# Free product of free groups and group presentations

Here is Problem 9-4(b) from Introduction to Topological Manifolds by John M. Lee:

Let $$S_1$$ and $$S_2$$ be disjoint sets, and let $$R_i$$ be a subset of the free group $$F(S_i)$$ for $$i=1,2$$. Prove that $$\langle S_1 \cup S_2 \mid R_1 \cup R_2 \rangle$$ is a presentation of the free product group $$\langle S_1 \mid R_1 \rangle * \langle S_2 \mid R_2 \rangle$$.

The proof is fairly straightforward if we stick to the universal properties of the free group, free product and quotient group. Here I will write out the full details of the proof, which is more fun to do than to read. There are three basic steps to showing that two things are isomorphic:

1. Set up the canonical maps and choose names for them. These maps are given to you by the definition of the free group, free product, etc.
2. Repeatedly use the existence part of the universal properties to derive two maps.
3. Use the uniqueness part of the universal properties to show that the maps are inverses to each other.

## Part 1

We first show that $$F(S_1 \cup S_2) \cong F(S_1)*F(S_2)$$. For $$i=1,2$$, let \begin{align}
j_i : S_i &\to S_1 \cup S_2, \\
k_i : S_i &\to F(S_i), \\
\ell_i : F(S_i) &\to F(S_1)*F(S_2), \\
j : S_1 \cup S_2 &\to F(S_1 \cup S_2)
\end{align} be the canonical injections. There exist homomorphisms $$m_i:F(S_i) \to F(S_1 \cup S_2)$$ such that $$m_i \circ k_i = j \circ j_i$$, and these maps induce a homomorphism $$\varphi : F(S_1)*F(S_2) \to F(S_1 \cup S_2)$$ satisfying $$m_i = \varphi \circ \ell_i$$. Since $$S_1$$ and $$S_2$$ are disjoint, there is a map $$k : S_1 \cup S_2 \to F(S_1)*F(S_2)$$ satisfying $$k \circ j_i = \ell_i \circ k_i$$. This induces a homomorphism $$\psi : F(S_1 \cup S_2) \to F(S_1)*F(S_2)$$ satisfying $$k = \psi \circ j$$. Now $$\varphi \circ \psi \circ j \circ j_i = \varphi \circ k \circ j_i = \varphi \circ \ell_i \circ k_i = m_i \circ k_i = j \circ j_i,$$ so $$\varphi \circ \psi \circ j = j$$ and $$\varphi \circ \psi = \mathrm{Id}_{F(S_1 \cup S_2)}$$ by uniqueness. Similarly, $$\psi \circ \varphi \circ \ell_i \circ k_i = \psi \circ m_i \circ k_i = \psi \circ j \circ j_i = k \circ j_i = \ell_i \circ k_i,$$ so $$\psi \circ \varphi \circ \ell_i = \ell_i$$ and $$\psi \circ \varphi = \mathrm{Id}_{F(S_1)*F(S_2)}$$ by uniqueness. This proves that $$\varphi$$ is an isomorphism.

## Part 2

Now we show that $$F(S_1 \cup S_2)/\overline{R_1 \cup R_2} \cong (F(S_1)/\overline{R_1})*(F(S_2)/\overline{R_2}),$$ where $$\overline{S}$$ denotes the normal closure of $$S$$. Let \begin{align}
\pi : F(S_1 \cup S_2) &\to F(S_1 \cup S_2)/\overline{R_1 \cup R_2}, \\
\pi_i : F(S_i) &\to F(S_i)/\overline{R_i}
\end{align} be the quotient maps and let $$n_i : F(S_i)/\overline{R_i} \to (F(S_1)/\overline{R_1})*(F(S_2)/\overline{R_2})$$ be the canonical injections. Since $$\overline{R_i}\subseteq\ker(\pi \circ m_i)$$, there are homomorphisms $$f_i : F(S_i)/\overline{R_i} \to F(S_1 \cup S_2)/\overline{R_1 \cup R_2}$$ such that $$f_i \circ \pi_i = \pi \circ m_i$$. Then there is a homomorphism $$f : (F(S_1)/\overline{R_1})*(F(S_2)/\overline{R_2}) \to F(S_1 \cup S_2)/\overline{R_1 \cup R_2}$$ such that $$f_i = f \circ n_i$$. There is a homomorphism $$p : F(S_1)*F(S_2) \to (F(S_1)/\overline{R_1})*(F(S_2)/\overline{R_2})$$ satisfying $$n_i \circ \pi_i = p \circ \ell_i$$. Since $$\overline{R_1 \cup R_2}\subseteq\ker(p \circ \psi)$$, there is a homomorphism $$g : F(S_1 \cup S_2)/\overline{R_1 \cup R_2} \to (F(S_1)/\overline{R_1})*(F(S_2)/\overline{R_2})$$ satisfying $$g \circ \pi = p \circ \psi$$. Now \begin{align}
g \circ f \circ n_i \circ \pi_i &= g \circ f_i \circ \pi_i = g \circ \pi \circ m_i \\
&= p \circ \psi \circ m_i = p \circ \psi \circ \varphi \circ \ell_i = p \circ \ell_i = n_i \circ \pi_i,
\end{align} so $$g \circ f \circ n_i = n_i$$ and $$g \circ f = \mathrm{Id}_{(F(S_1)/\overline{R_1})*(F(S_2)/\overline{R_2})}$$ by uniqueness. Similarly, \begin{align}
f \circ g \circ \pi \circ m_i &= f \circ p \circ \psi \circ m_i = f \circ p \circ \psi \circ \varphi \circ \ell_i \\
&= f \circ p \circ \ell_i = f \circ n_i \circ \pi_i = f_i \circ \pi_i = \pi \circ m_i
\end{align} implies that $$f \circ g \circ \pi \circ j \circ j_i = f \circ g \circ \pi \circ m_i \circ k_i = \pi \circ m_i \circ k_i = \pi \circ j \circ j_i,$$ so $$f \circ g \circ \pi = \pi$$ and $$f \circ g = \mathrm{Id}_{F(S_1 \cup S_2)/\overline{R_1 \cup R_2}}$$ by uniqueness. This proves that $$\langle S_1 \cup S_2 \mid R_1 \cup R_2 \rangle \cong \langle S_1 \mid R_1 \rangle * \langle S_2 \mid R_2 \rangle.$$

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