Cosets and Lagrange’s Theorem

Left and Right Cosets of Subgroups

Definition: Let $(G,\cdot )$ be a group and let $(H,\cdot )$ be a subgroup. Let $g\in G$. Then the Left Coset of $H$ with Representative $g$ is the set $gH=\{gh:h\in H\}$. The Right Coset of $H$ with Representative $g$ is the set $Hg=\{hg:h\in H\}$.

When the operation symbol “$+$” is used instead of $\cdot$ we often denote the left and right cosets of $H$ with representation $g$ with the notation $g+H$ and $H+g$ respectively.

For example, consider the group $(\mathbb {Z} ,+)$ and the subgroup $(3\mathbb {Z} ,+)$. Consider the element $2\in \mathbb {Z}$. Then the left coset of $3\mathbb {Z}$ with representative $2$ is:

${\begin{aligned}\quad 2+3\mathbb {Z} =\{2+h:h\in 3\mathbb {Z} \}=\{...,-1,2,5,...\}\end{aligned}}$

And the right coset of $3\mathbb {Z}$ with representative $2$ is:

${\begin{aligned}\quad 3\mathbb {Z} +2=\{h+2:h\in 3\mathbb {Z} \}=\{...,-1,2,5,...\}\end{aligned}}$

In this particular example we see that $2+3\mathbb {Z} =3\mathbb {Z} +2$. But in general, is $gH=Hg$ for a given subgroup $(H,\cdot )$ of $(G,\cdot )$ and for $g\in G$? The answer is NO. There are many examples when left cosets are not equal to corresponding right cosets.

To illustrate this, consider the symmetric group $(S_{3},\circ )$. Let $G=\{\epsilon ,(12)\}$. Then $(G,\circ )$ is a subgroup of $(S_{3},\circ )$ since $G\subset S_{3}$, $G$ is closed under $\circ$, and $\epsilon ^{-1}=\epsilon$, $(12)^{-1}=(12)$ (since $(12)$ is a transposition). Now consider the element $(13)\in S_{3}$. Then the left coset of $G$ with representative $(13)$ is:

${\begin{aligned}\quad (13)G=\{(13)\circ h:h\in G\}=\{(13)\circ \epsilon ,(13)\circ (12)\}=\{(13),(123)\}\end{aligned}}$

And the right coset of $G$ with representative $(13)$ is:

${\begin{aligned}\quad G(13)=\{h\circ (13):h\in G\}=\{\epsilon \circ (13),(12)\circ (13)\}=\{(13),(132)\}\end{aligned}}$

We note that $(123)\neq (132)$ and so $(13)G\neq G(13)$!

So, when exactly are the left and right cosets of a subgroup with representative $g$ equal? The following theorem gives us a simple criterion for a large class of groups.

Proposition 1: Let $(G,\cdot )$ be a group and let $(H,\cdot )$ be a subgroup. If $(G,\cdot )$ is abelian then for all $g\in G$, $gH=Hg$.

• Proof: Let $g\in G$. If $G$ is abelian then for all $h\in G$ (and hence for all $h\in H$) we have that $g\cdot h=h\cdot g$. So:

${\begin{aligned}\quad gH=\{g\cdot h:h\in H\}=\{h\cdot g:h\in H\}=Hg\quad \blacksquare \end{aligned}}$

Proposition 2: Let $(G,\cdot )$ be a group, $(H,\cdot )$ a subgroup, and $g_{1},g_{2}\in G$. Then the following statements are equivalent:

a) $g_{1}H=g_{2}H$.
b) $Hg_{1}^{-1}=Hg_{2}^{-1}$.
c) $g_{1}H\subseteq g_{2}H$.
d) $g_{1}\in g_{2}H$.

e) $g_{1}^{-1}g_{2}\in H$.

The Set of Left (Right) Cosets of a Subgroup Partitions the Whole Group

Recall that if $(G,\cdot )$ is a group, $(H,\cdot )$ is a subgroup, and $g\in G$ then the left coset of $H$ with representative $g$ is the set:

${\begin{aligned}\quad gH=\{gh:h\in H\}\end{aligned}}$

The right coset of $H$ with representative $g$ is the set:

${\begin{aligned}\quad Hg=\{hg:h\in H\}\end{aligned}}$

We will now look at a nice theorem which tells us that the set of all left cosets of a subgroup $(H,\cdot )$ actually partitions $(G,\cdot )$. The proof below can be mirrored to analogously show that the set of all right cosets of a subgroup $(H,\cdot )$ also partitions $(G,\cdot )$.

Theorem 1: Let $(G,\cdot )$ be a group and let $(H,\cdot )$ a subgroup. Then the set of all left cosets of $(H,\cdot )$ partitions $(G,\cdot )$.

Recall that a partition of a set $A$ is a collection of nonempty subsets of $A$ that are pairwise disjoint and whose union is all of $A$.

• Proof: We first show that any two distinct left cosets of $H$ are disjoint. Let $g_{1},g_{2}\in G$, $g_{1}\neq g_{2}$ and assume the left cosets $g_{1}H$ and $g_{2}H$ are distinct. Suppose that $g_{1}H\cap g_{2}H\neq \emptyset$. Then there exists an $x\in g_{1}H\cap g_{2}H$. So $x\in g_{1}H$ and $x\in g_{2}H$ and there exists $h_{1},h_{2}\in H$ such that:

${\begin{aligned}\quad x=g_{1}h_{1}\quad (*)\end{aligned}}$

${\begin{aligned}\quad x=g_{2}h_{2}\quad (**)\end{aligned}}$

• So using $(*)$ and $(**)$ we see that $g_{1}h_{1}=g_{2}h_{2}$. So $g_{1}=g_{2}h_{2}h_{1}^{-1}$. But $h_{2}h_{1}^{-1}\in H$ since $(H,\cdot )$ is a group and is hence closed under $\cdot$. So $g_{1}\in g_{2}H$. But this means that $g_{1}H=g_{2}H$, a contradiction since $g_{1}H$ and $g_{2}H$ distinct. So the assumption that $g_{1}H\cap g_{2}H\neq \emptyset$ was false. So:

${\begin{aligned}\quad g_{1}H\cap g_{2}H=\emptyset \end{aligned}}$

• Now if $i$ is the identity element for $(G,\cdot )$ then $i\in H$ since $(H,\cdot )$ is a subgroup and must contain the identity element. So $g=gi\in gH$ for all $g\in G$. So:

${\begin{aligned}\quad G=\bigcup _{g\in G}gH\end{aligned}}$

• Therefore the left cosets of $(H,\cdot )$ partition $(G,\cdot )$.

Licensing

Content obtained and/or adapted from:

• Left and Right Cosets of Subgroups, mathonline.wikidot.com under a CC BY-SA license