# The Unapologetic Mathematician

## Free products of groups

On Tuesday I talked about how to put a group structure on the Cartesian product of two groups, and showed that it satisfied a universal property. For free products I’m just going to start from the property. First, the picture.

This is the same diagram as before, but now it’s upside-down. All the arrows run the other way. So we read this property as follows: Given two groups $G$ and $H$, the free product is a group $G*H$ with homomorphisms $\iota_G:G\rightarrow G*H$ and $\iota_H:H\rightarrow G*H$ so that given any other group $X$ with homomorphisms $f_G:G\rightarrow X$ and $f_H:H\rightarrow X$ there is a unique homomorphism $f_G*f_H:G*H\rightarrow X$.

The exact same argument from Tuesday shows that if there is any such group $G*H$, it is uniquely determined up to isomorphism. What we need is to have an example of a group satisfying this property.

First of all, $\iota_G$ has to be an monomorphism. Just put $G$ itself for $X$, the identity on $G$ for $f_G$, and the trivial homomorphism (sending everything to the identity) for $f_H$. Now anything in the kernel of $\iota_G$ is automatically in the kernel of the composition of $\iota_G$ with the coproduct map. But this composite is the identity homomorphism on $G$, which has trivial kernel, and so must $\iota_G$. This means that a copy of $G$ must sit inside $G*H$. Similarly a copy of $H$ must sit inside $G*H$.

So how do those two copies interact? Let’s put $F_2$ in for $X$ and let $f_G$ send everything in $G$ to some power of the generator $a$ — effectively defining a homomorphism from $G$ to $F_1\cong\mathbb{Z}$ — while $f_H$ sends everything in $H$ to some power of the generator $b$. If there’s any relation between the copies of $G$ and $H$ sitting inside $G*H$, it won’t be respected by the function to $F_2$ required by the universal property. So there can’t be any such relation.

$G*H$ is actually very much like $F_2$, only instead of alternating powers of $a$ and $b$, we have alternating members of $G$ and $H$. An arbitrary element looks something like $g_1h_1g_2h_2...g_kh_k$. Of course it could start with an element of $H$ or end with an element of $G$. The important thing is that the entries from the two groups alternate. We compose by just sticking sequences together like for a free group. If one sequence ends with an element of $G$ and the next sequence starts with an element of $G$, we compose those elements in $G$ so the whole sequence is still alternating.

Does $G$ sit inside here? Of course! It’s just sequences with only an element of $G$ in them. The same goes for $H$. And given any group $X$ and homomorphisms $f_G$ and $f_H$, we can send $g_1h_1g_2h_2...g_k$ to $f_G(g_1)f_H(h_1)f_G(g_2)f_H(h_2)...f_G(g_k)$. That’s our $f_G*f_H$. As I said above, any other group that has this property is isomorphic to $G*H$, so we’re done.

If we compare the free product $G*H$ with the direct product $G\times H$, we see that the main difference is that elements of $G$ and $H$ don’t commute inside $G*H$, but they do inside $G\times H$. We can check that $(g,e_H)(e_G,h)=(ge_G,e_Hh)=(g,h)=(e_Gg,he_H)=(e_G,h)(g,e_H)$. In fact, take a presentation of $G$ with generators $X$ and relations $R$, and one of $H$ with generators $Y$ and relations $S$, and with $X$ and $Y$ sharing no elements. Then $G*H$ has generators $X\cup Y$ and relations $R\cup S$, while $G\times H$ has generators $X\cup Y$ and relations $R\cup S\cup\{xyx^{-1}y^{-1} (x\in X, y\in Y)\}$.

Since we’ve only added some relations to the presentation to get from $G*H$ to $G\times H$, the latter group is a quotient of the former. There should be some epimorphism from $G*H$ to $G\times H$. I’ll leave it to you to show that some such epimorphism does exist in two ways: once by the universal property of $G*H$ and once by the universal property of $G\times H$.

March 1, 2007 -

1. [...] of diagrams. Check that in this property is satisfied by disjoint unions. In coproducts are free products. In a preorder, coproducts are least upper bounds. And, of course, the coproduct defines a functor [...]

Pingback by Products and Coproducts « The Unapologetic Mathematician | December 5, 2007 | Reply

2. I thought the free product of groups did not necessarily yield a group. I read somewhere that the free product Z(2)*Z(2)(Z(2) = integers mod 2) is not a group. This was in the context of defining the free product from the “bottom up” – strings of representatives with concatenation and reduction. Does that give the same result as your top down method with morphisms etc?

Comment by GK | February 20, 2011 | Reply

3. Sorry. I was confusing free products of groups with free groups. Dnnnng!

Comment by GK | February 20, 2011 | Reply

4. Right: the product of free groups isn’t a free group. But the free product of free groups is a free group.

Comment by John Armstrong | February 20, 2011 | Reply

5. “Let’s put F_2 in for X and let f_G send everything in G to the generator a, while f_H sends everything in H to the generator b.”

Surely, then, f_G and f_H are not group homomorphisms? If we have an element g in G, by your definition we have f_G(g) = a, and f_G(g^2) = a, but we should have f_G(g^2) = a^2, which is not equal to a. Also, we can’t map the identity in G to a either. Similarly for H.

Comment by TC | May 29, 2012 | Reply

6. Good catch, TC; I was overly glib there. Really I was thinking of sending everything to some expression in $F_2$ involving only one of the generators.

Comment by John Armstrong | May 29, 2012 | Reply