Questions about cosets: "If $aH\neq Hb$, then $aH\cap Hb=\emptyset$"? Let $H$ be a subgroup of group $G$, and let $a$ and $b$ belong to $G$. Then, it is known that
$$
aH=bH\qquad\text{or}\qquad aH\cap bH=\emptyset 
$$
In other words, $aH\neq bH$ implies $aH\cap bH=\emptyset$. What can we say about the statement

"If $aH\neq Hb$, then $aH\cap Hb=\emptyset$" ?

[EDITED:]
What I think is that when $G$ is Abelian, this can be true since $aH=Ha$ for any $a\in G$. But what if $G$ is non-Abelian? How should I go on?
 A: The condition $aH=Ha$ characterizes a special kind of subgroup, called a normal subgroup. They play a very important role.
Theorem. Let $G$ be a group, and let $H$ be a subgroup of $G$. The following are equivalent:


*

*Every left coset of $H$ is also a right coset of $H$.

*$aH = Ha$ for every $a\in G$.

*$aHa^{-1} = H$ for every $a\in G$.

*$aHa^{-1}\subseteq H$ for every $a\in G$.

*The equivalence relations $a\equiv_H b$ and $a {}_H\equiv b$ are the same equivalence relation.

*There exists a group $K$ and a homomorphism $f\colon G\to K$ such that $H=\mathrm{ker}(f)$.


(Remember that $a\equiv_H b$ if and only if $ab^{-1}\in H$; and $a{}_H\equiv b$ if and only if $a^{-1}b\in H$).
If $H$ is a normal subgroup, then your implication holds, exactly the same way as it holds for abelian groups.
Conversely, suppose that $H$ is not a normal subgroup. Then there exists an $a$ such that $aH\neq Ha$; but $a\in aH\cap Ha$. Thus, $aH\neq Ha$ but $aH\cap Ha\neq\emptyset$, so the implication does not hold. That is: your desired implication is equivalent to $H$ being a normal subgroup of $G$.
So we could add a seventh point to the theorem: "if $aH\cap Hb\neq\emptyset$, then $aH=Hb$."
A: It is sometimes true and sometimes false.
For example, if $H$ is a normal subgroup of $G$, then it is true.
If $H$ is the subgroup generated by the permutation $(12)$ inside $G=S_3$, the symmetric group of degree $3$, then $(123)H\neq H(132)$, yet $(13)\in(123)H\cap H(132)$
