Define the function \(\phi:\mathbb{Z} \rightarrow 2\mathbb{Z}\) by \(\phi(n) = 2n\) for all \(n \in \mathbb{Z}\). We will first show that \(\phi\) is a bijection.
Assume that \(\phi(a) = \phi(b). \) Then \(2a = 2b \) which implies \(a = b.\) Hence, \(\phi\) is injective. Now, let \( n \in 2 \mathbb{Z}\). Since \(n\) is even, \(n=2k\) for \(k \in \mathbb{Z}.\) Now, \(\phi(k) = 2k = n\) and \(\phi\) is surjective. Since \(\phi\) is both injective and surjective, it is bijective.
Lastly, we have \[\phi(a+b) = 2(a+b) = 2a + 2b = \phi(a) + \phi(b) \]
Thus, \((2 \mathbb{Z}, +)\) and \((\mathbb{Z}, +)\) are isomorphic.
Q.E.D.