Re: Definition of a Z-Module

Basically because we are requiring $1\cdot a=a$ for all $a\in A$, and then the fact that $m\cdot a=a+\cdots+a$ (with $m$ summands) follows from the fact that $m=1+\cdots+1$ ($m$ summands) and applying the distributivity law.

Re: Definition of a Z-Module

Re: Definition of a Z-Module

Mhm. In other words, "Z-module" and "Abelian group" refer to the exact same objects. If one writes the product of an abelian group multiplicatively, then:

na = a^{n}.

In other words the action of Z on abelian groups was introduced to you as soon as you started to examine cyclic groups, disguised as "exponent rules".