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In the book Algebra of Serge Lang, the following is written: "The fact that Hom(G,X) is a group when G,X are commutative is of special significance." (Where Hom(G,X) is the set of homomorphisms from G into X) My question is: What is the binary operation of the group Hom(G,X)?

Thank you in advance

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I'm guessing it's composition. – jlv Nov 6 '12 at 19:09
It is the group operation in $X$: $f+g(x)=f(x)+g(x)$. – user641 Nov 6 '12 at 19:10
It can't be composition. Hom(G,X) may not be closed under this operation. – Amr Nov 6 '12 at 19:12
Not only "may not be closed": it can't be closed unless $\,X=G\,$ – DonAntonio Nov 6 '12 at 19:14
I guess now that we only need X to be commutative. We don't need G to be Abelian, do we? – Amr Nov 6 '12 at 19:18
up vote 1 down vote accepted

As Steve commented, $f+g:= \big( x\mapsto f(x)+g(x)\big)$, this is the operation (so called 'pointwise operation').

In general, for algebraic structures $(X,*)$ with one binary operation, then, the homsets $hom(X,Y)$ are naturally closed wrt. ($*$ defined pointwise) if and only if $$(a*b)*(c*d)=(a*c)*(b*d)$$ holds for all elements $a,b,c,d\in Y$. And, it fits perfectly for Abelian groups.

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