Let \(G\) be a topological group with operation \(\cdot\) and identity element \(x_0\). Let \(\Omega(G, x_0)\) denote the set of all loops in \(G\) based at \(x_0\). If \(f, g \in \Omega(G, x_0)\), let us define a loop \(f \otimes g\) by the rule
\[ (f \otimes g)(s) = f(s) \cdot g(s). \]
Show that this operation makes the set \(\Omega(G, x_0)\) into a group.
Show that this operation induces a group operation \(\otimes\) on \(\pi_1(G, x_0)\).
Show that the two group operations \(*\) and \(\otimes\) on \(\pi_1(G, x_0)\) are the same. [Hint: Compute \((f * e_{x_0}) \otimes (e_{x_0} * g)\).]
Show that \(\pi_1(G, x_0)\) is abelian.
Solution
Let \(e=x_0\) be the identity.
Pointwise multiplication of loops is associative because multiplication in \(G\) is associative. The constant loop \(e_e\) is the identity, and \[ f^{-1}(s)=f(s)^{-1} \] is a continuous loop and is the inverse of \(f\). Thus \((\Omega(G,e),\otimes)\) is a group.
If \(f\simeq f'\) and \(g\simeq g'\) through based homotopies \(F,G\), then \[ H(s,t)=F(s,t)G(s,t) \] is a based homotopy from \(f\otimes g\) to \(f'\otimes g'\). Hence \[ [f]\otimes[g]=[f\otimes g] \] is well defined on \(\pi_1(G,e)\).
In \(\pi_1(G,e)\), let \(*\) denote ordinary loop concatenation. Since \(f*e_e\simeq f\) and \(e_e*g\simeq g\), \[ [f]\otimes[g]=[f*e_e]\otimes[e_e*g]. \] For \(s\le1/2\), \(((f*e_e)\otimes(e_e*g))(s)=f(2s)\cdot e=f(2s)\), and for \(s\ge1/2\) it equals \(e\cdot g(2s-1)=g(2s-1)\). So \[ (f*e_e)\otimes(e_e*g)=f*g. \] Therefore \[ [f]\otimes[g]=[f]*[g]. \] So the two operations coincide.
The operation \(\otimes\) also satisfies the interchange law with concatenation: \[ (a*b)\otimes(c*d)=(a\otimes c)*(b\otimes d). \] This holds pointwise for loops, hence for their classes. Write \(1=[e_e]\), the identity for both operations. By part (c) and the interchange law, \[ [f]*[g] =[f]\otimes[g] =(1*[f])\otimes([g]*1) =(1\otimes[g])*([f]\otimes1) =[g]*[f]. \] Thus \(\pi_1(G,e)\) is abelian.