Show that if \(G_1\) and \(G_2\) have finite presentations, so does \(G_1 * G_2\).
Show that if \(\pi_1(U \cap V, x_0)\) is finitely generated and \(\pi_1(U, x_0)\) and \(\pi_1(V, x_0)\) have finite presentations, then \(\pi_1(X, x_0)\) has a finite presentation. [Hint: If \(N'\) is a normal subgroup of \(\pi_1(U, x_0) * \pi_1(V, x_0)\) that contains the elements \(i_1(g_i)^{-1} i_2(g_i)\) where \(g_i\) runs over a set of generators for \(\pi_1(U \cap V, x_0)\), then \(N'\) contains \(i_1(g)^{-1} i_2(g)\) for arbitrary \(g\).]
Solution
Write finite presentations \[ G_1=\langle A\mid R\rangle, \qquad G_2=\langle B\mid S\rangle, \] with \(A,B,R,S\) finite. Replacing generators by disjoint copies if necessary, the free product has presentation \[ G_1*G_2\cong \langle A\sqcup B\mid R\sqcup S\rangle. \] Indeed, the quotient of the free group on \(A\sqcup B\) by the normal closure of \(R\sqcup S\) has the universal property of the free product. Hence \(G_1*G_2\) is finitely presented.
Let \[ H=\pi_1(U\cap V,x_0),\quad G_U=\pi_1(U,x_0),\quad G_V=\pi_1(V,x_0), \] and let \[ i_1:H\to G_U,\qquad i_2:H\to G_V \] be induced by inclusion. By van Kampen, \[ \pi_1(X,x_0)\cong (G_U*G_V)/N, \] where \(N\) is the normal closure of all elements \[ i_1(h)^{-1}i_2(h),\qquad h\in H. \]
Choose a finite generating set \(h_1,\dots,h_r\) for \(H\), and let \(N'\) be the normal closure in \(G_U*G_V\) of the finitely many elements \[ i_1(h_j)^{-1}i_2(h_j),\qquad 1\le j\le r. \] We show \(N'=N\). In the quotient \((G_U*G_V)/N'\), the two homomorphisms \(i_1,i_2:H\to (G_U*G_V)/N'\) agree on the generating set \(h_1,\dots,h_r\), hence agree on all of \(H\). Therefore \[ i_1(h)^{-1}i_2(h)\in N' \] for every \(h\in H\). Thus \(N\subseteq N'\), while \(N'\subseteq N\) is immediate from the definition. Hence \(N=N'\).
By part (a), \(G_U*G_V\) has a finite presentation because both factors do. Quotienting by the normal closure of the finitely many additional relators \[ i_1(h_j)^{-1}i_2(h_j) \] still gives a finite presentation. Therefore \(\pi_1(X,x_0)\) is finitely presented.