From External Direct Product of Abelian Groups is Abelian Group it follows that $(M,+)$ is an abelian group. We need to show that: $\forall x, y, \in M, \forall \lambda, \mu \in R$: : $(1): \quad \lambda \circ \left({x + y}\right) = \left({\lambda \circ x}\right) + \left({\lambda \circ y}\right)$ : $(2): \quad \left({\lambda +_R \mu}\right) \circ x = \left({\lambda \circ x}\right) + \left({\mu \circ x}\right)$ : $(3): \quad \left({\lambda \times_R \mu}\right) \circ x = \lambda \circ \left({\mu \circ x}\right)$ Checking the criteria in order: === Criterion 1 === : $(1): \quad \lambda \circ \left({x + y}\right) = \left({\lambda \circ x}\right) + \left({\lambda \circ y}\right)$ Let $x = \left({x_i}\right)_{i\in I}, y = \left({y_i}\right)_{i\in I} \in M$. {{begin-eqn}} {{eqn | l = \lambda \circ \left({x + y}\right)       | r = \lambda \circ \left({ \left({x_i}\right)_{i\in I} + \left({y_i}\right)_{i\in I} }\right)       | c =  }} {{eqn | r = \lambda \circ \left({x_i+_iy_i}\right)_{i\in I}       | c =  }} {{eqn | r = \left({\lambda\circ_i x_i+_i \lambda\circ_i y_i}\right)_{i\in I}       | c =  }} {{eqn | r = \left({\lambda\circ_i x_i}\right)_{i\in I}+\left({\lambda\circ_i y_i}\right)_{i\in I}       | c =  }} {{eqn | r = \lambda\circ\left({x_i}\right)_{i\in I}+\lambda\circ\left({y_i}\right)_{i\in I}       | c =  }} {{eqn | r = \left({\lambda \circ x}\right) + \left({\lambda \circ y}\right)       | c =  }} {{end-eqn}} So $(1)$ holds. {{qed|lemma}} === Criterion 2 === : $(2): \quad \left({\lambda +_R \mu}\right) \circ x = \left({\lambda \circ x}\right) + \left({\mu \circ x}\right)$ Let $x = \left({x_i}\right)_{i\in I} \in M$. {{begin-eqn}} {{eqn | l = \left({\lambda +_R \mu}\right) \circ x       | r = \left({\lambda +_R \mu}\right) \circ \left({x_i}\right)_{i\in I}       | c =  }} {{eqn | r = \left({\left({\lambda +_R \mu}\right) \circ_i x_i}\right)_{i\in I}       | c =  }} {{eqn | r = \left({\lambda \circ_i x_i +_i \mu\circ_i x_i}\right)_{i\in I}       | c =  }} {{eqn | r = \left({\lambda \circ_i x_i}\right)_{i\in I} + \left({\mu\circ_i x_i}\right)_{i\in I}       | c =  }} {{eqn | r = \lambda \circ \left({x_i}\right)_{i\in I} + \lambda \circ \left({y_i}\right)_{i\in I}       | c =  }} {{eqn | r = \left({\lambda \circ x}\right) + \left({\mu \circ x}\right)       | c =  }} {{end-eqn}} So $(2)$ holds. {{qed|lemma}} === Criterion 3 === : $(3): \quad \left({\lambda \times_R \mu}\right) \circ x = \lambda \circ \left({\mu \circ x}\right)$ Let $x = \left({x_i}\right)_{i\in I} \in M$. {{begin-eqn}} {{eqn | l = \left({\lambda \times_R \mu}\right) \circ x       | r = \left({\lambda \times_R \mu}\right) \circ \left({x_i}\right)_{i\in I}       | c =  }} {{eqn | r = \left({\left({\lambda \times_R \mu}\right)\circ_i x_i}\right)_{i\in I}       | c =  }} {{eqn | r = \left({\lambda\circ_i\left({\mu\circ_i x_i}\right)}\right)_{i\in I}       | c =  }} {{eqn | r = \lambda \circ \left({\mu\circ_i x_i}\right)_{i\in I}       | c =  }} {{eqn | r = \lambda \circ \left({ \mu \circ \left({x_i}\right)_{i\in I} }\right)       | c =  }} {{eqn | r = \lambda \circ \left({\mu \circ x}\right)       | c =  }} {{end-eqn}} So $(3)$ holds. {{qed|lemma}} Thus all criteria are seen to hold. The result follows. {{Qed}}