# Intersection of cosets from possibly distinct subgroups is either empty or a coset of the intersection between the two subgroups

Let $G$ be a group and $H \leq G, K \leq G$.

The intersection of two left cosets (one from $H$ and the other from $K$) is either empty or a coset of $H\cap K$

I'm having some trouble proving this in a good way.

We know that, if $G$ is a group and $H \leq G, K \leq G$ then $(H\cap K) \leq G$. Let us a consider $H\cap K$ and some coset $a(H\cap K)$ of $H\cap K$ for a arbitrary $a\in G$

If $x \in a(H\cap K) \implies x =ab,$ for some $b \in H\cap K \implies x \in aH$ and $x \in aK \implies x \in aH\cap aK$. But this is not what I want, right? Or can this implication be done the other way as well?

One thought is to split it up in to cases.

1. $H\cap K = \{e \}$, where e is the identity.
2. $H = K$ or $H \subset K$ or $K \subset H$ or $H\cap K \neq \{e \}$. But I'm not sure if this would get me anywhere?

If I want to prove that the intersection must be empty I'm not sure where to start. I want to assume that some $x\in G$ is contained in the intersection of two cosets and then perhaps derive that it must be empty.

Should I then take $x \in aH\cap aK$ or $x \in aH\cap bK$ where $a,b$ may or may not be distinct?

If I take $x \in aH\cap aK$ then $x = ah, h\in H$ and $x = ak, k\in K \implies x = ah = ak \implies h = k$. And I'm stuck.

If I take $x \in aH\cap bK$ then $x = ah, h\in H$ and $x = bk, k\in K \implies x = ah = bk \implies a^{-1}bk \in H$ And I'm also stuck here.

Any hints on how to solve this?

$a$ and $b$ need not to be equal. You have to show that the intersection between a left coset $aH$ of $H$ and a left coset $bK$ of $K$ is a left coset of $H \cap K$, that is $aH \cap bK = c(H \cap K)$ for some $c \in G$.

As you hinted, suppose $aH \cap bK \neq \emptyset$ and let $x \in aH \cap bK$. Then $x = ah = bk$ for some $h \in H$, $k \in K$ and $h = xa^{-1}$, $k = xb^{-1}$.

We shall now prove that $aH = xH$. Let $g \in aH$, then $$g = ah' = xh^{-1}h' \in xH$$

for some $h' \in H$.

Conversely, if $g \in xH$, then $$g = xh' = ahh' \in aH$$ for some $h' \in H$.

Similarly, we have $bK = xK$.

Hence (explain why!) $aH \cap bK = xH \cap xK = x(H \cap K)$.

dani_s's answer has a typo in the second paragraph. It should state $$a=xh^{-1}, b=xk^{-1}$$ instead ($$h=xa^{-1}$$ would be wrong without assuming abelian group).

Also the middle section of the proof can be shortened: $$aH=xh^{-1}H=x(h^{-1}H)=xH \quad\text{ since }h^{-1} \in H$$ and $$bK=xk^{-1}K=x(k^{-1}K)=xK \quad\text{ since }k^{-1} \in K.$$