Show that \(K\) can not be a topological group.
Solution
1. The fundamental group of the Klein bottle \(K\) is nonabelian:\[ \pi_1(K)\cong\langle a,b\mid aba^{-1}=b^{-1}\rangle. \]
This is the standard presentation obtained from the square model of the Klein bottle. It is nonabelian, since the relation gives \(aba^{-1}=b^{-1}\) rather than \(b\); for example the group maps onto the nonabelian infinite dihedral group after imposing \(b^2=1\) and retaining the conjugation relation.
2. The fundamental group of every path-connected topological group is abelian.
Let \(G\) be a topological group with identity \(e\). On based loops at \(e\) there are two products: concatenation \(*\) and pointwise multiplication \[ (\alpha\cdot\beta)(t)=\alpha(t)\beta(t). \] They have the same unit and satisfy the interchange law \[ (\alpha*\beta)\cdot(\gamma*\delta) =(\alpha\cdot\gamma)*(\beta\cdot\delta). \] The Eckmann–Hilton argument therefore shows that the induced product on \(\pi_1(G,e)\) is commutative.
3. Hence the Klein bottle cannot admit a topological-group structure.
If \(K\) were a topological group, 2 would force \(\pi_1(K)\) to be abelian, contradicting 1.