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 $(G,\cdot )$ .

The Number of Elements in a Left (Right) Coset

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 defined as:

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

The right coset of $H$ with representative $g$ is defined as:

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

We will now look at a rather simple theorem which will tell us that the number of elements in a left (or right coset) will equal to the number of elements in $H$ . This seems rather obvious since $gH$ contains elements of the form $gh$ where we range through all of $h$ . So $gH$ has at most the same number of elements in $H$ . Of course, $gH$ may have less elements if $gh_{1}=gh_{2}$ for distinct $h_{1},h_{2}\in H$ . Of course this cannot be since by cancellation we would then have that $h_{1}=h_{2}$ . We make this argument more rigorous below.

Theorem 1: Let $(G,\cdot )$ be a group, $(H,\cdot )$ a subgroup, and let $g\in G$ . Then the number of elements in $gH$ equals the number of elements in $H$ , i.e., $\mid gH\mid =\mid H\mid$ . Similarly, $\mid Hg\mid =\mid H\mid$ .

We only prove the case of this theorem for left cosets. The case for right cosets is analogous.

Proof: Define a function $f:H\to gH$ for all $h\in H$ by:

{\begin{aligned}\quad f(h)=gh\end{aligned}}

We will show that $f$ is bijective. First we show that $f$ is injective. Let $h_{1},h_{2}\in H$ and assume that $f(h_{1})=f(h_{2})$ . Then we have that:

{\begin{aligned}\quad gh_{1}=gh_{2}\end{aligned}}

By left cancellation this implies that $h_{1}=h_{2}$ and so $f$ is injective. We now show that $f$ is surjective. Let $gh\in gH$ . Then $f(h)=gh$ (somewhat trivially) so $f$ is surjective.

Since $f$ is bijective we have that $\mid H\mid =\mid gH\mid$ so the number of elements in the left coset $gH$ is equal to the number of elements in $H$ . $\blacksquare$

The Index of a Subgroup

If $(G,\cdot )$ is a group and $(H,\cdot )$ is a subgroup then we might want to know the number of left cosets and the number of right cosets of $H$ . As the following theorem will show - the number of left cosets of $H$ will always equal the number of right cosets of $H$ .

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

Proof: Let $L_{H}$ denote the set of all left cosets of $H$ and let $R_{H}$ denote the set of all right cosets of $H$ . Define a function $f:L_{H}\to R_{H}$ for all $gH\in L_{H}$ by

{\begin{aligned}\quad f(gH)=Hg^{-1}\end{aligned}}

If we can show that $f$ is bijective then $\mid L_{H}\mid =\mid R_{H}$ . We first show that $f$ is injective. Let $g_{1}H,g_{2}H\in L_{H}$ and suppose that $f(g_{1}H)=f(g_{2}H)$ . Then:

{\begin{aligned}\quad Hg_{1}^{-1}=Hg_{2}^{-1}\end{aligned}}

But then by the theorem of equivalent statements presented on the <a href="/left-and-right-cosets-of-subgroups">Left and Right Cosets of Subgroups</a> page we must have that:

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

Hence $f$ is injective. We now show that $f$ is surjective. Let $Hg\in R_{H}$ . Then we have that for $g^{-1}H\in L_{H}$ that:

{\begin{aligned}\quad f(g^{-1}H)=H(g^{-1})^{-1}=Hg\end{aligned}}

So $f$ is indeed surjective. Since $f$ is a function from a finite set to a finite set that is bijective we must have that $\mid L_{H}\mid =\mid R_{H}\mid$ , i.e., the number of left cosets of $H$ equals the number of right cosets of $H$ . $\blacksquare$

With the result above, we can unambiguously define the index of subgroup $(H,\cdot )$ in a group $(G,\cdot )$ .

Definition: Let $(G,\cdot )$ be a group and let $(H,\cdot )$ be a subgroup. The Index of $H$ in $G$ denoted $[G:H]$ is defined as the number of left (or right) cosets of $H$ .

For example, consider the symmetric group $(S_{3},\circ )$ and let $H=\{\epsilon ,(12)\}$ . Let's find the index $S_{3}:H]$ . Note that $S_{3}=\{\epsilon ,(12),(13),(23),(123),(132)\}$ . So the left cosets of $H$ in $S_{3}$ are:

{\begin{aligned}\quad \epsilon H=\{\epsilon \circ \epsilon ,\epsilon \circ (12)\}=\{\epsilon ,(12)\}\\\end{aligned}}

{\begin{aligned}\quad (12)H=\{(12)\circ \epsilon ,(12)\circ (12)\}=\{(12),\epsilon \}\end{aligned}}

{\begin{aligned}\quad (13)H=\{(13)\circ \epsilon ,(13)\circ (12)\}=\{(13),(123)\}\end{aligned}}

{\begin{aligned}\quad (23)H=\{(23)\circ \epsilon ,(23)\circ (12)\}=\{(23),(132)\}\end{aligned}}

{\begin{aligned}\quad (123)H=\{(123)\circ \epsilon ,(123)\circ (12)\}=\{(123),(13)\}\end{aligned}}

{\begin{aligned}\quad (132)H=\{(132)\circ \epsilon ,(132)\circ (12)\}=\{(132),(23)\}\end{aligned}}

So the set of left cosets of $H$ is:

{\begin{aligned}\quad \{\{\epsilon ,(12)\},\{(13),(123)\},\{(23),(132)\}\}\end{aligned}}

There are three such cosets so $[S_{3}:H]=3$ .

