Difference between revisions of "Coset"

Latest revision as of 00:28, 3 September 2008

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 $H$ . $\blacksquare$

@@ Line 31: / Line 31: @@
 == See also ==
-* [[Lagrange's theorem]]
+* [[Lagrange's Theorem]]
 * [[Quotient group]]
 * [[Orbit]]
 [[Category:Group theory]]

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Difference between revisions of "Coset"

Latest revision as of 00:28, 3 September 2008

Cosets and compatible relations

See also