# Group Theory/Cosets and Lagrange's theorem

Definition (left coset):

Let ${\displaystyle G}$ be a group, and ${\displaystyle H\leq G}$ a subgroup, and ${\displaystyle g\in G}$. Then the left coset of ${\displaystyle H}$ represented by ${\displaystyle g}$ is the set

${\displaystyle gH:=\{gh|h\in H\}}$.

Right cosets are defined in an analogous fashion:

Definition (right coset):

Let ${\displaystyle G}$ be a group, and ${\displaystyle H\leq G}$ a subgroup, and ${\displaystyle g\in G}$. Then the right coset of ${\displaystyle H}$ represented by ${\displaystyle g}$ is the set

${\displaystyle Hg:=\{hg|h\in H\}}$.

For both of these, we have the following proposition:

Proposition (being in the same left coset is an equivalence relation):

Let ${\displaystyle G}$ be a group, and define a relation on ${\displaystyle G}$ by

${\displaystyle g\sim g':\Leftrightarrow \exists g''\in G:g\in g''H\wedge g'\in g''H}$.

Then ${\displaystyle \sim }$ is an equivalence relation, and we also have the formula

${\displaystyle g\sim g'\Leftrightarrow g\in g'H}$.

Proof: We first prove the alternative formula for being in the same coset. If ${\displaystyle g\in g''H}$ and ${\displaystyle g'\in g''H}$, we find from the latter equation an ${\displaystyle h\in H}$ so that ${\displaystyle g'=g''h}$, so that ${\displaystyle g''=g'h^{-1}}$ and hence ${\displaystyle g\in g'h^{-1}H=g'H}$, the latter identity because ${\displaystyle H}$ is a group and in particular closed under inversion. On the other hand, if ${\displaystyle g\in g'H}$, then ${\displaystyle g=g'h}$ for some ${\displaystyle h\in H}$, and hence ${\displaystyle g'\in g'H}$ (because the identity is in ${\displaystyle H}$) and then ${\displaystyle g'\in g'H=gh^{-1}H=gH}$.

Hence, the two formulae for the relations coincide, and it remains to check that we're dealing with an equivalence relation. Indeed, suppose ${\displaystyle g\sim g'}$ and ${\displaystyle g'\sim g''}$. Then there exist ${\displaystyle h\in H}$ so that ${\displaystyle g=g'h}$ and ${\displaystyle h'\in H}$ so that ${\displaystyle g'=g''h'}$. Then ${\displaystyle g=g''h'h}$, so that ${\displaystyle g\in g''H}$, ie. ${\displaystyle g\in g''H}$, proving transitivity. Reflexivity follows since the identity is in ${\displaystyle H}$, and symmetry follows because ${\displaystyle g\in g'H}$ implies ${\displaystyle g=g'h}$ and hence ${\displaystyle g'\in g'H=ghH=gH}$. ${\displaystyle \Box }$

Analogously, we have the following proposition:

Proposition (being in the same right coset is an equivalence relation):

Let ${\displaystyle G}$ be a group, and define a relation on ${\displaystyle G}$ by

${\displaystyle g\sim 'g':\Leftrightarrow \exists g''\in G:g\in Hg''\wedge g'\in Hg''}$.

Then ${\displaystyle \sim }$ is an equivalence relation, and we also have the formula

${\displaystyle g\sim 'g'\Leftrightarrow g\in Hg'}$.

Proof: Consider the opposite group ${\displaystyle G^{o}}$ of ${\displaystyle G}$. There is a bijection ${\displaystyle G\to G^{o}}$, given by ${\displaystyle g\mapsto g}$, under which two elements belonging to the same right coset of ${\displaystyle H\leq G}$ correspond to elements belonging to the same left coset of ${\displaystyle H^{o}\leq G^{o}}$. But the relation defined by the latter was seen to be an equivalence relation. ${\displaystyle \Box }$

Definition (index):

Let ${\displaystyle G}$ be a group, and let ${\displaystyle H\leq G}$ be a subgroup. Then the index of ${\displaystyle H}$ is defined to be the number

${\displaystyle [G:H]:=|\{gH|g\in G\}|}$.

That is, the index is precisely the number of left cosets.

Proposition (Lagrange's theorem):

Let ${\displaystyle G}$ be a group, and let ${\displaystyle H\leq G}$ be a finite subgroup. Then

${\displaystyle [G:H]=|G|/|H|}$.

In particular, the order of ${\displaystyle H}$ divides the order of ${\displaystyle G}$.

Proof: We have seen that being in the same left coset is an equivalence relation, so that the equivalence classes partition ${\displaystyle G}$. Moreover, every equivalence class (ie. coset) has the same cardinality as ${\displaystyle H}$ via the bijection ${\displaystyle H\to gH,h\mapsto gh}$. ${\displaystyle \Box }$

Proposition (number of right cosets equals number of left cosets):

Let ${\displaystyle G}$ be a group, and ${\displaystyle H\leq G}$ a subgroup. Then the number of right cosets of ${\displaystyle H}$ equals the number of left cosets of ${\displaystyle H}$.

Proof: By Lagrange's theorem, the number of left cosets equals ${\displaystyle |G|/|H|}$. But we may consider the opposite group ${\displaystyle G^{o}}$ of ${\displaystyle G}$. Its left cosets are almost exactly the right cosets of ${\displaystyle G}$; only the orders of the products are interchanged. But in particular, the number of left cosets of ${\displaystyle G^{o}}$, which, by Lagrange's theorem equals ${\displaystyle |G^{o}|/|H^{o}|=|G|/|H|}$, is equal to ${\displaystyle [G:H]}$, which is what we wanted to prove. ${\displaystyle \Box }$

Hence, we may also use the notation ${\displaystyle [G:H]}$ for the number of right cosets.

Proposition (degree formula):

Let ${\displaystyle G}$ be a group, and let ${\displaystyle L\leq H\leq G}$. Then

${\displaystyle [G:L]=[G:H][H:L]}$.

Proof: We may partition ${\displaystyle H}$ into a family of left cosets ${\displaystyle (h_{\lambda }L)_{\lambda \in \Lambda }}$, where for all ${\displaystyle \lambda \in \Lambda }$ we have ${\displaystyle h_{\lambda }\in H}$. Moreover, ${\displaystyle G}$ may be partitioned into a family ${\displaystyle (g_{\alpha }H)_{\alpha \in A}}$ of left cosets of ${\displaystyle H}$. Then ${\displaystyle (g_{\alpha }h_{\lambda }L)_{(\alpha ,\lambda )\in A\times \Lambda }}$ is a family of left cosets of ${\displaystyle L}$ that partitions ${\displaystyle G}$ (since each element ${\displaystyle j\in G}$ is in one left coset ${\displaystyle g_{\alpha }H}$ of ${\displaystyle H}$, and then ${\displaystyle g_{\alpha }^{-1}j\in H}$ is in a unique coset ${\displaystyle h_{\lambda }L}$, and then ${\displaystyle g_{\alpha }h_{\lambda }L}$ is the unique coset in which ${\displaystyle j}$ is), and the cardinality of this family, which is ${\displaystyle |A\times \Lambda |=|A||\Lambda |}$, is the number of left cosets of ${\displaystyle L}$ in ${\displaystyle G}$. ${\displaystyle \Box }$

## Exercises

1. Prove that ${\displaystyle g\in g'H\Leftrightarrow gH=g'H}$ , thus establishing another formula for the equivalence relation of being in the same coset.
2. Formulate Lagrange's theorem for right cosets, without using index notation.