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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $\displaystyle aH=bH$ or $\displaystyle aK=bK$ implies $\displaystyle aH\cap bK$ is $\displaystyle b(H\cap K)$ or $\displaystyle a(H\cap K)$, respectively, then we can assume $\displaystyle aH\cap bH=aK\cap bK=\emptyset$, which means

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

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

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