Prove distinct right cosets of $H$ in $G$ form a partition in $G$.

Let $H$ be a subgroup of the group $G$. Prove that if two right cosets $Ha$ and $Hb$ are not disjoint, that is $$Ha~\cap~ Hb \neq \emptyset$$ then we have $$Ha=Hb$$ that is, the distinct right cosets of $H$ in $G$ form a partition in $G$.

My attempt:

Let $H$ be a subgroup of the group $G$. Let $a\neq b \in G$ be arbitrtary.

Suppose now that $$Ha~\cap~ Hb \neq \emptyset$$ thus, there exists some $x \in Ha~\cap~ Hb$. Therefore we must have that $x \in Ha$ and $x\in Hb$. From this we can deduce that $$x=h_1 a ~~ \text{and}~~ x=h_2b,$$ for some $h_1, h_2 \in H$.

Equality thus yields that \begin{align}h_1a &= h_2b \\ h_1^{-1}h_1a &=h_1^{-1}h_2b \\ a &= (h_1^{-1}h_2)b\end{align}

This is where I am stuck. Can anyone please show me how to continue from here? Or am I on the wrong track completely?

• You last line shows that $a\in Hb$. Can you use it to show that $Ha\subset Hb$? – uniquesolution Oct 17 '15 at 11:36
• @uniquesolution - Can you please show me how I can show that? Or give me some hint? :) – user860374 Oct 17 '15 at 11:38
• @DJS Pick any $ha\in Ha$. Since $a=h'b$ for some $h'\in H$, we have $ha=h(h'b)=(hh')b$. Can you see what to do now? – Daniel Oct 17 '15 at 11:39

If $a\in Hb$, then $Ha\subset H(Hb)=(HH)b= Hb$. For $Hb\subset Ha$, swap $a$ and $b$.