For another example, consider the group $(\mathbb {Z} ,+)$ and the subgroup $(3\mathbb {Z} ,+)$ . Then the left cosets of $(3\mathbb {Z} ,+)$ are:

{\begin{aligned}\quad 0+3\mathbb {Z} =\{...,-3,0,3,...\}\end{aligned}}

{\begin{aligned}\quad 1+3\mathbb {Z} =\{...,-2,1,4,...\}\end{aligned}}

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

Note that if $m\equiv n{\pmod {3}}$ then $m+3\mathbb {Z} =n+3\mathbb {Z}$ . So in this example we have that $[\mathbb {Z} :3\mathbb {Z} ]=3$ .

Lagrange's Theorem

We now have all of the tools to prove a very important and astonishing theorem regarding subgroups. This theorem is known as Lagrange's theorem and will tell us that the number of elements in a subgroup of a larger group must divide the number of elements in the larger group.

Theorem 1 (Lagrange's Theorem): Let $(G,\cdot )$ be a finite group and let $(H,\cdot )$ be a subgroup. Then the number of elements in $H$ must divide the number of elements in $G$ .

Proof:The set of left cosets of $H$ partition $G$ , that is for all $g_{1},g_{2}\in G$ with $g_{1}\neq g_{2}$ we have that, $g_{1}H\cap g_{2}H=\emptyset$ and:

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

The number of left cosets of $H$ is the index $[G:H]$ and the number of elements in $gH$ is equal to the number of elements in $H$ . So:

{\begin{aligned}\quad \mid G\mid =[G:H]\mid H\mid \end{aligned}}

Therefore $\mid H\mid$ divides $\mid G\mid$ , i.e., the number of elements in any subgroup $(H,\cdot )$ of a finite group $(G,\cdot )$ must divide the number of elements in $(G,\cdot )$ . $\blacksquare$

Corollaries to Lagrange's Theorem

Recall that if $(G,\cdot )$ is a finite group and $(H,\cdot )$ is a subgroup then the number of elements in $H$ must divide the number of elements in $G$ and moreover:

{\begin{aligned}\quad \mid G\mid =[G:H]\mid H\mid \end{aligned}}

We will now prove some amazing corollaries relating to Lagrange's theorem.

Corollary 1: Let $(G,\cdot )$ be a finite group and let $(H,\cdot )$ be a subgroup. Then the index of $H$ is $G$ is given by $\displaystyle {[G:H]={\frac {\mid G\mid }{\mid H\mid }}}$ .

Proof: Rearrange the formula in the proof of Lagrange's theorem. $\blacksquare$

Corollary 2: Let $(G,\cdot )$ be a finite group and let $g\in G$ . Then the order of the element $g$ must divide $\mid G\mid$ .

The order of an element $g$ is the smallest positive integer $n$ such that $g^{n}=e$ (where of course, $e$ is the identity element for the group.

Proof: The order of the element $g$ is the smallest positive integer $n$ such that $g^{n}=e$ . The generated subgroup $(<g>,\cdot )$ is such that $\mid <g>\mid =n$ . So by Lagrange's theorem $n=\mid <g>\mid$ must divide $\mid G\mid$ , i.e., the order of $g$ must divide $\mid G\mid$ .

Corollary 3: Let $(G,\cdot )$ be a finite group of order $\mid G\mid =p$ where $p$ is a prime number. Then $G$ is a cyclic group.

Proof: Let $g\in G$ be any non-identity element in $G$ (which exists since $p\geq 2$ ). By Lagrange's Theorem, the subgroup $(<g>,\cdot )$ must be such that $\mid <g>\mid$ divides $\mid G\mid =p$ . But the only positive divisors of $p$ are $1$ and $p$ .

So if $\mid <g>\mid =1$ then $g=e$ which is a contradiction since we assumed that $g$ is not the identity for the group. So $\mid <g>\mid =p$ , i.e., $G=<g>$ . So $G$ is a cyclic group.

Corollary 4: Let $(G,\cdot )$ be a finite group and let $(H,\cdot )$ and $(I,\cdot )$ be subgroups of $G$ such that $I\subseteq H\subseteq G$ . Then $[G:I]=[G:H][H:I]$ .