# Difference between revisions of "Coset"

A coset is a subset of a group.

Specifically, let $G$ be a group, and let $H$ be a subgroup of $G$. The left cosets modulo $H$ are the subsets of $G$ of the form $aH$, for $a\in G$. Note that for any coset $aH$, the mapping $x \mapsto ax$ is a bijection from $H$ to $aH$. Hence for any $a\in G$, $|aH| = |H| = |Ha|$.

The image of a left coset $aH$ under the mapping $x \mapsto x^{-1}$ is the right coset $Hx^{-1}$. This mapping induces a bijection from the set of left cosets of $H$ to the set of right cosets of $H$.

The cardinality of the set of left cosets of $H$ is called the index of $H$ with respect to $G$; it is denoted $(G:H)$. This is also the cardinality of the set of right cosets of $H$.

Proposition. The relations $x^{-1}y \in H$, $xy^{-1} \in H$ are equivalence relations.

Proof. We prove that the first relation is an equivalence relation; the second then follows by passing to the opposite law on $G$.

We abbreviate $x^{-1}y \in H$ as $R(x,y)$. For any $x$, note that $x^{-1}x \in H$, so $R(x,x)$. If $x^{-1}y \in H$, then $y^{-1}x = (x^{-1}y)^{-1} \in H$, so $R(x,y)$ implies $R(y,x)$. Finally, if $x^{-1}y \in H$ and $y^{-1}z \in H$, then $x^{-1}z = (x^{-1}y)(y^{-1}z) \in H$; hence $R(x,y)$ and $R(y,z)$ together imply $R(x,z)$. Hence $R(x,y)$ is an equivalence relation. $\blacksquare$

## Cosets and compatible relations

We call a relation $R(x,y)$ left compatible with the group structure of $G$ if $x \equiv y \pmod{R}$ implies $zx \equiv zy \pmod{R}$, for all $z\in G$. Similarly, we say $R$ is right compatible with the group structure of $G$ if $x \equiv y \pmod{R}$ implies $xz \equiv yz \pmod{R}$. Note that $R$ is compatible with the group law on $G$ if and only if it is both left- and right-compatible with the structure.

Theorem. An equivalence relation $R(x,y)$ on a group $G$ is left (resp. right) compatible with $G$ if and only if it is of the form $x^{-1}y \in H$ (resp. $y^{-1}x \in H$), for some subgroup $H$ of $G$. In this case, $H$ is the equivalence class of $e$, the identity, and the equivalence classes are the left (resp. right) cosets of $H$.

Proof. We will consider only the case for $R(x,y)$ left compatible with $G$; the other case follows from symmetry.

Let $H$ be the equivalence class of $e$. Note that $x \equiv y \pmod{R}$ if and only if $e \equiv x^{-1}x \equiv x^{-1}y \pmod{R}$, which is true if and only if $x^{-1}y \in H$. It thus remains to show that $H$ is a subgroup of $G$.

To this end, we note that evidently $e\in H$; also, if $x\in H$, then $(x^{-1})^{-1}x = x^2 \equiv xe \pmod{R}$, so $x^{-1} \in H$. Finally, if $x,y$ are in $H$, then $xy \equiv xe \in H$. Thus $H$ is a subgroup of $G$.

Conversely, suppose $H$ is a subgroup of $G$, and define $R(x,y)$ as $x^{-1}y \in H$. We have proven that $R(x,y)$ is an equivalence relation; evidently $e\equiv x \pmod{R}$ if and only if $x = e^{-1}x \in H$. Now, if $x \equiv y \pmod{R}$, then $(zx)^{-1}(zy) = x^{-1}z^{-1}zy = x^{-1}y \in H$, so $R(x,y)$ is left-compatible with the group structure of $G$.

Now, $x^{-1}y \in H$ if and only if $y \in xH$;. Hence the set of $y$ equivalent to $x$ (mod $R$) is the set $xH$. Thus the equivalence classes of $R$ are the left cosets mod $H$. $\blacksquare$