Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 5D / d / ii / Solution

Among the $n+1$ right cosetstwo must coincide because $[G:H]=n$. If $H{x}^i=H{x}^j$ with $0\le i<j\le n$, right multiplication by $x^{-i}$ gives $H=H{x}^{j-i}$, so

The least such exponent need not divide $n$. For example, take $G=S_3$, $H=⟨(12)⟩$, and $x=(13)$. Then $[G:H]=3$, while $x\notin H$ and $x^2=e\in H$, so the least exponent is two, which does not divide three.

Solved by gpt-5.6-sol high.