Compatibility of direct product and quotient in group theory This question came to me when I tried comparing direct product and quotients of groups with products and quotients of natural numbers. When we divide a number by another and multiply the result with the divisor we get back the original number. That is, if $b|a$ $$\frac{a}{b} \cdot b = a$$ However this is not true in case of groups. For example $$\mathbb{Z}/n\mathbb{Z} \times n\mathbb{Z} \not\cong \mathbb{Z}$$ I don't know how to make my question precise but is there a product-quotient operator pair that mimics the above mentioned property of product-quotient on numbers?
 A: I don't know if this is the answer you are looking for, but your question is not really precise (as you say) so I would try this.
Well let's say I have the set of groups $\mathcal{G}$ (assume It can be indeed defined). On this set, there is the relation of isomorphism. Let's call $G$ the quotient :
$$G:=\mathcal{G}/\text{ isomorphism}$$
Define $(G,\times)$ where $\times$ denotes the direct product of groups. I claim (straightforward verification) that $\times$ is an internal law for $G$ (in particular it is well defined), is  commutative ($A\times B$ is isomorphic to $B\times A$), associative and has a neutral element (the trivial group $\mathbb{1}$). 
This commutative monoid with a unit $(G,\times)$ is not a group. You can check that :
$$\frac{\mathbb{Z}}{2\mathbb{Z}}^{\mathbb{N}}\times\mathbb{1}\text{ is isomorphic to }\frac{\mathbb{Z}}{2\mathbb{Z}}^{\mathbb{N}}\times \frac{\mathbb{Z}}{2\mathbb{Z}}  $$
If $\frac{\mathbb{Z}}{2\mathbb{Z}}^{\mathbb{N}}$ had an inverse then :
$$\mathbb{1}\text{ would be isomorphic to }\frac{\mathbb{Z}}{2\mathbb{Z}}$$
Now what can we do about this ? Well check this : http://en.wikipedia.org/wiki/Grothendieck_group. The Grothendieck group construction allows you to make a group out of a commutative monoid with a unit (even if the monoid in question has no such thing as a cancellation property). 
Well I claim now that the Grothendieck group associated to $G$ is the trivial group. To show this :  this relies on the fact that for any group $H_1,H_2$ we have that :
$$(H_1^{\mathbb{N}}\times H_2^{\mathbb{N}})\times H_1\text{ is isomorphic to }(H_1^{\mathbb{N}}\times H_2^{\mathbb{N}})\times H_2 $$
A first statement as an answer to your question is : you cannot have this (in a natural way) because the most natural "groupification" is trivial (see the wikipedia page to see what this associated group must verify).
However If we take $G$ to be the finite groups (or finitely generated groups) up to isomorphism. I am not so sure of what happens, you still have $(G,\times)$ is a commutative monoide with a unit. So the Grothendieck group associated to this still makes sens and obviously the argument I have given cannot work when we restrict to finite groups (nor for finitely generated groups).
Edit : If $G$ is the set of finite groups up to isomorphism then $A\times B$ is isomorphic to $A\times C$ implies $B$ isomorphic to $C$, see this  :
http://en.wikipedia.org/wiki/Krull%E2%80%93Schmidt_theorem
In this case our monoid $(G,\times)$ satisfies the cancellation property and the Grothendieck group $\mathcal{G}(G)$ associated contains $(G,\times)$ as a submonoid.  The group is infinitely generated by all the classes of undecomposable finite groups. Given a bijection between classes of undecomposable finite groups and $\mathbb{N}$ we see that :
$$\mathcal{G}(G)\text{ is isomorphic to } \mathbb{Z}^{(\mathbb{N})} $$
And with this isomorphism :
$$G\text{ is given by } \mathbb{N}^{(\mathbb{N})} $$
For the case where $G$ is the set of finitely generated groups up to isomorphism it is not so clear what happens.
