Let\(G = G_1 * G_2\). Given\(c \in G\), let\(cG_1c^{-1}\)denote the set of all elements of the form\(cxc^{-1}\), for\(x \in G_1\). It is a subgroup of\(G\); show that its intersection with\(G_2\)consists of the identity alone.
Solution
Let \(y\in cG_1c^{-1}\cap G_2\). We prove \(y=1\). Write \[ y=cxc^{-1} \] with \(x\in G_1\). If \(x=1\), then \(y=1\), so assume \(x\ne1\).
Choose a reduced word for \(c\). Cancel from its right end, if necessary, against \(x\), and rewrite the conjugate in cyclically reduced form. Concretely, there is a reduced word \(d\) and a nonidentity element \(x'\in G_1\) such that \[ cxc^{-1}=d x' d^{-1}, \] and either \(d\) is empty or the last letter of \(d\) lies in \(G_2\). In the latter case the displayed word is reduced: the last letter of \(d\) lies in \(G_2\), then \(x'\in G_1\setminus\{1\}\), then the first letter of \(d^{-1}\) lies in \(G_2\). Thus its reduced length is at least \(3\), so it cannot represent an element of the factor \(G_2\), whose nonidentity elements have reduced length \(1\).
If \(d\) is empty, then \(y=x'\in G_1\cap G_2\). By the reduced-word normal form for a free product, the two factors intersect only in the identity. Hence again \(y=1\).
Therefore \[ cG_1c^{-1}\cap G_2=\{1\}. \]