If \(m > 1\), show the fundamental group of the \(m\)-fold projective plane is not abelian. [Hint: There is a homomorphism mapping this group onto the group \(\mathbb{Z}/2 * \mathbb{Z}/2\).]
Solution
For the \(m\)-fold projective plane, \[ \pi_1(P^{2\# m})\cong \left\langle \alpha_1,\ldots,\alpha_m\ \middle|\ \alpha_1^2\alpha_2^2\cdots\alpha_m^2=1\right\rangle. \] Let \[ D_\infty\cong C_2*C_2=\langle a,b\mid a^2=b^2=1\rangle. \] For \(m>1\), send \[ \alpha_1\mapsto a,\qquad \alpha_2\mapsto b,\qquad \alpha_i\mapsto1\quad(i\ge3). \] The surface relator maps to \(a^2b^2=1\), so this defines a surjective homomorphism \[ \pi_1(P^{2\# m})\twoheadrightarrow C_2*C_2. \] The group \(C_2*C_2\) is nonabelian: its reduced words \(ab\) and \(ba\) are distinct. Therefore the source cannot be abelian. Hence \(\pi_1(P^{2\# m})\) is nonabelian for every \(m>1\).