Group theory/Left and right cosets/Examples/Introduction/Section

Definition

Let ${}G$ be a group, and let ${}H\subseteq G$ be a subgroup. We set ${}x\sim _{H}y$ (and say that ${}x$ and ${}y$ are equivalent) if

${}x^{-1}y\in H$.

This is indeed an equivalence relation: From ${}x^{-1}x=e_{G}\in H$ we get that this relation is reflexive. From ${}x^{-1}y\in H$ we get immediately ${}y^{-1}x=(x^{-1}y)^{-1}\in H$, and from ${}x^{-1}y\in H$ and ${}y^{-1}z\in H$ we get ${}x^{-1}z\in H$.

Two group elements ${}x$ and ${}y$ are equivalent if and only if there exists an element ${}h\in H$ of the subgroup with ${}y=xh$. In accordance with example, we can interpret this situation in the sense that the subgroup ${}H$ provides a set of possible moves, and two elements are equivalent if and only if they can be moved to each other by a movement from ${}H$.

Example

In an (additively written) commutative group like ${}\mathbb {Z}$ or a vector space ${}V$, and of a given subgroup ${}H$, the equivalence relation ${}x\sim _{H}y$ means ${}y-x\in H$, that is, there exists a ${}h\in H$ such that

${}y=x+h\,.$

The equivalence classes are of the form ${}x+H={\left\{x+h\mid h\in H\right\}}$. In case ${}H=\mathbb {Z} d\subseteq \mathbb {Z}$ with a fixed ${}d$, the equivalence classes have the form

$H=\mathbb {Z} d,\,1+H=\{\ldots ,1-d,1,1+d,1+2d,\ldots \},\,2+H=\{\ldots ,2-d,2,2+d,2+2d,\ldots \},\ldots .$

These classes encompass those integer numbers that have, upon division by ${}d$, the remainder ${}0$, or ${}1$, or ${}2$, etc. They form a partition of ${}\mathbb {Z}$.

The equivalence classes of a linear subspace.

If ${}H=U\subseteq V$ is a linear subspace, then the equivalence classes have the form ${}v+U={\left\{v+u\mid u\in U\right\}}$ for a vector ${}v\in V$. This is ther affine space with starting point ${}v$ and the translating space ${}U$ (in the sense of definition). The equivalence classes form a family of affine subspaces parallel to each other.

Definition

Let ${}G$ be a group, and let ${}H\subseteq G$ be a subgroup. For every ${}x\in G$, the subset

${}xH={\left\{xh\mid h\in H\right\}}\,$

is called the left coset of ${}x$ in ${}G$ with respect to ${}H$. Every subset of this form is called a left coset. Accordingly, a set of the form

${}Hx={\left\{hx\mid h\in H\right\}}\,$

is called the right coset

(of ${}x$).

The equivalence classes to the equivalence relation defined above are, because of

${}{\begin{aligned}{[}x{]}&={\left\{y\in G\mid x\sim y\right\}}\\&={\left\{y\in G\mid x^{-1}y\in H\right\}}\\&={\left\{y\in G\mid {\text{there exists }}h\in H{\text{ with }}x^{-1}y=h\right\}}\\&={\left\{y\in G\mid {\text{there exists }}h\in H{\text{ with }}y=xh\right\}}\\&=xH,\end{aligned}}$

the left cosets. The coset to the neutral element is the subgroup ${}H$ itself. Therefore, the left cosets form a disjoint decomposition (a partition) of ${}G$. This holds for the right cosets as well. In the commutative case, one does not have to distinguish between left cosets and right cosets.

Lemma

Let ${}G$ be a group, and let ${}H\subseteq G$ be a subgroup. Let ${}x,y\in G$ be elements. Then the following statements are equivalent.

1. ${}x\in yH$.
2. ${}y\in xH$.
3. ${}y^{-1}x\in H$.
4. ${}x^{-1}y\in H$.
5. ${}xH\cap yH\neq \emptyset$.
6. ${}x\sim _{H}y$.
7. ${}xH=yH$.

Proof

The equivalence between ${}(1)$ and ${}(3)$ (and between ${}(2)$ and ${}(4)$) follows by multiplication with ${}y^{-1}$ and with ${}y$. The equivalence between ${}(3)$ and ${}(4)$ follows by going to the inverse elements. From ${}(1)$ we get ${}(5)$ because of ${}1\in H$. If ${}(5)$ holds, then this means that ${}xh_{1}=yh_{2}$ holds with certain ${}h_{1},h_{2}\in H$. Therefore, ${}x=yh_{2}h_{1}^{-1}$, and ${}(1)$ is satisfied. (4) and (6) are equivalent due to the definition. Since the left cosets are the equivalence classes, the equivalence between (5) and (7) follows.

$\Box$