Theorem 16.1. Let $H$ denote a subgroup of a group $G,$ and define a relation $\sim$ on $G$ as follows: \[ a\sim b\quad\text{ iff }\quad ab^{-1}\in H. \] Then $\sim$ is an equivalence relation on $G.$

Example. (Page 80) Recall that if $n\in\Z,$ then $\gen{n}$ is the subgroup consisting of multiples of $n.$ Therefore, congruence modulo $n$ may be defined like so: \[ a\equiv b \pmod{n}\quad\text{ iff }\quad a-b \in\gen{n}. \] Theorem 16.1 generalizes this relation by replacing $\Z$ by an arbitrary group $G,$ $\gen{n}$ by an arbitrary subgroup $H$ of $G,$ and $a-b$ by the corresponding expression $ab^{-1}$ in our general multiplicative notation.

Definition. (Page 81) The equivalence classes for the equivalence relation defined in Theorem 16.1 are called the right cosets of $H$ in $G.$

Example. (Page 81) The right cosets of $\gen{n}$ in $\Z$ are the congruence classes mod $n.$

Definition. (Page 81) If $H$ is a subgroup of $G$ and $a\in G,$ define \[ Ha=\{ha\mid h\in H\}. \] It will be shown in Lemma 16.1 that $Ha$ so defined is the right coset of $H$ to which $a$ belongs. Therefore, \[ Ha=[a]_\sim \] where $\sim$ is the equivalence relation of Theorem 16.1. If the group operation is $+,$ then $H+a$ is written in place of $Ha;$ similarly for other operations.

Example 16.1. Let $G=\Z$ and $H=\gen{7}.$ Then \begin{align*} H+3 &=\gen{7} + 3\\ &=\{ \ldots,-14,-7,0,7,14,\ldots \} + 3\\ &=\{ \ldots,-11,-4,3,10,17,\ldots \}\\ &=[3]\pmod{7} \end{align*} That is, $H3=\gen{7} + 3 = [3]_7.$

Example 16.2. Let $G=S_3$ and $H=\{(1),(1\ 2)\}.$ Then \begin{align*} H(1)&=\{(1)(1),(1\ 2)(1)\}\\ &=\{(1),(1\ 2)\}\\ H(1\ 2\ 3)&=\{(1)(1\ 2\ 3),(1\ 2)(1\ 2\ 3)\}\\ &=\{(1\ 2\ 3),(2\ 3)\}\\ H(1\ 3\ 2)&=\{(1)(1\ 3\ 2),(1\ 2)(1\ 3\ 2)\}\\ &=\{(1\ 3\ 2),(1\ 3)\} \end{align*} Notice that these three sets partition $G.$ In fact, by Lemma 16.1 next, they are the right cosets of $H$ in $G.$

Lemma 16.1. If $H$ is a subgroup of a group $G,$ and $a,b\in G,$ then the following four conditions are equivalent.

  1. $ab^{-1}\in H$
  2. $a=hb\ $ for some $h\in H.$
  3. $a\in Hb.$
  4. $Ha=Hb$
As a consequence, $Ha$ is the right coset of $H$ to which $a$ belongs.

Steps. To compute all of the right cosets of a subgroup $H$ in a finite group $G.$

  1. Write $H=He.$ This is your first right coset of $H$ in $G.$
  2. Choose any element $a\in G$ such that $a\not\in H.$
  3. Compute $Ha.$ This is your second right coset of $H$ in $G.$
  4. Choose any element $b\in G$ such that $b\not\in H\cup Ha.$
  5. Compute $Hb.$ This is your third right coset of $H$ in $G.$
  6. Continue in this way until the elements of $G$ have been exhausted. At which time, you will have found all the right cosets of $H$ in $G.$

Example 16.3. Let $G=\Z_{12}$ and $H=\gen{[4]}.$ The right cosets of $H$ in $G$ are \begin{align*} H &=\{[0],[4],[8]\}\\ H\oplus[1] &=\{[1],[5],[9]\}\\ H\oplus[2] &=\{[2],[6],[10]\}\\ H\oplus[3] &=\{[3],[7],[11]\} \end{align*}

Lemma 16.2. If $H$ is a finite subgroup of $G,$ and $a\in G,$ then $\abs{H}=\abs{Ha}.$

Definition. Left cosets are defined in the same as right cosets by replacing $ab^{-1}\in H$ in Theorem 16.1 by $a^{-1}b\in H,$ and replacing "right" with "left" in the Definition of right coset. The left cosets then have the form \[ aH=\{ah\mid h\in H\}. \] and \[ aH=[a]_\sim \]