Cosets and Lagrange’s Theorem

Cosets

Coset

Let $G$ be a group, $H\le G$ a subgroup and $a\in G$. The following subsets are called cosets of $H$ in $G$: $\begin{array}{rl}aH& =\{ah\mid h\in H\}(\mathrm{left}\mathrm{coset}),\\ Ha& =\{ha\mid h\in H\}(\mathrm{right}\mathrm{coset}).\end{array}$

Proposition

Let $H$ be a subgroup of a group $G$. Then the left (or right) cosets form a partition of $G$. That is $G={\coprod }_{a\in G}aH$ and for $a,b\in G$ either $aH=bH$ or $aH\cap bH=\varnothing$.

Lagrange’s Theorem

Let $H$ be a subgroup of a finite group $G$. Then $|H|$ divides $|G|$.

Proof Let $a_{1}H,...,a_{m}H$ be the distinct cosets. Then $|G|=|a_{1}H|+\cdot \cdot \cdot +|a_{m}H|=m|H|$. $\square$

Corollary

Let $G$ be a finite group and $a\in G$. Then $o(a)\mid |G|$. It follows that $x^{|G|}=e$ for all $x\in G$. In particular, if there exists $g\in G$ such that $o(g)=|G|$ then $G$ is cyclic.

Proof Consider $⟨a⟩$. By Lagrange’s Theorem, we have $|⟨a⟩|\mid |G|$. Since $⟨a⟩$ is cyclic, $|⟨a⟩|=o(a)$, which implies that $o(a)\mid |G|$. Moreover, if $o(g)=|G|$, then $⟨g⟩$ will have order $|G|$, which means $G=⟨g⟩$. $\square$

Index

Index

The index of a subgroup $H$ in a group $G$, denoted $[G:H]$, is defined as the cardinality of the set of (left) cosets of $H$ in $G$.
In particular, if $G$ is a finite group, then $[G:H]=|G|/|H|$.

Corollary

A group of prime order is cyclic.

Proof Suppose group $G$ has prime order. Then for all subgroup $H\le G$, by corollary, for all $a\in G$,we have $o(a)\mid |G|$, thus $o(a)=1$ or $o(a)=|G|$, indicating that $G$ is a cyclic group. $\square$

Proposition

A subgroup whose index is the smallest prime dividing the order of $G$ is normal. In, particular, any subgroup of index 2 in $G$ is automatically normal.