Free Product (Coproduct, or Categorical Sum)
Each multiplicand is represented by a 'word' consisting of a sequence of generators. We apply the multiplication by putting the words one after the other. This is the most general type of product that contains the two multiplicands as subgroups.
We can 'reduce' the result by:
- removing the identity element
- replacing and sequence that is within one of the multiplicands
but unless we apply any specific reductions the result will always be infinite for non trivial multiplicands.
In order to try to understand this product of two groups lets try multiplying two very simple groups together, the simplest groups I can think of are C2 and C3
|<m | m²>||
|<r | r³>||
Free Product C3 × C2
So what would be the elements of this group? Lets assume they are the same as for the direct product, that is: 1,r,r²,m,rm and r²m. Using these as words would give the table:
We can then apply any constrains for the individual groups, that is: m²=1 and r³=1 which gives:
But we still get elements that are not members of the group (the group is not closed). If we make these elements of the group, then we have to make them rows and columns, in which case more elements get added to the body of the table. So the group is infinite. The elements of the group are alternating r's and m's such as:
where r may be replaced by r² at any place.
|<m,r | m²,r³>|