Generator of the quotient group $\mathbb{Z}/a\mathbb{Z}$ for $a>0$

Question

Let $a$ be an integer $>0$. The additive group $\mathbb{Z}/a\mathbb{Z}$ is a cyclic group. I would like to show that $$\mathbb{Z}/a\mathbb{Z}=\{X\ |\ (\exists n)(n\in\mathbb{Z}\ \land X=n\cdot(a\mathbb{Z}+1))\}.$$

Attempt:

Let $X\in\mathbb{Z}/a\mathbb{Z}$: there exists $m\in\mathbb{Z}$ such that $X=a\mathbb{Z}+m$. I read somewhere that $$m(a\mathbb{Z}+1)=m(a\mathbb{Z})+m=a\mathbb{Z}+m.$$ But is this true? For example, let $m=0$: then this would imply that $$0(a\mathbb{Z}+1)=0(a\mathbb{Z})+0\cdot1=0\ne a\mathbb{Z}.$$

What am I doing wrong? I would like to show that $a\mathbb{Z}+1$ generates $\mathbb{Z}/a\mathbb{Z}$.

Answer 1

I guess you implicitly suppose $a$ is a positive integer.

What you misunderstood is how $\;m\cdot(1+a\mathbf Z)\;$ is defined:

Suppose first $m=2$: by definition $$2(1+a\mathbf Z)=(1+a\mathbf Z)+(1+a\mathbf Z)=(1+1)+a\mathbf Z=2+a\mathbf Z,$$ and more generally: $$m(1+a\mathbf Z)=m\cdot 1+a\mathbf Z=m+a\mathbf Z,$$ not $\;m+ma\mathbf Z$.