Constructing an isomorphism of group products To introduce some symbolism for clarity's sake, here is the problem I'm faced with in an assignment:

Find an example of four groups $A,B,C,D$ such that
  $$A \times B \cong C \times D$$
  but at the same time neither of $A,B$ are isomorphic to either of $C,D$.

I feel like I touched on a possible means of solving this, but I'm not sure it's sufficient/correct, and even if it is constructing the isomorphism is proving ... not too easy.

Initial Motivation:
So, we know: for two groups to be isomorphic, a necessary - but not sufficient! -condition (as noted and proved here) is that their underlying sets have the same cardinality. I took this as a sort of starting point to start finding such groups. So, in this problem, I now seek $A,B,C,D$ such that
$$|A \times B| = |C \times D| = |A|\cdot |B| = |C| \cdot |D|$$
So it seems sufficient to think of $|A\times B|=|C \times D| = n$ for $n$ finite, and to find factors $a,b,c,d$ of $n$ such that
$$n=ab=cd$$
where $a,b,c,d$ are the factors of $n$, and the orders of groups $A,B,C,D$. In my case, $n=12$ seems a convenient example, as
$$12 = 2\cdot 6 = 3 \cdot 4$$
Some of the first groups that come to mind of the corresponding orders are, each with modular addition,
$$A = \Bbb Z / 2 \Bbb Z \;\;\;\;\;B = \Bbb Z / 6 \Bbb Z \;\;\;\;\;C = \Bbb Z / 3 \Bbb Z \;\;\;\;\;D = \Bbb Z / 4 \Bbb Z$$

Problems in What Remains:
So what remains is to construct an isomorphism $f : (\Bbb Z / 2 \Bbb Z \times \Bbb Z / 6 \Bbb Z) \to (\Bbb Z / 3 \Bbb Z \times \Bbb Z / 4 \Bbb Z)$ and demonstrate it, but I'm running into problems there, nothing seems to really work out, at least nicely. I also tried to just show each product was isomorphic to $\Bbb Z / 12 \Bbb Z$ but ran into the same fundamental problem.
So I'm guessing this product isn't the easiest to deal with. Indeed I also tried a slightly simpler
$$A = \Bbb Z / 2 \Bbb Z \;\;\;\;\;B = \Bbb Z / 2 \Bbb Z \;\;\;\;\;C = \Bbb Z / 4 \Bbb Z \;\;\;\;\;D = \Bbb Z / 1 \Bbb Z = \langle e \rangle$$
but I couldn't construct a nice function that wouldn't rely on a bunch of cases in proving it's an isomorphism. Is there perhaps a function that I'm overlooking, or an easier set of products to work with?
 A: Why not adjust your try, to $C_1×\Bbb Z_{12}\cong \Bbb Z_3×\Bbb Z_4$, now that you know it was incorrect.
There are lots of examples like this, you can see after the other answer.  
A: It is easier to find examples of numbers having two different factorizations, with both 'different from each other'.
For example take 4 distinct prime numbers $p,q,r,s$. Let $n$ be their product.
Now define $a=pq$ and $b=rs$. So $n=ab$. Now define $c=pr$ and $d= qs$. We still have $n=cd$. Clearly both are two distinct factorizations. $ab=cd$ with $a\neq c,d$ and $b\neq c,d$.
Now replace every integer above by cyclic groups of that order, and replace each factorization of two numbers, by direct product of corresponding groups, and we will get examples you are looking for.
Specific example: for the four prime numbers, $2,3,5,7$,  their product is 
$210$. Using the notation $C_n$ for a cyclic group of order $n$, we see that
$$ C_{210}\cong  C_6\times C_{35}\cong C_{10}\times C_{21}$$
We are using the fact that for two finite cyclic groups of coprime orders, their direct product is again a cyclic group.
All this is elementary number theory. Examples  involving non-abelian groups would be a bit more challenging and would be real group theory.
A: The problem is that $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$ is not isomorphic to $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$, and only the latter isomorphic to $\mathbb{Z}/12\mathbb{Z}$. The only time that $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/m\mathbb{Z} \cong \mathbb{Z}/(nm)\mathbb{Z}$ is when $n$ and $m$ are relatively prime. This fact should help you come up with an example that works.
