the free product of two presentations is isomorphic to a third presentation using UP of free product.

Question

Here is the question that I want an answer to it using commutative diagrams (as small number of them as possible):

Prove that the free product of $ \langle g_1, \dots ,g_m | r_1, \dots ,r_n \rangle$ and $ \langle h_1, \dots ,h_k| s_1, \dots ,s_l \rangle$ is isomorphic to $\langle g_1, \dots ,g_m,h_1, \dots ,h_k|r_1, \dots ,r_n,s_1, \dots ,s_l \rangle.$

I had an incomplete proof for it where I ignored the relations, which is off course a mistake. I showed that the free group of a disjoint union $F(A\amalg B)$ is isomorphic to the free product of the corresponding free groups $F(A)*F(B)$.

Now, my question is:

How can I prove this problem using the Universal property of free product?

The Universal property of free product is:

Note that the free product is the "coproduct in the category of groups." The universal property is:

If $G_1$ and $G_2$ are two groups,then for any group $G$ and for any pair of homomorphisms $\varphi_1:G_1 \to G$ and $\varphi_2:G_2 \to G$ there is a unique homomorphism $G_1 * G_2$ that on words representatives' of length one agrees with $\varphi_1$ and $\varphi_2.$

Any help will be greatly appreciated!

Answer 1

Probably you want to use a different definition of presentation: A group $G$ with presentation $\langle S| R \rangle$ is the quotient $F/N(R)$ where $F$ is the free group generated on $S$ and $N(R)$ is the smallest normal subgroup generated by $R$.

Now for $\langle F(S_i) \mid N(R_i)\rangle:=G_i$ you want to figure out their coproduct.

Here's a sketch of how I think the argument should go. No promises, but the flavor is probably something similar.

Well note that there are maps $F(S_i) \to (F(S_1 \cup S_2)/N(R_1 \cup R_2):=H)$ induced by just sending $S_i \to S_i$ and since $N(R_i)$ is killed in the image, we get induced maps $G_i \to H$ so by the universal property of coproducts a unique map $G_1*G_2 \to H$. First show that this map is surjective and $N(R_1 \cup R_2)$ is contained in the kernel. Then use uniqueness to deduce that it's exactly the kernel

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since you tagged this algebraic topology:

Let $G_i$ have presentation complex given by the wedge of $S_i$ circles with a two cell glued in for each element of $R_i$. Then the coproduct in pointed spaces is the wedge sum and use Van Kampen's theorem (Pushouts are sent to pushouts in good circumstances) to deduce the result.

Answer 2

The two (not necessarily finitely) presented groups $G_i:=\langle S_i\mid R_i\rangle$ ($i=1,2$) come with maps $\sigma_i:S_i\to G_i$, and $(G_i,\sigma_i)$ is characterized (up to isomorphism) by the following universal property:

• $\sigma_i^*(R_i)=1$, i.e. $\forall s_1^{n_1}\dots s_j^{n_j}\in R_i\quad\sigma_i(s_1)^{n_1}\dots\sigma_i(s_j)^{n_j}$ is the identity element of $G_i$;
• for every group $H$ and every map $f:S_i\to H$ such that $f^*(R_i)=1$, there is a unique homomorphism $F:G_i\to H$ such that $F\circ\sigma_i=f$.

The free product $G_0:=G_1*G_2$ comes with two homomorphisms $\psi_i:G_i\to G_0$ ($i=1,2$), and $(G_0,\psi_1,\psi_2)$ is characterized (up to isomorphism) by the following universal property (correcting your formulation, since there is no categorical notion of `words representatives of length one'):

• for any group $G$ and any pair of homomorphisms $\phi_i:G_i\to G$ ($i=1,2$), there is a unique homomorphism $\varphi:G_0→G$ such that $\varphi∘ψ_i=\varphi_i$ ($i=1,2$).

Let $S:=S_1\sqcup S_2$ and $R:=R_1\sqcup R_2$. In order to prove that $G_0$ is isomorphic to $\langle S\mid R\rangle$, it will suffice to define a map $\sigma:S\to G_0$ such that:

• $\sigma^*(R)=1$
• for every group $H$ and every map $f:S\to H$ such that $f^*(R)=1$, there is a unique homomorphism $F:G_0\to H$ such that $F\circ\sigma=f$.

First define $\sigma:=(\psi_1\circ\sigma_1)\sqcup(\psi_2\circ\sigma_2).$

• The first property is easy to check: for every $r\in R_i$ ($i=1,2$), $\sigma^*(r)=(\psi_i\circ\sigma_i)^*(r)=\psi_i(\sigma_i^*(r))=\psi_i(1_{G_i})=1_{G_0}$.
• Let now $H$ be a group and $f:S\to H$ be a map such that $f^*(R)=1$.
  • For $i=1,2$, the map $f_i:=f_{|S_i}:S_i\to H$ satisfies $f_i^*(R_i)=1$ hence there is a (unique) homomorphism $\varphi_i:G_i\to H$ such that $\varphi_i\circ\sigma_i=f_i$. Then, there is a (unique) homomorphism $\varphi:G_0\to H$ such that $\varphi_i=\varphi\circ\psi_i$ ($i=1,2$). Such a $\varphi$ satisfies $$\varphi\circ\sigma=(\varphi\circ\psi_1\circ\sigma_1)\sqcup(\varphi\circ\psi_2\circ\sigma_2)=(\varphi_1\circ\sigma_1)\sqcup(\varphi_2\circ\sigma_2)=f_1\sqcup f_2=f.$$
  • Any other homomorphism $F:G_0\to H$ such that $F\circ\sigma=f$ is equal to $\varphi$, since $F\circ\psi_i\circ\sigma_i=f_i=\varphi_i\circ\sigma_i$ ($i=1,2$) $\implies F\circ\psi_i=\varphi_i=\varphi\circ\psi_i$ ($i=1,2$) $\implies F=\varphi$.