# Free products of groups

Hi all! Today I’ll talk about free products of groups.

Let $G$ and $H$ be groups. Then a word in $G$ and $H$ is a concatenation $abc...n$ where each letter is an element of either $G$ or $H$. Now suppose this concatenation does not have elements alternating from $G$ and $H$. Then there must be at least one pair of elements from either $G$ or $H$ which are right next to each other. So let’s take their product in the group $G$ or $H$ (this includes the identity next to some other element). We can do this so that we get a word which has elements alternating from $G$ and $H$. A word like this is called reduced. So if $G$ and $H$ are groups, we can define their free product, denoted $G\ast H$ to be the collection of all reduced words of $G$ and $H$. This is actually a group under concatenation (and reduction, obviously).

For example, suppose $G=H=\mathbf{Z}$. Then $\mathbf{Z}$ is cyclic, so is the free group on one generator. Thus $\mathbf{Z}\ast\mathbf{Z}$ is the free group on two generators. However, note that it’s not abelian. This is because it’s not necessarily true that $ab=ba$ where $G=\langle a\rangle$ and $H=\langle b\rangle$. The free abelian group on two generators is the quotient of $\mathbf{Z}\ast\mathbf{Z}$ by the normal subgroup generated by $aba^{-1}b^{-1}$. Another example would be the free product $\mathbf{Z}/2\mathbf{Z}\ast\mathbf{Z}/2\mathbf{Z}$. This is isomorphic to the infinite dihedral group.

Alright, so suppose $G$ is a group, and $S$ is a generating set of $G$. Then we can consider the free group $\mathscr{F}_S$ over $S$ (the collection of all unreduced words over $S$). This has a natural map $\mathscr{F}_S\to G$ determined by $g\mapsto g$ for all $g\in S\cup S^{-1}\cup \{1\}$ (where $S^{-1}$ is the set of all inverses of elements of $S$). Let $N$ be the kernel of this map. Let $T$ be a set which generates $N$ such that it is normal, i.e., such that $\{ghg^{-1}|g\in G,h\in H\}$ is a generating set of $G$.

OK, so now $\mathscr{F}_S/N$ is isomorphic to $G$, by the isomorphism theorems. The compact way of writing this is as $\langle S|T\rangle$, and is called a presentation of $G$. Every group has a presentation. If $S$ and $T$ can be chosen to be finite, then the group $G$ is said to be finitely presented. Cool.

Now, suppose $G,H,K$ are groups. Write down presentations of $G$ and $H$ as follows: $\langle S|T\rangle=G$, and $H=\langle U|V\rangle$. Also suppose $\phi:K\to G$ and $\psi:K\to H$ are homomorphisms. Then define $G\ast_K H$ to be $\{S\cup U|T\cup V\cup\{\phi(a)=\psi^{-1}(a)|a\in K\}\}$. This is called the amalgamated product of $G$ and $H$ (associated to the homomorphisms, etc.). If $K$ is the trivial group, then $G\ast_K H$ is simply the free product of $G$ and $H$. (Do you see why?) So the free product $G\ast H$ is $\langle S\cup U|T\cup V\rangle$.

This is really all you need to know to understand a future post on algebraic topology. We’ll conclude this post with one simple remark. The idea behind free groups is that they are the coproduct in the category of groups. However, they do not form the coproduct in the category of abelian groups, because the free product of abelian groups need not be abelian (see the above example of $\mathbf{Z}\ast\mathbf{Z}$). (The coproduct in the category of abelian groups is the direct sum.)

Have fun,
SKD