Proof of a simple isomorphism 
I'm having trouble with part b. I know I have to show that a map from a to da is an isomorphism from Z to dZ, but I don't know the general way to prove that a map is an isomorphism. I think I'm struggling a bit with the actual definition of an isomorphism. Any help or advice would be appreciated. 
 A: An isomorphism is a bijective homomorphism.
A homomorphism from $(G,*_G)$ ti $(H,*_H)$ is a function $\phi:G\to H$ which satisfies the following property: $$\phi(g_1*_Gg_2)=\phi(g_1)*_H\phi(g_2),$$ where $*_G$ and $*_H$ are the corresponding group operations and $g_1,g_2 \in G.$
As you and other people have noted, the "obvious" candidate function for our isomorphism is $$\phi(a)=d\times a.$$ We need to show this is a homomorphism first. Note that when $a,b\in \mathbb{Z}$ and $*$ is the group operation on both $\mathbb{Z}$ and $d\mathbb{Z}$ (i.e. addition), $$\phi(a*b)=d\times (a*b)=d\times (a+b)=d\times a+d\times b=\phi(a)*\phi(b).$$
So our function indeed is homomorphic.
Now we need to show it is bijective, but that shouldn't be too hard as it has a natural inverse map $\phi^{-1}(da)=a,$ so I'll leave this up to you.
A: An isomorphism is a bijective homomorphism, while a homomorphism between groups $G$ and $H$ is a map $\phi:G\to H$ such that $\phi(ab)=\phi(a)\phi(b)$.
The ("obvious") natural map is $\phi:\mathbb{Z}\to d\mathbb{Z}$ defined by $\phi(a)=da$. You just need to show that is is injective, surjective, and a homomorphism.
