The free group \(F(a,b)\) on two generators is the fundamental group of the space \(Z=S^1\vee S^1\), using the wedge point as basepoint, with \(a\) and \(b\) corresponding to each of the loops \(S^1\). Construct the covering space of \(Z\) corresponding to the subgroup \(H=\langle a^2b,aba,ab^2,b^2a\rangle\), determine the index of \(H\) in \(F(a,b)\), and decide if \(H\) is a normal subgroup of \(F(a,b)\).
Solution
Let \(\phi\colon F(a,b)\to\ZZ/3\) be the surjective homomorphism with \(\phi(a)=\phi(b)=1\), and \(K=\ker\phi\), of index \(3\).
1. \(H\le K\).
Each of \(a^2b\), \(aba\), \(ab^2\), \(b^2a\) has total exponent sum \(3\).
2. \(K\) is free on \(s_1=ba^{-1}\), \(s_2=aba^{-2}\), \(s_3=a^3\), \(s_4=a^2b\).
These are the nontrivial Reidemeister–Schreier generators for the Schreier transversal \(\{1,a,a^2\}\): for \(a\) they are \(1\), \(1\), \(a^3\), and for \(b\) they are \(ba^{-1}\), \(aba^{-2}\), \(a^2b\).
3. \(K\le H\).
Put \(u=a^2b\), \(w=aba\), \(x=ab^2\), \(y=b^2a\). Then \(u=s_4\), \(w=s_2s_3\), \(x=s_2s_4\), and \(y=s_1s_2s_3=s_1w\). Hence \(s_4=u\), \(s_2=xu^{-1}\), \(s_3=s_2^{-1}w\), and \(s_1=yw^{-1}\) lie in \(H\), and by step 2 so does \(K\).
4. Q.E.D.
By steps 1 and 3, \(H=K\), so \([F(a,b):H]=\boxed{3}\), and \(H\) is normal as the kernel of \(\phi\). The covering space is the graph with vertices \(v_0,v_1,v_2\) indexed by \(\ZZ/3\), basepoint \(v_0\), and for each label \(c\in\{a,b\}\) oriented \(c\)-edges \(v_0\to v_1\to v_2\to v_0\), mapped to \(Z\) by sending each \(c\)-edge around the loop \(c\). Each vertex has one incoming and one outgoing edge of each label, so this is a connected \(3\)-sheeted covering, and a word lifts to a loop at \(v_0\) exactly when \(\phi\) vanishes on it, so its subgroup is \(H\).