# Free Product of Groups with Presentations

There is a highly believable theorem:

Let $A, B$ be disjoint sets of generators and let $F(A), F(B)$ be the corresponding free groups. Let $R_1 \subset F(A)$, $R_2 \subset F(B)$ be sets of relations and consider the quotient groups $\langle A | R_1 \rangle$ and $\langle B | R_2 \rangle$.

Then $\langle A \cup B | R_1 \cup R_2 \rangle \simeq \langle A | R_1 \rangle \ast \langle B | R_2 \rangle$, where $\ast$ is the free product.

I can supply a horrible proof using normal closures and elements and such, but surely there is a reasonable diagram-chase proof using the universal properties of free groups and the free product. However, I have been unable to construct it; I think there is probably some intermediary but basic categorical statement needed about presentations.

There is a hint given for the problem (it is coming from Lee, Topological Manifolds, problem 9-4) that is:

If $f_1 : G_1 \rightarrow H_1$ and $f_2 : G_2 \rightarrow H_2$ are homomorphisms and for $j = 1,2$ we have $i_j, i_j'$ the injections of $G_j, H_j$ into $G_1 \ast G_2$ and $H_1 \ast H_2$ respectively, there is a unique homomorphism $f_1 \ast f_2 : G_1 \ast G_2 \rightarrow H_1 \ast H_2$ making the square $(i_j, f_j, i_j', f_1 \ast f_2)$ commute for $j = 1,2$.

This is almost trivial to prove with the UMP for free products, but I can't wrangle it into a proof of the top statement. Also, anything legitimately trivial like $F(A) \ast F(B) \simeq F(A \cup B)$ or $F(F(S)) \simeq F(S)$ is fair game.

I can get a whole lot of arrows, but I am having trouble conjuring up any arrows with $\langle A | R_1 \rangle \ast \langle B | R_2 \rangle$ as their domain. We also know that the free product is the coproduct in $\textbf{Grp}$ but admittedly I haven't tried exploiting this yet.

Can anyone help with a hint? Thanks a lot!

• This seems straightforward, and I am not really sure where you are stuck. I generally use the Universal Property as the definition of the free product, and then prove its existence by showing that the presentation $\langle A \cup B \mid R_1 \cup R_2 \rangle$ satisfies the Universal Property, which I think is what you are saying is almost trivial. So what's the problem exactly? – Derek Holt Nov 30 '15 at 21:03
• How do you show that group satisfies the UMP? What I am saying is almost trivial is that statement about commuting squares. – John Samples Nov 30 '15 at 21:07
• Actually, what is your UMP for free product? The one I am familiar with requires the canonical injections from its product terms. We don't have such a construction for the single group $\langle A \cup B | R_1 \cup R_2 \rangle$. So I think we might have different UMP's? – John Samples Nov 30 '15 at 21:14
• There's a pretty nice topological proof if you build spaces with the corresponding fundamental groups via the usual 'wedge of circles with disks glued in to kill relators' construction. The wedge of two of these spaces then clearly has the fundamental group of the left hand side, and the Van Kampen theorem gives you the right hand side. This might not be as rigorous as you'd like though. – Dan Rust Nov 30 '15 at 21:14
• Sadly SVK is the next chapter haha – John Samples Nov 30 '15 at 21:14

The Universal Property of the free product $G*H$ of groups $G$ and $H$ is that there exist homomorphisms $i_G:G \to G*H$ and $i_H:H \to G*H$ such that, for any group $K$ and any homomorphisms $\tau_G:G \to K$ and $\tau_H:H \to K$, there is a unique homomorphism $\phi:G*H \to K$ with $\phi i_G=\tau_G$ and $\phi i_H=\tau_H$.
It is not hard to prove uniqueness of $G*H$ up to isomorphism directly from this definition.
To prove existence, let $G = \langle A \mid R \rangle$ and $H = \langle B \mid S \rangle$ be presentations of $G$ and $H$ , and let $P = \langle A \cup B \mid R \cup S \rangle$. Then by basic properties of group presentations, the identity maps on $X$ induce homomorphisms $i_G:G \to P$ and $i_H:H \to P$.
Also, given $K,\tau_G,\tau_H$ as above, define $\psi:F(A \cup B) \to K$ by $\psi(a) = \tau_G(a)$ and $\psi(b) = \tau_H(b)$ for $a \in A,b \in B$. Then, since $\tau_G$ and $\tau_H$ are homomorphisms of $G$ and 0f $H$, $\psi$ maps all elements of $R \cup S$ to the identity, and so $\phi$ induces a homomorphism $\phi:P \to H$ with $\phi i_G=\tau_G$ and $\phi i_H=\tau_H$. Hence $P \cong G*H$.