Group Theory/Cosets and Lagrange's theorem

Definition (left coset):

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

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

Right cosets are defined in an analogous fashion:

Definition (right coset):

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

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

For both of these, we have the following proposition:

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

Let $G$ be a group, and define a relation on $G$ by

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

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

$g\sim g'\Leftrightarrow g\in g'H$.

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

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

Analogously, we have the following proposition:

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

Let $G$ be a group, and define a relation on $G$ by

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

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

$g\sim 'g'\Leftrightarrow g\in Hg'$.

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

Definition (index):

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

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

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

Proposition (Lagrange's theorem):

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

$[G:H]=|G|/|H|$.

In particular, the order of $H$ divides the order of $G$.

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

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

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

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

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

Proposition (degree formula):

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

$[G:L]=[G:H][H:L]$.

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

Exercises

1. Prove that $g\in g'H\Leftrightarrow gH=g'H$, 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.