/sci/ - Science & Math

>>11204199
His definition boils down to calling it the pushout in the category of groups. This tells you why it's important, but not how to construct it, nor is it obvious that it even exists a priori. Terrible definition IMO.

Here's an explicit construction. Let's assume [math]K[/math] is the trivial group first for simplicity, so we're going to make the free product [math]G \ast H[/math]. This is supposed to be to most general group generated by the disjoint union of [math]G[/math] and [math]H[/math]; i.e. making a group out of it by doing the least amount of work. Since I need to be able to multiply all the elements of my group, I need products like [math]gh[/math] to be in there. This is fine when [math]g,h[/math] are both in either [math]G[/math] or [math]H[/math], but what about when [math]g \in G[/math] and [math]h \in H[/math]? Well, the easiest thing to do is just to add [math]gh[/math] as an entirely new element, treated as a formal product. The same goes for [math]hg[/math]. But now I need to be able to multiply these things as well. So now I have elements like [math]hghg[/math], [math]ghgh[/math], [math]hg^2h[/math], [math]gh^2g[/math], etc. (note that I can "simplify" the last two, since [math]g^2[/math] is some element in [math]G[/math] and likewise for [math]h^2[/math]).

Continuing this line of reasoning, you get that [math]G \ast H[/math] consists of all "words" (e.g. [math]g_1 h_1 ... g_n h_n[/math]) in the elements of [math]G[/math] and [math]H[/math], with multiplication being string concatenation. Words can "simplified" when there are two elements from the same group next to each other, and the identity elements of [math]G[/math] and [math]H[/math] are identified as the common identity element of [math]G \ast H[/math].

As a simple example, [math]\mathbb{Z} \ast \mathbb{Z}[/math] is [math]F_2[/math], the free group on two generators. In general, if [math]F_m,F_n[/math] are free groups, then [math]F_m \ast F_n \cong F_{m+n}[/math].