# Thread: [SOLVED] intersection of cosets of different subgroups

1. ## [SOLVED] intersection of cosets of different subgroups

Let $H$ and $K$ be subgroups of the group $G$, and let $a,b\in G$. Show that either $aH\cap bK=\emptyset$ or else $aH\cap bK$ is a left coset of $H\cap K$.

I've been working on this one for over a day, now, and I'm still stumped. Assistance would therefore be much appreciated!

EDIT: I finally solved it, and it is indeed extremely simple. I can't believe I missed it for so long.

Proof: Suppose $aH\cap bK\neq\emptyset$, and let $x\in aH\cap bK$. Then there are $h\in H$, $k\in K$ with $x=ah=bk$. So $a=bkh^{-1}$, and therefore $aH\cap bK=bkh^{-1}H\cap bK=bkH\cap bK=bkH\cap bkK=bk(H\cap K)$. The conclusion follows. $\blacksquare$

2. Suppose $\emptyset\neq aH\cap bK$

First: show that $c(H\cap K) = cH\cap cK$.

Secondly: Observe that $ah=bk \Longleftrightarrow h = (a^{-1}b)k$

Let $c = (a^{-1}b)$ and show that $c(H\cap K) = aH\cap bK$

3. Originally Posted by Dinkydoe
Suppose $\emptyset\neq aH\cap bK$

First: show that $c(H\cap K) = cH\cap cK$.

Secondly: Observe that $ah=bk \Longleftrightarrow h = (a^{-1}b)k$

Let $c = (a^{-1}b)$ and show that $c(H\cap K) = aH\cap bK$
Thanks, I suspected $a^{-1}b(H\cap K)$ might be the coset in question, but unfortunately I cannot prove this.

To show that $cH\cap cK=c(H\cap K)$ is quite straightforward: Let $x\in cH \cap cK$. Then there are $h\in H,k\in K$ with $x=ch=ck$. So $h=k=c^{-1}x$, and therefore $x\in c(H\cap K)$. Now let $x'\in c(H\cap K)$. Then there is $h'\in H\cap K$ with $x'=ch'$ and therefore $x'\in cH\cap cK$. So $cH\cap cK=c(H\cap K)$. The conclusion follows.

However, I am unable to show that $a^{-1}bH\cap a^{-1}bK= aH\cap bK$.

If anyone has a solution, or the outline of a solution, it would be much appreciated!

4. Perhaps my last post didn't really clarify that much:

I hope this will help:

Observe that $aH\cap bK = H\cap a^{-1}bK$. This is easy to prove.

Let $c= a^{-1}b\neq e$
One can show that $H\cap cK = [H\setminus (H\cap K)]\cap c[K\setminus (H\cap K)]$.

This follows from the fact that different cosets are disjunct. (for any $H$, we have $H \cap cH = \emptyset$)

I'll try to give a better explanation later.

5. Originally Posted by Dinkydoe
Perhaps my last post didn't really clarify that much:

I hope this will help:

Observe that $aH\cap bK = H\cap a^{-1}bK$. This is easy to prove.

Let $c= a^{-1}b\neq e$
One can show that $H\cap cK = [H\setminus (H\cap K)]\cap c[K\setminus (H\cap K)]$.

This follows from the fact that different cosets are disjunct. (for any $H$, we have $H \cap cH = \emptyset$)

I'll try to give a better explanation later.
Thank you very much for your assistance, but unfortunately I'm still stuck.

It helps to recall that cosets partition a group, which I had forgotten. Since $aH=bH$ or $aK=bK$ implies $aH\cap bK$ is $b(H\cap K)$ or $a(H\cap K)$, respectively, then we can assume $aH\cap bH=aK\cap bK=\emptyset$, which means

$H\cap a^{-1}bK=(H\setminus K)\cap a^{-1}bK=(H\setminus[H\cap K])\cap a^{-1}b(K\setminus [H\cap K])$,

as you have already noted. However, I don't see how this implies $aH\cap bK = H\cap a^{-1}bK$. I worked on it for several hours this morning without further progress.

6. So, just to recap, we may assume that $a^{-1}b\notin H\cup K$, because if $a^{-1}b$ is in either subgroup, the result is easy to prove.