Given :
$\langle G,\star\rangle$ is a finite group and $H$ is a finite subgroup of $G$. For $a \in G$, define $aH = \{ah\mid h \in H\}$
i.e. $aH$ is a left coset.
a.
The order of a coset is defined as its cardinality.
any $2$ sets have the same cardinality if and only if there is a bijection between them.
So to show any two left cosets have the same cardinality, it suffices to demonstrate a bijection between them.
Let $aH$ and $bH$ are $2$ left cosets of $H$ in $G$ such that
$aH = \{ah|h \in H\}$ and $bH = \{bh|h \in H\}$ where $a,b \in G$
Let $f$ be a mapping defined from $aH$ to $bH$ such that
$f(ah) = bh\ \forall\ h \in H$
Let $ah_1 =ah_2$ where $h_1,h_2 \in H$
$\implies a^{-1}ah_1 =a^{-1}ah_2$
$\implies h_1 =h_2$
$\implies bh_1 =bh_2$
$\implies f(ah_1) =f(ah_2)$
$\therefore$ $f$ is well defined.
Let $f(ah_1) =f(ah_2)$ where $h_1,h_2 \in H$
$\implies bh_1 =bh_2$
$\implies b^{-1}bh_1 =b^{-1}bh_2$
$\implies h_1 =h_2$
$\implies ah_1 =ah_2$
$\therefore$ $f$ is one-one mapping.
Let $y \in bH$ then $y=bh$ where $h \in H$
$\exists ah \in aH$ such that
$f(ah) =hb =y$
$\implies y$ has a pre-image in $aH$
$\implies $ Every element of $bH$ has a pre-image in $aH$
$\therefore$ $f$ is an onto mapping.
$\because\ f$ is a well-defined one-one and onto mapping
$\implies f$ is a bijective function
$\implies $ the Cosets $aH$ and $bH$ have the same cardinality.
$\therefore\ |aH| = |bH|$
b.
For $aH$ and $bH$
Case 1. $aH$ and $bH$ are disjoint i.e. they have no elements in common. i.e. $(aH \cap bH)=\phi$
Case 2. . $aH=bH$ i.e. all the elements of $aH$ and $bH$ are common(they are identical) i.e. $(aH \cap bH)=(aH \cup bH)$
Case 3. We can't have a case such that $aH$ and $bH$ have only some common elements and other elements are not common.
i.e We can have (either Case 1. or Case 2. ) and Case 3. should also be satisfied.
Let $aH$ and $bH$ be two left cosets of $H$ in $G$ where $a,b \in G$
If we get $aH \cap bH = \phi$
then Case 1. is satisfied. ($\because$ we assumed it ) $\implies$ Case 3. so is also satisfied ($\because $ they will have no common elements)
Now let us check for Case 2.
Let $aH \cap bH \neq \phi$ where $a,b \in G$
$\implies$ At least one element of $aH$ must be equal to an element of $bH$
Let $\alpha$ be one such common element
$\implies$ $\alpha \in aH \cap bH$ such that $\alpha =ah_1=bh_2$ for some $h_1,h_2 \in H$.
$\implies$ $ah_1 = bh_2$
$\implies$ $ah_1h_{1}^{-1} = bh_2h_{1}^{-1}$
$\implies a = b h_2 h_{1}^{-1}$
$\implies a = b h_2 h_{1}^{-1} \in bH$ ($\because$ $h_1,h_2 \in H \implies h_{1}^{-1},h_2 \in H \implies h_{1}^{−1}h_2 \in H$ by closure property)
$\therefore$ $a \in bH$
So $ a = bh_3$ (where $h_3 = h_2 h_{1}^{-1} \in H$)
$\implies aH = bh_3H$
$\implies aH = bH$ ($\because kH = H$ if $k \in H$)
So for every pair of elements $a,b \in G$, $aH$ and $bH$ are either disjoint else if they are not disjoint then they will have all the element same.
c.
From $a.$ and $b.$ we can see that
The cosets partition the entire group $G$ into mutually disjoint subsets i.e.
$G = a_1H \cup a_2H \cup .....a_kH $ ($\because$ $2$ left cosets are either identical or disjoint)
$\implies o(G) = o( a_1H) + o(a_2H) + .....o(a_kH) $
$\implies o(G) = o(H)+ o(H) + ....k\ times $ ($\because o(aH)=o(bH) = o(H)$)
$\implies o(G) = k*o(H)$
$\therefore$ the order of $H$ must divide the order of $G$.
This is also known as Lagrange's Theorem.
NOTE :-
$kH = H$ if $k \in H$
Proof : -
Suppose $k \in H$ and $x \in kH$
$\therefore x = kh$ where $h \in H$
$\implies xh^{-1} = k $
$\because k \in H \implies xh^{-1} \in H\implies x \in H $(closure property)$\implies kH \subseteq H \quad \to I$
Let $ x \in H$ then
$x = ex = k k^{-1} x = k(k^{-1} x) \in kH$ ($\because$ $k,x \in H \implies k^{-1},x \in H \implies k^{−1}x \in H$ by closure property)
$\implies H \subseteq kH\quad \to II$
Combining $I$ and $II$ we get $H = kH$