Show, using covering spaces, that the fundamental group of the Klein bottle is not abelian.
Solution
Let \(K\) be the Klein bottle, and let \(\Gamma\) be the group of homeomorphisms of \(\RR^2\) generated by \(A(x,y)=(x+1,y)\) and \(B(x,y)=(-x,y+1)\).
1. The quotient map \(p\colon\RR^2\to\RR^2/\Gamma\cong K\) is a universal covering, and \(A\), \(B\) are deck transformations.
The unit square is a fundamental region for \(\Gamma\). \(A\) identifies \((0,y)\) with \((1,y)\), and \(A\circ B\) sends \((x,0)\) to \((1-x,1)\), so the bottom and top sides are identified with reversed orientation; these are the square identifications of the Klein bottle. Since \(BA=A^{-1}B\), every element of \(\Gamma\) has the form \(A^mB^n\), which sends \((x,y)\) to \((\pm x+m,\,y+n)\). If \(n\neq0\) it moves every point vertically by a nonzero integer, and if \(n=0\), \(m\neq0\) it moves every point horizontally by a nonzero integer; so a ball of radius \(\tfrac14\) is disjoint from all its nontrivial translates, and the action is properly discontinuous. Hence \(p\) is a covering map whose deck group contains \(\Gamma\), and it is universal because \(\RR^2\) is simply connected.
2. \(AB\neq BA\).
\(B^{-1}(x,y)=(-x,y-1)\), so \(BAB^{-1}(x,y)=BA(-x,y-1)=B(1-x,y-1)=(x-1,y)=A^{-1}(x,y)\). Since \(A(0,0)=(1,0)\neq(-1,0)=A^{-1}(0,0)\), \(BAB^{-1}\neq A\).
3. Q.E.D.
The deck transformation group of a universal covering is isomorphic to the fundamental group of the base. By steps 1 and 2 this group contains the noncommuting elements \(A\) and \(B\), so \(\pi_1(K)\) is not abelian.