Let $H$ be a subgroup of $G$ and let $a$ and $b$ belong to $G$. Then $aH=bH$ or $aH \bigcap bH =\emptyset$

Question

Let $H$ be a subgroup of $G$ and let $a$ and $b$ belong to $G$. Then :

$aH=bH$ or $aH \bigcap bH =\emptyset$ ...... (1)

Gallian gives the following proof which i have a little trouble understanding :

By a previous result, we have : $aH = bH $ if and only if $a \in bH$ ....(2)

It says $(1) $ follows directly from $(2)$ for if there is an element c in $aH \bigcap bH$, then $cH = aH $ and $cH = bH$ .

I don't understand the above statement, so i have made the following attempt :

Attempt: If $aH \neq bH$ , then $a \notin bH$ by $(2)$

So , if $a \notin bH$, what can i say about $aH \bigcap bH$?

Thank you for the help

Answer 1

More generally, given a subgroup $H \leq G$, one can define an equivalence relation on $G$ by $g \sim h \iff g^{-1}h \in H$. The equivalence classes are precisely the left cosets of $H$ in $G$ (i.e., $[g] = gH$). Since two equivalence classes are either equal or disjoint, two cosets must either be equal or disjoint.

Answer 2

In general to prove a statement of the type $A \cup B$ to be true, you can suppose that $B$ is not true, and you have to prove that $A$ is true. In the above question, the statement $A$ is: $aH = bH$, and the statement $B$ is: $aH \bigcap bH = \emptyset$. So if $B$ is not true, then it means that $aH \bigcap bH \neq \emptyset$. This means that there is an element $c \in aH \bigcap bH$, then you would have $aH = cH = bH$, and this implies that $aH = bH$ which in turn makes $A$ to be true, and completes the proof.

Answer 3

I think i got the an answer from jobrien929's answer here Cosets of a subgroup do not overlap

Let $H$ be a subgroup of $G$ and let $a$, $b$, and $c$ be elements of $G$ such that $b\not\in aH$ and $c$ is in both $aH$ and $bH$. Then there are elements $h$ and $h^\prime$ in $H$ such that $c=ah$ and $c=bh^\prime$. So $ah=bh^\prime$ and $b=ahh^{\prime -1}$. But $hh^{\prime -1}\in H$ so $b\in aH$. This contradicts our original assumption. Therefore, there can be no element in more than one distinct coset.

Thanks !