Find the right cosets of $H$ in $G$ simple example

Question

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?

Answer 1

Any two distinct right cosets of any subgroup $H$ of any group $G$ are disjoint. This is why: Assume the two cosets $Hx$ and $Hy$ have one element $g$ in common (i.e. they are not disjoint). Since $g \in Hx$, there is a $h_1\in H$ such that $g = h_1x$ (that's what $g\in Hx$ means). At the same time there is an $h_2 \in H$ such that $g = h_2y$. We therefore have $h_1x = h_2y$, or said differently, $x = h_1^{-1}h_2y$.

Now any element $h'x \in Hx$ can be written as $h'h^{-1}_1h_2y \in Hy$, so we have $Hx\subseteq Hy$. Since also $y = h_2^{-1}h_1x$, we get $Hy \subseteq Hx$, so the two cosets are in fact equal.

Of course, the moment you know that there are two cosets of $H$, they are disjoint, and that every element of $G$ is in some coset of $H$, it's not very difficult to conclude that with $H$ being one coset, $G\setminus H$ must be the other.

Side note: If $|G:H| = 2$, then the right and left cosets coincide. This is not necessarily the case otherwise, unless $G$ is abelian.

Answer 2

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.