Prove Poincaré’s theorem: if \(H\le G\) has finite index \(n\), then there is \(N\trianglelefteq G\) with \(N\subseteq H\) and \[ [G:N]\mid n!. \]
Solution
Let \(G\) act by left multiplication on the \(n\) left cosets \(G/H\). This gives a homomorphism \[ \rho:G\longrightarrow S_n. \] Let \[ N=\ker\rho. \] Then \(N\trianglelefteq G\). Moreover, \[ g\in N \iff gxH=xH\text{ for every }x\in G \iff x^{-1}gx\in H\text{ for every }x\in G, \] so \[ N=\bigcap_{x\in G}xHx^{-1}\subseteq H. \] By the first isomorphism theorem, \[ G/N\cong\operatorname{im}\rho\le S_n. \] Hence, by Lagrange, \[ [G:N]=|G/N|\mid |S_n|=n!. \]