Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 6 / a (Coset partitions and counting)

Each left coset $gH$ is nonempty, since it contains $g=ge$, and every $g\in G$ lies in its own left coset. Suppose $aH$ and $bH$ intersect, say $a{h}_1=b{h}_2$ with $h_1,h_2\in H$. ThenSince $H$ is a subgroup, $h_2{h}_1^{-1}\in H$, and $aH\subseteq bH$. The reversed relation $b=a{h}_1{h}_2^{-1}$ gives $bH\subseteq aH$. Thus any two left cosets are either equal or disjoint, and together they cover $G$. Therefore $G/H$ is a partition of a set $G$. No finiteness or normal subgroup assumption is needed.

For $W=aH$ and $W^{\prime }=bH$, defineIf $w=ah$ then $f(w)=bh\in bH$, so this function is well defined. Its inverse function is $w^{\prime }\mapsto a{b}^{-1}{w}^{\prime }$, since successive left translation in a group by $b{a}^{-1}$ and $a{b}^{-1}$ cancels. ThusIn particular every left coset has the same cardinality as $H$.

Let the finite set $G/H$ contain $r$ distinct left cosets. By i, they form a disjoint partition of a set $G$, and by ii each has $|H|$ elements. If $H$ is also finite, counting this finite disjoint set union givesThus $G$ is a finite group. This is the counting argument behind Lagrange's theorem.