Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 5 / a

For a subgroup $H$ of $G$, the left coset represented by $g$ isSuppose two left cosets intersect. There are $h_1,h_2\in H$ such that $g_1{h}_1=g_2{h}_2$, soBecause $H$ is a subgroup, $h_2{h}_1^{-1}\in H$. Thus, for every $h\in H$, $g_{1}h=g_2(h_2{h}_1^{-1}h)\in g_{2}H$, proving $g_{1}H\subseteq g_{2}H$. Interchanging the two representatives gives the reverse inclusion. Therefore two left cosets are equal or disjoint.

Lagrange's theorem states that if $H$ is a subgroup of a finite group $G$, thenwhere the index of a subgroup $[G:H]$ is the number of distinct left cosets.

Every $g\in G$ lies in $gH$, since $g=ge$. By i, distinct left cosets are disjoint, so these sets form a partition of a set $G$. Each has cardinality $|H|$: the function $h\mapsto gh$ from $H$ to $gH$ is a bijection, with inverse $gh\mapsto h$, by cancellation in a group. Counting the elements in this finite partition of a set gives $|G|=[G:H]|H|$, proving the theorem.

The order of a group element $g$ is the smallest positive integer $d$ with $g^d=e$. Such an integer exists in a finite group: two powers in $e,g,g^2,\dots$ must agree, and cancellation gives a positive power equal to $e$.

By minimality, $e,g,\dots ,g^{d-1}$ are distinct. Every integer power of $g$ is one of them, by division of its exponent by $d$. They form the cyclic subgroupApplying Lagrange's theorem to this subgroup gives