Past exam of the mathematics course of the University of Oxford / 2014 / Prelims / Mathematics I / 5 / a

Among the $|G|+1$ powers $e,g,g^2,\dots ,g^{|G|}$ in the finite group $G$, the pigeonhole principle forces two to be equal: $g^i=g^j$ with $0\le i<j\le |G|$. Multiply by $g^{-i}$ to obtainThis proves the existence of a positive power equal to the identity element, without presupposing the order of a group element.

The order of a group element $o(g)$ is the least positive integer $n$ for which $g^n=e$, whose existence follows from part (i). If $g^k=e$, use the Euclidean division to write $k=qn+r$, where $0\le r<n$. ThenMinimality of $n$ forces $r=0$. ThusThe argument also works for negative $k$, using integer powers and the same remainder convention.

Lagrange's theorem states that for a subgroup $H$ of a finite group $G$,where $[G:H]$ is the index of a subgroup, the number of its left cosets. In particular, $|H|$ divides $|G|$.

The cyclic subgroup $⟨g⟩$ consists of the distinct elements $e,g,\dots ,g^{n-1}$, where $n=o(g)$. Every power reduces to one of these by the Euclidean division, and a repetition among them would contradict minimality of $n$. Thus $|⟨g⟩|=o(g)$. Applying Lagrange's theorem gives