Refer to Lang's "Algebra" second paragraph from top, p. 122. Let $A$ be a ring and $X$, $X^'$ be $A$-modules. Let $Hom_A(X,X')$ be the set of $A$-homomorphisms from $X$ to $X'$. It is mentioned that if $A$ is commutative, then we can make $Hom_A(X,X')$ into an $A$-module by defining $(\alpha f)(x) = \alpha f(x)$ for $\alpha \in A$, $f \in Hom_A(X,X')$ and $x \in X$, whereas if $A$ is not commutative, then we can only regard $Hom_A(X,X')$ as an abelian group. The question is: why do we need the commutativity property to have a well-defined operation of $A$ on $Hom_A(X,X')$ as above? What goes wrong if we remove commutativity?
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For the function $\alpha f$ to be a module homomorphism, we need it to satisfy $$(\alpha f)(x+y) = (\alpha f)(x) + (\alpha f)(y)$$ and $$(\alpha f)(\lambda x) = \lambda (\alpha f)(x).$$ The first condition is not a problem, but the second condition may fail if $R$ is not commutative: let $\alpha$ and $\lambda$ be two elements of $R$ such that $\alpha \lambda\neq \lambda\alpha$. Then $$(\alpha f)(\lambda x) = \alpha \Bigl(f(\lambda x)\Bigr) = \alpha\Bigl(\lambda f(x)\Bigr) = \alpha\lambda f(x);$$ but $$\lambda(\alpha f)(x) = \lambda\Bigl(\alpha f(x)\Bigr) = \lambda\alpha f(x).$$ We have no warrant for asserting that $\alpha\lambda f(x) = \lambda\alpha f(x)$ for all $x$, so $\alpha f$ need not be a module homomorphism when $R$ is not commutative.