Showing $n\Bbb{Z}$ is a group under addition Let $n$ be a positive integer and let $n\Bbb{Z}=\{nm\mid m \in\Bbb{Z}\}$. I need to show that $\left< n\Bbb{Z},+ \right>$ is a group. And I need to show that $\left< n\Bbb{Z},+ \right>\cong\left< \Bbb{Z},+ \right>$. 
Added: If $n\mathbb Z$ is a subgroup of $\mathbb Z$ then it must be closed under "+". The identity element 0 is an element of the subgroup. And for any $n$ in $n\mathbb Z$, it's inverse $-n$ must be in $n\mathbb Z$... 

Since every element of $n\mathbb Z$ is an element of $\mathbb Z$, we can do an easier proof that it is a group by showing that it is a subgroup of $\mathbb Z$.
It happens to be true that if $H\subset G$, where $G$ is a group, then $H$ is group if it satisfies the single condition that if $x,y\in H$, then $x\ast y^{-1}\in H$, where $\ast$ is the group operation of $G$ and $y^{-1}$ is the inverse of $y$ in $G$.
In order to find an isomorphism $\varphi:\mathbb Z\to n\mathbb Z$, ask yourself what $\varphi(1)$ should be. Can you go from here?
