Let \(F\) be a free group on two free generators \(\alpha\) and \(\beta\). Let \(H\) be the subgroup generated by \(\alpha\). Show that \(H\) has infinite index in \(F\).
Solution
Consider the right cosets \[ H\beta^n,\qquad n\in\mathbb Z, \] where \(H=\langle\alpha\rangle\).
If \[ H\beta^m=H\beta^n, \] then \[ \beta^{m-n}\in H=\langle\alpha\rangle. \] Hence for some \(k\in\mathbb Z\), \[ \beta^{m-n}=\alpha^k. \] In the free group \(F(\alpha,\beta)\), a nonzero power of \(\beta\) is a reduced word involving only the letter \(\beta^{\pm1}\), while a nonzero power of \(\alpha\) involves only \(\alpha^{\pm1}\). By uniqueness of reduced words, equality is possible only when \[ m-n=0\quad\text{and}\quad k=0. \] Thus the cosets \(H\beta^n\) are pairwise distinct as \(n\) varies. There are infinitely many of them, so \[ \boxed{[F:H]=\infty.} \]