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

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

For $a\in A$, the element $a=ae$ belongs to both $A$ and $aA$. The assumed equal-or-disjoint property therefore gives $aA=A$.

If $a,b\in A$, then $ab\in aA=A$, so $A$ is closed under multiplication. Also $e\in A=aA$, so some $c\in A$ satisfies $ac=e$. Therefore $c=a^{-1}\in A$. Along with the given identity element, closure and existence of inverse elements establish that $A$ is a subgroup of $G$. This is the subgroup test. The proof even works without assuming that $G$ is finite.

Choose $b_0\in B$ and put $H=b_0^{-1}B$. Then $e\in H$. For any $g_1,g_2\in G$,so these translates are equal or disjoint by the assumed property of $B$. Applying i shows that $H$ is a subgroup. ThusThis proves that $B$ is a left coset. Together with i, it gives the equal-or-disjoint translates characterize cosets criterion.