How are $2\mathbb{Z}\ncong3\mathbb{Z}$ different as rings? How are $2\mathbb{Z}\ncong3\mathbb{Z}$ different as rings? What interesting properties does one have that the other doesn't?
 A: Hint: Due to the additive structure of the underlying groups $2\mathbb{Z}$ and $3\mathbb{Z}$, we see that $2\mathbb{Z} = \left\langle 2\right\rangle$ and $3\mathbb{Z} = \left\langle 3\right\rangle$. A theorem of cyclic groups says that any group isomorphism (which ring isomorphisms must also be) between cyclic groups must send a generator of one to a generator of the other, so we have two options: $2 \mapsto 3$ or $2 \mapsto -3$. Try playing around with these maps and see if you can get any contradictions with the mulitplicative structure of $2\mathbb{Z}$ and $3\mathbb{Z}$.
Proof: Let $\phi: 3\mathbb{Z} \rightarrow 2\mathbb{Z}$ by $3 \mapsto 2$ (this is sufficient definition due to the group structure of the rings in question). Note that $\phi(6) = \phi(3+3) = \phi(3)+\phi(3) = 2 + 2 = 4$ and $\phi(9) = \phi(3\cdot 3) = \phi(3)\cdot\phi(3) = 2\cdot 2 = 4$, so this $\phi$ is not an injection and thus not an isomorphism.
Now take $\psi:3 \mathbb{Z} \rightarrow 2\mathbb{Z}$ by $3 \mapsto -2$ (again, this is sufficient). Note that $\phi(6) = \phi(3) + \phi(3) = -2 + -2 = -4$ and $\phi(-9) = \phi(-3\cdot 3) = \phi(-3)\cdot \phi(3) = 2 \cdot (-2) = -4$. Again, this map is not an injection.
We need not consider $3\mathbb{Z}$ generated by $-3$ instead of $3$ as the already given isomorphisms cover those cases (as they say where $-3$ goes as well).
A: Directly and, perhaps, simpler: 
Suppose there's a ring isomorphism $\;\phi:2\Bbb Z\to3\Bbb Z\;$ , then for sure $\;3=\phi(2n)\;$ , for some $\;n\in\Bbb Z\;$ , but then:
$$\phi(2\cdot2n)=\begin{cases}\phi(2)\phi(2n)=3\phi(2)\\{}\\\phi(2n+2n)=2\phi(2n)=6\end{cases}$$ 
Comparing, we get that it must be
$$3\phi(2)=6\implies\phi(2)=2$$
which, of course, is absurd as $\;2\notin3\Bbb Z\;$
