Proving right cosets are equal or disjoint Let G be a group, H is a subgroup of G.  Let $Hx = \{hx | h\in H\}$.  Show that given $a, b \in G$, then $Ha \cap Hb = \emptyset$ or $Ha=Hb$.
Suppose we have a $y \in Ha$.  This means that $y=ha$ for some $h \in H$.  Now we have two cases.
Suppose a=b.  Then $y=ha=hb$.  Since $hb \in Hb$, $y \in Hb$.  Thus, $Ha\subseteq Hb$ and $Hb\subseteq Ha$
Now suppose $a\neq b$, then I know from a lemma in the textbook that $ha\neq hb$.  Thus, $Ha \nsubseteq Hb$.  Thus, $Hb \nsubseteq Ha$.  Thus,  $Ha \cap Hb = \emptyset$ 
 A: *

*$ha\neq hb$ does not imply $Ha\not\subseteq Hb$, it might just be that for all $h\in H$, there is $h'\neq h$ in $H$ such that $ha = h'b$.

*$Ha\not\subseteq Hb$ and $Hb\not\subseteq Ha$ does not imply $Ha\cap Hb =\emptyset$. 
(Ok, to be fair, it does if you already know what you want to prove, but that would be circular argument. But, what I mean is $\{0,1\}\not\subseteq \{1,2\}$ and $\{1,2\}\not\subseteq \{0,1\}$, but $\{0,1\}\cap\{1,2\} \neq \emptyset.$)


*Premise that $a\neq b$ will lead to $Ha\cap Hb = \emptyset$ is wrong. If $b = ha$ for $e\neq h\in H$, then $Hb = Hha = Ha$. Notice that I've used another counterexample in the previous one: $e\neq h\in H$, but $He = H = Hh$.


Hint: $a\sim b$ iff $ab^{-1}\in H$ is equivalence relation where equivalence class of $a$ is right coset $Ha$. 
A: Suppose
$k \in Ha \cap Hb; \tag 1$
then there must exist $h_1, h_2 \in H$ with
$k = h_1a; \; k = h_2b; \tag 2$
thus
$h_1a = h_2b, \tag 3$
whence
$a = h_1^{-1}h_2b, \tag 4$
which clearly implies
$a \in Hb, \tag 5$
and so
$Ha \subset Hb; \tag 6$
interchanging he roles of $a$ and $b$, we see that
$Hb \subset Ha, \tag 7$
so
$Ha = Hb. \tag 8$
We have shown that if $Ha$ and $Hb$ have at least one element in common, then they are the same.  Therefore
$Ha \cap Hb \ne \emptyset \Longrightarrow Ha = Hb; \tag 9$
and by contraposition,
$Ha \ne Hb \Longrightarrow Ha \cap Hb = \emptyset. \tag{10}$
A: Your proof is wrong. Firstly, as noted by yanko in comments, if $a=b$ then of course $Ha=Hb$. Your mistake is: $ha\neq hb\implies Ha\not\subseteq Hb$. I give you a somehow trivial counterexample: take $h,h'\in H$ such that $h\neq h'$. You can prove that $Hh=Hh'=H$.
To prove your assertion, note that you have $G=\bigcup_{x\in G}Hx$. What you're asked to prove is simply that $\{Hx\mid x\in G\}$ is a partition of $G$, so you can think of the $Hx$ as equivalence classes of the equivalence relation:
$$\forall x,y\in H,\,x\sim y\iff xy^{-1}\in H.$$
So, prove that this is an equivalence relation and that the class of $x$ is $Hx$.
