Find the right cosets of $H$ in $G$ simple example Question: Let $G$ be a group and $H<G$ a subgroup with $|G:H|=2$
Show that the right cosets of $H$ in $G$ are $H$ and $G\backslash H$
Answer given: There are two right cosets, they are disjoint and their union is $G$. One coset is $H$ and so the other is $G\backslash H$
My interpretation: I understand that the two right cosets comes from $|G:H|=2$.
Why are they disjoint?
The definition I'm working with for right cosets is: Let $H\leq G$ and let $x\in G$. Then the subset $Hx=\{hx\mid h\in H\}\subseteq G$ is called a right coset of $H$ in $G$.
From this I can see that $H$ is a right coset of $G$ from letting $x=e$
How is $G\backslash H$ the other right coset?
I can see that $G\backslash H\leq G$ and is disjoint to $H$ by definition, but how did we arrive at this result? Is this just a standard result?

A nice way to see that distinct right cosets are disjoint is to do the following: Let $H\leq G$, and for $a,b \in G$, define the relation $a \sim b$ if and only if $a \in Hb$. If we can show this is an equivalence relation, we will have right cosets of $H$ as equivalence classes, which shows they partition $G$ into disjoint subsets.
Reflexivity is obvious, as $a=ea \in Ha$ for all $a \in G$. Now if $a \in Hb$, $a=hb$, so $b=h^{-1}a \in Ha$, which shows the relation is symmetric. Finally, if $a \in Hb$ and $b \in Hc$, we have $a=hb$ and $b=h'c$, so $a=(hh')c \in Hc$, so the relation is transitive.
On a somewhat related note, for finite groups, Lagrange's Theorem is a trivial consequence of this construction, as one need only show that a typical coset of $H$ can be bijectively identified with $H$. The map from $H$ into $Ha$ that sends $h\mapsto ha$ suffices. Cosets of $H$ partition $G$ into equal sized chunks, which is a very nice property.
