Isomorphism between groups defined by set builder notation How is it possible to show that for a set $G = \{x + y\sqrt{3}| x, y\in\mathbb{Z}\}$ and a set $H = \{2^i*5^j| i,j\in\mathbb{Z}\}$ that the group $[A, +]$ is isomorphic to $[B, *]$ (let $+$ be standard addition and $*$ be standard multiplication between reals)? In particular I am struggling to find a valid function that holds for the homomorphism property.
 A: Yes, because both groups are isomorphic to $(\mathbb Z \times \mathbb Z, +)$.
This gives an isomorphism $G \to \mathbb Z \times \mathbb Z \to H$:
$$
x + y\sqrt{3} \mapsto (x, y) \mapsto 2^x 5^y
$$
To see that this is a homomorphism, compute:
$$
(x + y\sqrt{3})+(x' + y'\sqrt{3})
=(x + x') + (y+y')\sqrt{3}
\mapsto 2^{x+x'} 5^{y+y'} = \cdots
$$
A: Since by definition any number g in G can be represented by (x,y) such that g=x+y*sqrt(3), all you have to show is that the mapping of elements from G to ZXZ is one-one and from there you can use those guaranteed unique x's and y's to go to 2^x*5^y and the isomorphism will be complete. While it would be interesting to find a way to extract the x and y from just the numerical value of the element in G you're already given their existence so the one-one mapping is all you need.
A: The homomorphism $\varphi : G \to H$ is simply defined 
$$(\forall x,y \in \Bbb Z): \quad \varphi (x+y\sqrt{3})=2^x 5^y$$
and its inverse is
$$(\forall x,y \in \Bbb Z): \quad \varphi^{-1} (2^x5^y) = x+y\sqrt{3}.$$
Sure, it seems a bit weird to define functions like this, but after looking closely you should realize that these are proper functions, that is, they are mapping every element in their domain and we never have the case that one element is mapped to two or more.
We just need to make sure these are homomorphisms, which if fairly straightforward. We have $\forall x_1,y_1,x_2,y_2 \in \mathbb Z:$
$\begin{array}
\ \varphi\bigg((x_1+y_1 \sqrt{3})+(x_2+y_2 \sqrt{3})\bigg) &= \varphi \bigg((x_1+x_2) + (y_1+y_2)\sqrt{3}\bigg) \\
&= 2^{x_1+x_2}5^{y_1+y_2} \\
&=2^{x_1}2^{x_2}5^{y_1}5^{y_2} \\
&=2^{x_1}5^{y_1}\cdot2^{x_2}5^{y_2} \\
&= \varphi(x_1+y_1\sqrt{3})\varphi(x_2+y_2\sqrt{3})
\end{array}$
thus $\varphi$ is a homomorphism, could you should that $\varphi^{-1}$ is a homomorphism then, completing the proof?
Note: Also, it should be clear that these are both $\varphi$ and $\varphi^{-1}$ bijective. But as others have stated, you can use the argument that $G\cong \Bbb Z^2$ and $H \cong \Bbb Z^2$ and that isomorphisms are equivalence relations providing:
$$G \cong \Bbb Z^2 \text{ and } H \cong \Bbb Z^2 \implies G \cong \Bbb Z^2 \text{ and } \Bbb Z^2 \cong H \implies G \cong \Bbb Z^2 \cong H \implies G \cong H.$$
