplus\ mathrm{g})(b)=\mathrm{f}(a)+\mathrm{g}(a)+\mathrm{ f}(b)+\mathrm{g}(b)$

Hence $(\mathrm{f}\oplus\mathrm{g})(a+b)=(\mathrm{f}\oplu s\mathrm{g})(a)+(\mathrm{f}\oplus\mathrm{g})(b)\ \Rightarrow\ \mathrm{f}\oplus\mathrm{g}\in\mathrm{End}(A)$.