# Coset problem

• Sep 3rd 2008, 05:48 AM
Coset problem
Let H be a subgroup of G, and b a fixed element of G.

Prove the following statements are equviaent.

1. bH = Hb
2. If $c \in G$ such that $cH=bH$, then $Hc = Hb$
3. If $f \in G$ such that $Hf=Hb$, then $fH=bH$

My Proof.

1 to 2:

Let x be in cH, then there exist an element u in H such that $x=cu$. Since $cH=bH$, there exist an element v in H such that $x=cu=bv$. Since $bH=Hb$, there exist an element w in H such that $x=cu=bv=wb$.

Now, suppose that y is in Hc, then there exist an element d in H such that $y=dc$.

I want to get y to look like Hb, but can't really get it...

Thanks.
• Sep 3rd 2008, 07:10 AM
Laurent

Assume 1 (i.e. $bH=Hb$). Let $c$ be such that $cH=bH$. Hence $cH=bH=Hb$.

Notice that $c\in cH=bH=Hb$, so that there are $h,g\in H$ such that $c=bh=gb$.

Now, if $y\in Hc$, $y=dc$ for some $d\in H$, hence $y=d(gb)=(dg)b\in Hb$ (because $H$ is a subgroup), and $Hb=cH$, so that $y\in cH$. We've proved that $Hc\subset cH$. Let's prove the reverse inclusion. If $y\in cH$, then $y=db$ for some $d\in H$ (because $cH=Hb$), hence $y=db=d(g^{-1}c)=(dg^{-1})c\in Hc$ (because $H$ is a subgroup, again). This proves that $cH\subset Hc$. Finally, $Hc=cH$.

Laurent.
• Sep 4th 2008, 09:10 AM
Thanks for the tips, and I have proved from i to iii similiarly.

Now, I'm trying to prove from ii to i:

Let c be in G such that cH=bH, and we have Hc=Hb.

c is in cH and Hc, so there exist elements x,y in H such that c=bx=yb. Implies that $b=cx^{-1}=y^{-1}c$

Suppose that u is in bH, then there exist h in H such that u=bh, I'm trying to get this to look like (xxx)b, am I still on the right track?
• Sep 5th 2008, 12:37 AM
Laurent
Don't forget that (ii) is "for every c such that ..." and not "there exists c such that...", so you'll probably need to apply (ii) to some well chosen c.
The most obvious choice would be to try applying (ii) to the element $u$ you introduced. You can check that it indeed works, and conclude the proof.
• Sep 6th 2008, 01:18 PM
Here is what I have so far:

Let $u=bh$ then $uH=(bh)H=bH$, then I have $Hu=Hb$

I want to show that $uH = Hu$, which would mean that $bH=Hb$

Claim: $uH \subseteq Hu$

Let $x \in uH$, then there exist an element $a \in H$ such that $x = ua = (bh)a$

Am I on the right track here?

Thanks.
• Sep 6th 2008, 05:14 PM
Laurent
Remember you let $u\in bH$, so you want to prove that $u\in Hb$.

Once you have proved $Hu = Hb$, the trivial identity $u\in Hu$ shows you that $u\in Hb$. Hence $bH\subset Hb$.

Two remarks:
1) There is something we should have mentioned first: if $xH\subset yH$, then $xH=yH$. We already proved this once or twice: we have $x=yh$ for some $h\in H$, hence $y=xh^{-1}$ and $yH=xh^{-1}H\subset xH$. Using this, the proof could have been shortened a bit.

2) More important: I wonder if the statement of your problem is correct. The reason is: we proved that (ii) implies $bH\subset Hb$. There remains to prove that $Hb\subset bH$. However, if you look at the proof of (i) implies (ii), we only made use of the inclusion $bH\subset Hb$. Hence we have proved that:
$bH\subset Hb$ is equivalent to: for every $c\in G,$ $cH=bH\Rightarrow Hc=Hb.$
The same way, there is an equivalence between the inverse inclusion and (iii).
So if your statement were exact, (ii) and (iii) would be equivalent, which means that: $bH\subset Hb$ is equivalent to $Hb\subset bH$, or $bHb^{-1}\subset H$ is equivalent to $bHb^{-1}=H$.
If the subgroup $H$ is finite, this is OK ( $bH$ and $Hb$ have same cardinality). Otherwise I feel like this is not always true but I can't find a counter-example. I'm going to ask the question in a new thread.
• Sep 7th 2008, 02:03 AM
Laurent
As a conclusion:

The answer (by NonCommAlg) to this thread proves that it is possible to have $bH\subset Hb$ and not $Hb\subset bH$.

For that reason, (ii) and (iii) can't be equivalent.

A possible correct statement would be: For any $b\in G$, the following properties are equivalent:
(i') $bH\subset Hb$;
(ii) For any $c\in G$, if $cH=bH$, then $Hc=Hb$.

And the same with reverse inclusion and (iii).

I spent a bunch of time yesterday trying to prove (i) from (ii) before I realized this was possibly false... Where did you get this exercise from? (or did you forget to mention that $H$ is supposed to be finite?)
• Sep 7th 2008, 12:39 PM