Coproduct of groups Can anyone explain why the coproduct of groups are the free product?
For finite groups, the products are direct products which are also finite. But the free groups are infinite? So the coproduct is larger than the product?
 A: Well, neither for positive integers, addition is not always bigger than mulitplication: just think about $1\cdot 1\cdot 1 < 1+1+1$. So we shouldn't expect anything like $X\times Y \ge X+Y$  in general (whatever these symbols would mean).
By a (group) presentation we mean a syntactic entity: a pair $\langle X,\Gamma\rangle$ where $X$ is a set (of 'variables') and elements are $\Gamma$ are relations $(\tau,\sigma)$ (interpreted as '$\tau=\sigma$') using words $\tau,\ \sigma$ from the free group on alphabet $X$. 
Note. For groups, a relation $\tau=\sigma$ of course can be written up using only one term, stating $\tau\sigma^{-1}=1$, but the same construction works for general algebraic structures as well.
Now, consider the category ${\bf Pres}$ of presentations with functions $f:X\to Y$ as arrows $\langle X,\Gamma\rangle\to \langle Y,\Delta\rangle$ which satisfy $(\tau,\sigma)\in\Gamma\ \implies (f(\tau),\,f(\sigma))\in\Delta$.
Coproduct in this category is simply disjoint union, and the 'free' functor ${\bf Pres}\to{\bf Grp}$ (sending  $\langle X,\Gamma\rangle\ \mapsto \ F(X)/(\Gamma)\,$) has a right adjoint $G\mapsto \langle G,\{$all valid equations of $G\}\rangle$, so it preserves coproduct.
