Previous: [[A brief intro to Cayley graphs]] The free product is a construction that generalizes free groups. It is best to think of groups via generators and relations when we talk about groups below. If we have two groups given by $G = \langle S_1|R_1 \rangle$, $H = \langle S_2|R_2 \rangle$, then their free product $G \ast H$ is given by $G\ast H = \langle S_1 \sqcup S_2|R_1 \sqcup R_2 \rangle$ Simple enough...? Let's unpack what this actually looks like. In a free group, we had some generators, and we considered words formed from the letters of those generators. Here, we have two groups, and in the free product we consider words formed from letters *in those groups.* Then, when we juxtapose $g$ from $G$ and $h$ from $H$, we can't really "move them past each other," so we just stick them next to each other like we would two separate generators from the free group. We can construct this formally via the construction above: we just throw in all the generators and all the relations of either group and call it a day^[we paid the price when constructing free groups, of making reduced words and whatnot. Now we know that it is perfectly valid to present a group with generators and relations]. Why do we care? The real reason is that it shows up in a lot of natural constructions, but even now we have some interesting reasons to think about this. For one, this generalizes the construction of free groups with relations included. Second, it has the following universal property: <iframe class="quiver-embed" src="https://q.uiver.app/#q=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&embed" width="451" height="432" style="border-radius: 8px; border: none;"></iframe> We know that $G$ and $H$ embed into the free product via the obvious inclusions: just send an element in $G$ to itself. This universal property, which is very similar to the universal property for free groups, says that if there are any maps from $G$ and $H$ into some other group $X$, then there is a unique map from the free product to that group that commutes with the inclusions. In this sense it is the "universal group containing $G$ and $H$." In a few posts we will see that it is the "coproduct in the category of groups." For now we can just see it concretely: if $\phi$ is the name of the induced map then on an arbitrary word $g_1h_1g_2h_2\ldots g_n h_n$ (this is the most we can "condense" a word because we can't move elements from different groups past each other). Then $\phi(g_1h_1g_2h_2\ldots g_n h_n)= \phi_G(g_1)\phi_H(h_1)\cdots \phi_G(g_n) \phi_H(h_n)$ If the groups originally had no relations, then of course neither does the free product. Thus the free product of two free groups is free, and by looking at the number of generators we have $F_n\ast F_m = F_{n+m}$ and in particular $F_2 = \mathbb Z\ast\mathbb Z$ since $\mathbb Z = F_1$. This also lets us see a slight subtlety here: the free product of two abelian groups is *not* abelian because elements from different groups don't automatically commute with each other. They only commute internally. We will later see that this means that the "free product is not the coproduct in the category $\mathcal{Ab}$ of abelian groups" Next: [[Not really an intro to categorical language]]