Let \(s\in G\). Define \(C_G(s)\), prove that it is a subgroup of \(G\), and show that \(\langle s\rangle\normal C_G(s)\).
Solution
The centralizer of \(s\) in \(G\) is \[ C_G(s)=\{g\in G:gs=sg\}. \]
It is a subgroup. The identity commutes with \(s\). If \(g,h\in C_G(s)\), then \[ (gh)s=g(hs)=g(sh)=(gs)h=(sg)h=s(gh), \] so \(gh\in C_G(s)\). If \(g\in C_G(s)\), then \(gs=sg\) implies \[ g^{-1}s=sg^{-1}, \] so \(g^{-1}\in C_G(s)\).
Also \(s\in C_G(s)\), hence \(\langle s\rangle\le C_G(s)\). For \(g\in C_G(s)\) and any integer \(k\), \[ gs^kg^{-1}=s^kgg^{-1}=s^k. \] Thus every element of \(C_G(s)\) normalizes \(\langle s\rangle\), and in fact conjugation acts trivially on it. Therefore \[ \langle s\rangle\normal C_G(s). \]